The Transistor Amplifier
is available as a .pdf but this file is not updated as
fast as the web page.
New items are added on a daily basis as we get a lot of
requests from readers to help design a circuit and explain
how a circuit works.
We have not opted for covering transistor circuit design as
found in most text books because there are already many available on
the web for free download.
We have decided to cover this topic in a completely
different way, with a circuit to cover each explanation.
This way you will pick up all the pointers that the text
It's only after you start designing a circuit that you find
out how little you have been supplied via conventional
teaching and that's why our approach is so important.
If you look at some magazines you will find faults and
poor descriptions in almost every one of their circuits.
No only is the designer poorly informed but the technical
editor of the magazine is unaware of the mistakes and the
readers do not reply with corrections. It's total ignorance
The Transistor Amplifier article will help you understand
some of the faults and how to avoid them.
It's pointless learning about "LOAD LINES" etc and producing
equations for all sorts of results when most transistors
have gain-values that can be 200% more or 80% less than the
The gain of a transistor falls 90% as the current increases
and you have absolutely no idea what value to use until you
build the circuit. By that time you have already solved your
Secondly, it's pointless learning about equations until you
have looked at the hundreds of different transistor circuits
so you know what circuit to use.
The DIFFERENTIAL AMPLIFIER is also
called the "Difference Amplifier" or long-tailed pair
(LTP), or emitter-coupled pair, because it
amplifies the difference between the voltages on Input 1 and
Input 2. It is called a Long Tailed Pair because the
emitter resistor has a high value. The circuit has the advantage
of ONLY amplifying the signals on the Inputs. Any noise on the
power rail is not detected on the output as both transistors
will see this fluctuation and both outputs will either rise
or fall and thus the output will not change.
Since the Long Tailed Pair does not pick up noise from
the supply, it is ideal as a pre-amplifier as shown in the 60
watt amplifier in Fig 71ae:
Fig 71a Constant-Current Circuits
The three circuits above provide a constant
current through the LED (or LEDs) when the supply rises to 15v
and higher. The second and third circuits can be turned on and
off via the input line.
Fig 71b Constant-Current Circuit
The first circuit in
Fig 71b is a constant-current arrangement,
providing a fixed current to the LEDs, no matter the supply
This is done by turning on the top transistor via the
2k2 resistor. It keeps turning on until the voltage-drop across
is 0.65v. At this point the lower transistor starts to turn on
and current flows through the collector-emitter terminals and it
"robs" the top transistor of current from the 2k2 resistor. The
top transistor cannot turn on any more and the current flowing
though R is the same as the current flowing through the LEDs and
does not increase.
The second diagram in Fig 71b is also a constant-current circuit
with the base fixed at:
0.7v + 0.7v = 1.4v via the two diodes.
The transistor is turned on via the 2k2 resistor and a voltage
is developed across resistor R. When this voltage is
0.7v, the emitter is 0.7v above the 0v rail and the base is
1.4v. If the transistor turns on more, the emitter will be 0.8v
above the 0v rail and this will only give 0.6v between base and
emitter. The transistor would not be turned on with this
voltage-drop, so the transistor cannot be turned on any more
than 0.65v across the resistor R.
Fig 71ba Constant-Current Circuit
shows two more constant current circuits "sourcing" the LEDs.
The 7 constant current circuits give you the choice of either
sourcing or sinking the LED current.
Fig 71bab Constant-Current
Circuit for high voltage supply
If the supply voltage is high, the transistor
controlling the current (BC547) will get hot and alter the
Fig 71bab uses a POWER TRANSISTOR to dissipate the losses and
the current-controlling transistor remains cold.
When the circuit turns ON, the current through R is zero and the
voltage on the base of the BC547 turns it on fully. The voltage
between collector and emitter is about 0.2v and this means the
emitter of the power transistor is below the base of the BC547.
The base of the power transistor is 0.7v above the base of the
BC547 and the power transistor also turns on fully.
Current increases through R and when the voltage across R
reaches 0.7v, The BC547 starts to turn OFF. The collector
voltage rises and this starts to turn OFF the power transistor.
This is how the current through the LOAD is limited by the value
THE CURRENT MIRROR
Fig 71bac Current Mirror
This is not a constant current circuit. It
is a CURRENT SOURCE circuit. A constant current
circuit means the current will not change if the supply voltage is
increased or decreased. This circuit simply supplies a DC signal (in the
form of a voltage) to another circuit so that the current in the
original circuit is available in the second circuit and this is called a
current mirror arrangement.
We start with diagram A.
The transistor is turned on because the base is connected to the
collector. The collector can only rise to about 0.7v because it is
connected to the base so that most
of the supply-voltage appears across the load. This means the current
through the load is known.
It can be determined by Ohm's Law:
Here's how the circuit works: When the circuit is turned ON, current
flows through the resistor and through the base-emitter junction. This
turns the transistor ON very hard and the current through the
collector-emitter circuit increases. This reduces the voltage on the
collector and as it decreases, the voltage on the base decreases
and the transistor starts to turn OFF. In the end, the transistor
is turned on to allow 10mA to flow through the collector-emitter
junction due to the 10v supply and 1k resistor.
Suppose we instantly change the 1k for 100 ohms.
The transistor is only lightly turned ON and current though the
collector-emitter is only 10mA. But the 100R will deliver 100mA and the
extra current will flow into the base and turn the transistor ON harder.
This will increase the current thorough the collector-emitter junction
and rob the base of the extra current, however the current into the base
will be higher than before because the transistor has to be turned on
more to allow about 100mA to flow through the collector-emitter
If we take a lead from the base of the transistor, as shown in fig
B we can connect it to the base of
an identical transistor and the second transistor will allow the same
current to flow though the collector-emitter junction.
The result is circuit C. The current
through the 100R resistor will be 10mA (normally it would be 100mA). The
second transistor is only lightly turned on and allows 10mA to flow.
CURRENT POWER SUPPLY A reader requested a circuit for an
Adjustable-Current 5v Power Supply.
In other words he wanted a power supply with CURRENT LIMITING.
This type of power supply is very handy so you can test an
unknown circuit and prevent it being damaged.
For this design we will make the current adjustable from 100mA
to 1 amp.
This circuit can be added to any power supply with an output of
more than 7v. Our circuit requires at least two volts
"head-room" for the voltage across the regulating transistor
(the transistor that delivers the voltage and current ) and
about 0.5v for the current-detecting resistor.
The maximum current is set by the 100R pot and
This circuit delivers 5v when no current is flowing and the
voltage gradually reduces. When the set value of current as
selected by the 100R pot is reached the output voltage will have
dropped by 0.6v. This is the voltage developed across the
current-sensing resistor and this voltage is detected by the
BC547 to to start to reduce the output voltage. As soon as the
maximum current is reached, the voltage falls at a faster rate
and if the output is short-circuited, the current-flow will be
as set by the pot.
CURRENT POWER SUPPLY
The output voltage
of this power supply can be increased by changing the voltage of
the zener diode. The voltage of the plug pack must be at least
3v above the output voltage to allow the regulator transistor
and current-detector resistor to function.
As soon as the load reaches the point where it takes the full
current, the circuit turns into a CONSTANT CURRENT power
Before we go to the 2-transistor Voltage Regulator, we will
explain how a voltage regulator works.
The basis of all voltage regulators is a diode.
A diode has a voltage characteristic. When a voltage is placed
across its terminals, and the voltage starts at zero, no
current flows through the diode until the voltage reaches 0.65v. As
soon as it reaches 0.65v, current flows and as you increase the
voltage, more current flows but the voltage across the diode remains
at 0.65v. If the voltage is increased further, the current increases
enormously and the diode will be destroyed.
This characteristic does not apply to a resistor. The voltage across
a resistor will increase when the supply voltage increases and thus
a resistor cannot be used as a Voltage Regulator.
We have selected 0.65v for this discussion as this is the
characteristic voltage-drop for a normal silicon diode.
However germanium diodes and Schottky diodes have different
characteristic voltage drops. On top of this, special diodes can be
produced with higher voltages. These are called ZENER DIODES.
They all have the same characteristic. As soon as the specified
voltage appears across the terminals of the diode, current starts to
flow and if the voltage is increased too much, the diode will be
To prevent this, a resistor must be placed in series with the diode.
This is the basis of all voltage regulators.
Fig 71be The Unregulated Voltage
is regulated by the diode (zener)
In Fig 71be, the supply voltage is called the
UNREGULATED VOLTAGE and it is connected to resistor R and
a diode. The voltage at the top of the diode is called the
REGULATED VOLTAGE. The diode produces a fixed 0.65v and
the zener produces a fixed 6v1 or 12v.
This circuit is called a SHUNT REGULATOR because the
regulator is shunted (placed across) the load. [A Shunt is a
load - generally a low-value resistor - placed across a
component in a circuit to take a high current to either protect
the other components or to test the circuit under high-current
That's exactly what the diode or zener diode does.
It takes ALL THE CURRENT from the unregulated supply and and
feeds it to the 0v rail. During this condition the circuit is
100% wasteful. All the wattage is being lost in heating resistor
R and heating the diode.
The circuit is providing a fixed voltage at the top of the
When a load is added to the circuit, it takes (or draws) current
and this current comes from the current flowing though the
The load-current can increase to a point where it takes nearly
all the current from the zener.
If it takes more current than the zener, two things happen.
Current stops flowing though the zener and the voltage on the
top of the zener drops to a lower value. This is the point where
the zener has dropped out of regulation andthe
circuit is no longer regulating.
In other words: A current is flowing into the regulator circuit
and it is being divided into two paths: The zener path and the
load path. The load path cannot be more than 95% or the
regulator will drop out of regulation (the output voltage goes
below the zener voltage).
Here's how the diode (or the zener) works: The zener is just
like a bucket with a large hole in the side. As you fill the
bucket, the water (the voltage ) rises until it reaches the
hole. It then flows out the hole (through the zener) and does
not rise any further. When you draw current from the circuit it
is the same as a tap at the bottom of the bucket and the water
flows out the tap and not the hole. The pressure out the tap is
the voltage of the zener.
The only disadvantage of this circuit is the voltage across the
zener changes a small amount when the current through it
The SHUNT REGULATOR
is limited to small currents due to the fact that the load is
taking the current from the zener.
The current can be increased by adding a buffer transistor to
produce a BUFFERED SHUNT REGULATOR as shown in
Fig 71bf. This circuit actually becomes a PASS TRANSISTOR
Fig 71bf Buffered Shunt
called a PASS TRANSISTOR Regulator
with the transistor in an
The transistor operates as an amplifier and if the DC
gain of the transistor is 100, the output current of a Buffered
Shunt Regulator can be 100 times more than a Shunt regulator.
See more circuits on the Zener Regulator and the Transistor Shunt Regulator and
Pass Transistor Regulator in
101-200 Transistor Circuits. A very clever circuit to reduce
ripple is called the
The 3 PNP transistors in the Power Supply circuit
above are called PASS TRANSISTORS, because they pass or
CONDUCT the current. They pass the current from the
bridge to the output.
These transistors are AMPLIFIERS - CURRENT
AMPLIFIERS - as they amplify the current entering
the base and deliver a higher current through the
To see how these transistors work, we will simplify the
Only 3 components are involved in the explanation,
the transistor, the base resistor and the rectangle
containing the voltage regulator IC.
The 3 emitter resistors are only needed to make sure
each power transistor delivers the same current.
It is called CURRENT SHARING.
The IN voltage is about 18v and the voltage OUT is fixed
at 12v by the 3-terminal regulator.
The regulator produces the 12v OUT. The transistor does
not have any control over the output voltage.
At the moment there is NO LOAD. The transistor is not
turned on AT ALL and the output voltage is produced by
the regulator. The regulator takes about 10mA to provide
Now put a LOAD on the output.
This load creates a voltage drop across the 15R and as
soon as it is more than 0.6v, the transistor starts to
As the load increase (the resistance of the load
decreases) by say adding more globes, the current
increases and the voltage-drop across R increases and
this turns the transistor ON more.
The 3-terminal regulator (in the rectangle) delivers
almost no current to the load during any of these
conditions (only about 50mA !! ). It is the transistor
that delivers all the current.
The current PASSES through the transistor and that’s why
it is called a PASS TRANSISTOR.
The purpose of the 15R resistor is SOLELY to allow the
transistor(s) to turn ON.
As the load takes more current, the transistor will be
"asked" for this current, but it does not deliver.
So, the regulator delivers the extra current and this
current flows through the emitter-base junction. The 15R
is added so that some of the current flows through the
The extra current requested by the regulator flows
through the emitter-base junction of the PNP transistor
and turns it ON more. This allows the transistor(s) to
pass the current from the bridge to the output. The
current taken by the 3-terminal regulator mainly flows
through the emitter-base junction.
I don't know why the 15R is rated at 10 watts.
The voltage across this resistor cannot rise to more
than 1v and this means the resistor can only pass 1/15
amp = 60mA !!
Obviously the circuit has never been tested.
The whole concept of a regulator (removing the ripple while
maintaining the required voltage) revolves around the
voltage-drop across a diode and in Fig 71bb, the diode is
replaced with the voltage-drop across the base-emitter junction
of a transistor. This voltage-drop is fairly constant when a
small current flows and this is the basis of the Two
If we take the Constant-Current Circuit shown in
Fig 71b above, and split resistor R into Ra and Rb,
we produce an identical circuit with a completely different name. It
is called a TWO TRANSISTOR REGULATOR.
The circuit will produce a smooth voltage on the output, even though
the rail voltage fluctuates AND even if the current required by the
output increases and decreases.
That's why it is called a REGULATOR CIRCUIT.
The current through Ra and Rb is "wasted current" so it does not
have to be more than 1mA - enough to turn on the lower NPN
Ra and Rb form a voltage divider and when the join of the two
resistor reaches 0.7v, the lower transistor turns ON.
lower transistor forms a voltage-divider with the 2k2 to pull the
top BC547 transistor DOWN so the voltage
on the output is kept at the "design voltage" (the top transistor is
an emitter follower). If the device connected to the output requires
more current, the top transistor will not be able to provide it and
the output voltage will drop. This will reduce the voltage on the
base of the lower transistor and it will turn OFF slightly.
The voltage on the base of the top transistor will rise and since
this transistor is an emitter-follower, the emitter will rise too
and increase the output voltage to the original "design value."
Regulation is also maintained if the supply decreases (or
If the supply decreases, the voltage on the base of the top
transistor will fall and the output voltage will also fall.
The voltage on the base of the lower transistor will also fall and
it will turn off slightly.
This will increase the voltage on the base of the top transistor and
Vregulated will rise to the design value. Both the supply and
the load can change at the same time and the circuit will
All we have to do is re-draw the circuit as a standard 2-Transistor
Regulator as shown in Fig 71bc and you have covered the principle of
2-Transistor Voltage Regulator
THE TRANSISTOR AS AN AF
AND RF DETECTOR
A transistor can be
used as a "detector" in a radio circuit. The Detector stage in a radio
(such as an AM receiver), is usually a crystal, but can be the
base-emitter junction of a transistor.
It detects the slowly rising and falling audio component of an RF
signal. This signal is further amplified and delivered to a speaker. A
single transistor will perform both "detection" and amplification.
In Fig 71bd, the first transistor provides these two functions
and the output is passed to the second transistor via direct-coupling.
The first two transistors provide enormous gain and a very high input
impedance for the tuned circuit made up of the 60t aerial coil and 415p
tuning capacitor. The signal generated in the "tuned circuit" is
prevented from "disappearing out the left end" by the presence of the 10n capacitor
as it holds the left end rigid.
Fig 71bd 5-TRANSISTOR RADIO
THE COUPLING CAPACITOR
We have shown the coupling
capacitor transfers very little energy when it does not get
fully discharged during part of the cycle and this means it
cannot receive a lot of energy to charge it during the
"charging" part of the cycle.
This is a point that has never been discussed in any text books.
It is the energy (actually the current - due to the difference
in voltage between the two terminals of the capacitor) that
flows into the capacitor that creates the flow of energy from
one stage to the other. It is the "magnet on the door" analogy
But the question is:
1. How much energy will a capacitor pass under ideal conditions?
2. How do you work out if a capacitor needs to be: 100n, 1u, 10u
Without going into any mathematics, we will explain how to
select a capacitor.
Many text books talk about the capacitive reactance of a
capacitor. This is its "resistance" at a particular
But an audio circuit has a wide range of frequencies and the
lowest frequency is generally selected as the capacitor will have the
highest resistance at the lowest frequency.
We will select 200Hz as the lowest frequency for an amplifier.
A 100n will have a "resistance" of about 10k at 200Hz
A 1u will
have a "resistance" of about 1k at 200Hz
A 10u will have a "resistance" of about 100R at
A 100u will have a "resistance" of about 10R at 200Hz
A 100n capacitor at 200Hz is like putting a 10k resistor
between one stage and the next.
capacitor at 200Hz is like putting a 1k resistor between one
stage and the next.
10u capacitor at 200Hz is like putting a 100R resistor between
one stage and the next and a 100u capacitor at 200Hz is like putting a 10R resistor between
one stage and the next.
In other words, the resistor transfers approximately the same amount of energy
as the capacitor but the capacitor separates the DC voltages
- the capacitor allows the naturally-occurring voltages to be
The capacitive reactance of the 100u ranges from 10R to
than 1R (depending on the frequency being processed).
In Fig 71d
you can see the "resistance" of a capacitor is very small
compared to the LOAD resistance (the main component that
determines the amount of energy that can be transferred from one
stage to another and the impedance of the receiving stage
- the component that determines the discharging of the
capacitor). The "resistance" of a capacitor decreases as the
Thus the "capacitive reactance" of a capacitor has very little
effect on the transfer of energy from one stage to the next
(when it is correctly selected). The
major problem is not discharging the capacitor. It only transfers
the maximum amount of energy when it is completely discharged.
When it is completely discharged, it acts like a "zero-ohm"
resistor during its initial charging-cycle. This is called
INRUSH CURRENT and can be ENORMOUS. This is the "plop"
you hear from some amplifiers when they are turned ON. It
is also the inrush current to a power supply. To reduce this
enormous in-rush current, a small-value resistor is included in
series with the input of the electrolytic(s) in the circuit (or
Let's go over this again:
The transfer of energy from one stage to another depends on 3
1. The value of the LOAD resistor of the first stage. This
resistor charges the capacitor. Its resistance should be
as low as possible to transfer the maximum energy.
2. The value of the capacitor. It should be as high as
possible to transfer the maximum energy.
3. The value of the input impedance of the receiving stage. It
should be as low as possible to discharge the capacitor.
Let's take a 100n capacitor:
In the following circuit, a 100n capacitor separates an electret
microphone from the input of a common-emitter stage.
The waveform on the output of the electret microphone is 20mV
This amplitude passes through the 100n capacitor, which we have
drawn as a 10k resistor, (to represent the capacitive reactance
of the capacitor at 200Hz). The input impedance of the
common-emitter amplifier is about 500 ohms to 2k. (500 ohms when
the base current is a maximum and 2k when the base current is
The capacitor and the input impedance form a simple
voltage-divider, as shown in Fig 71f. When a 20mV signal appears
on the input of the voltage divider, the voltage at the join of
the two resistors will be about 3.3mV.
This is 3.3mV ON TOP of the 630mV provided by the 1M base-bias
about 16% of the waveform gets transferred to the base of the
transistor. A common-emitter
stage will have a gain of about 70, so 3.3mV input will create
230mV output. It's called a "swing" of 230mV or 230mV P-P
(peak-to-Peak) or 230mV AC signal.
But most signals have a frequency of about 2kHz and the
capacitive reactance of the capacitor will be about 1k. In this case
the transfer will be 66% or 13mV and the output of the stage
will be nearly 1v.
This is an ideal situation where the capacitor is being fully
The actual transfer of energy from one stage to another is much
more complex than we have described, however you can see it
involves the LOAD resistor, the size of the capacitor and the
efficiency of discharging the capacitor.
The only way to see the actual result is to view the waveforms on
a CRO (Cathode ray Oscilloscope).
TRANSFER OF ENERGY THROUGH THE COUPLING CAPACITOR
It is very difficult to work out
how much energy will be transferred by the capacitor coupling
the two stages in the circuit above.
As we have mentioned before, the transistor does not transfer
very much energy. The first transistor merely pulls the
capacitor "down" and this turns OFF the second transistor and
discharges the capacitor. But the discharge path is through the
base emitter junction of the second transistor and when the base
of the second transistor drops below 0.55v, the junction
exhibits a very high resistance and does not discharge the
capacitor. The base needs to go about 6v negative to exhibit
reverse-bias and the 5v circuit does not allow this to occur.
So, some slight discharge is done via the 1M resistor. The only
other discharge is done when the capacitor is being pulled down
and it is still feeding energy into the base of the second
This loss of energy is not very much and when the capacitor
charges again, ONLY the pervious losses can be added to the
capacitor and thus it transfers very little energy on the second
and future cycles of its operation.
The secret to transferring a lot of energy is to get the
capacitor discharged fully during part of the cycle.
The 200uA in the circuit above is only a theoretical value for
the first cycle, when the capacitor is completely discharged.
The second cycle will transfer much LEES than this.
The coupling capacitor passes CURRENT not VOLTAGE The next
point is this: The coupling capacitor does not pass VOLTAGE from
one stage to the next but CURRENT.
The voltage on the base of the second transistor may be 0.55v
when the capacitor is not passing any energy and rise to 0.65v
when passing the maximum energy. This may be equivalent to a 3v
signal. The resistance of the LOAD resistor in the first stage
determines how much current will delivered to the capacitor to
charge it. In other words, the capacitor converts the voltage
(the amplitude of the signal) to CURRENT and the second
transistor converts the current to a voltage across its LOAD
But the energy passed by the capacitor depends on how much it is
discharged, as this is the energy (called the charging energy)
that will be transferred. There is no way to determine how much
energy is discharged on each cycle.
Everyone looks at how much energy (or charge, or current)
the capacitor will pass during the "active" part of the cycle or
the "forward" part of the cycle, but the most important part of
the cycle is the DISCHARGING of the capacitor and when the base
of the transistor falls below 0.5v, the capacitor is NOT
discharged via the base. This means you have to look-at what is
discharging the capacitor. This is the main reason why a
particular stage is not amplifying. A diode between base and 0v
rail will become forward-biased when the voltage drops below 0v
and this can be used to discharge the capacitor.
Now you can see why mathematical calculations in Lecture Notes
are SO INACCURATE. Everything is completely different to what
you have been told.
INPUT AND OUTPUT IMPEDANCE
shows each transistor stage has an input and output
impedance. This really means an input and output
resistance, but because we cannot measure the value with
a multimeter, we have to find the value of resistance by
measuring other things such as "waveform amplitudes" and
then create a value of resistance, we call IMPEDANCE.
The values shown are only approximate and apply to
transistors called SMALL SIGNAL DEVICES. The values are
really just a comparison to show how the different
stages "appear" to input and output devices, such as
when connecting stages together.
The input impedance of a common-emitter stage ranges
from 500R to 2k. This variation depends on the type of
transistor and how much the stage is being turned ON. In
other words, the amount of current entering the base.
The value of 2k2 for the emitter-follower depends on the
current entering the base.
These values are all approximate and are just to give an
idea of how to describe the various values of impedance.
There are so many discussions in text books on the INPUT
IMPEDANCE of a transistor and they are all complex and
confusing for the beginner.
Here is a simple explanation.
The transistor is like a FORK-LIFT TRUCK. You move the
lever UP and the fork lifts a pallet of bricks.
The same with a transistor, you deliver a small current
into the base and the collector delivers a higher
current at the collector terminal. This higher current
passes through a load (such as a resistor) placed
between the collector and the positive rail.
But a transistor can only amplify the current about 100
times. A fork-lift truck can amplify the lever about
So, don't worry about terms such as IMPEDANCE. All you
have to remember is this: The transistor will amplify
the current delivered to the base, about 100 times.
Of course you also have to know how much current is
being delivered to the base from a previous stage, but
that is a problem for another time.
When you place two transistors "on top of each other" to
produce a "transistor" called a super-alpha transistor
or DARLINGTON transistor, the first transistor amplifies
the current by 100 and the second amplifies this current
another 100 times, making a total of 100 x 100 = 10,000
and it is as strong as a fork-lift truck !!
Because the first transistor is helping the second
transistor, we also say the input impedance is increased
by a factor of 100.
But it really all boils down to the fact that the
current capability is INCREASED.
There are other facts such as voltage, and amplification
will decrease as the current increases, but you
can study this AFTER you realise the transistor is
simply INCREASES THE CURRENT.
THE TIME DELAY
Also called the
TRANSISTOR TIME DELAY
RC Delay Circuit
A Delay Circuit is made with a capacitor and resistor in series:
The TIME DELAY circuit
the two components that create the TIME DELAY. No other parts
are needed. When the value of the capacitor and resistor
are multiplied together the result is called the TIME CONSTANT
and when the capacitor value is in FARADS and resistor in OHMs,
the result is SECONDS
To detect when the capacitor has reached about 63% of its final
voltage, we need some form of detecting device, such as a
But the detecting device cannot "steal" any of the current
entering the capacitor, otherwise the voltage on the capacitor
will never increase or take longer to increase.
We know a transistor requires current for it to operate but a
Darlington Pair (or Darlington) requires very little current, so the
detecting device must be something like a Darlington.
The transistor plays no part in the timing (or TIME DELAY)
of the circuit. It is just a detector.
The main secret behind a good TIME DELAY circuit is to allow the capacitor
to charge to a high voltage and use a large timing resistor.
This reduces the size of the capacitor (electrolytic) and
produces a long time delay.
There are lots of chips (Integrated Circuit) especially
made for timing operations (time delays). Transistors (of the
"normal" type - called Bipolar Junction) are not suited for long
Field Effect Transistors, Programmable Uni Junction transistors
and some other types are more suited.
However a normal transistor can be used, as shown in Fig 71h.
The normal detection-point is 63% but you can make the circuit
"trigger" at any voltage-level. The value "63%" has been chosen
because the voltage on the capacitor is increasing very little
(each second) when it is nearly fully charged and waiting for it
to reach 65% may take many seconds. Trying to detect an extra 10%
or 25% is very hard to do and since it takes a long time for the
voltage to rise, the circuit becomes very unreliable and very
inaccurate. That's why 63% has been chosen.
See also Integration and Differentiation.
The same two components (a resistor and capacitor) can be used for a
completely different purpose. That's the intrigue of electronics.
Fig 71h shows
a TIME DELAY circuit. This circuit does not wait for the
capacitor to charge to 63% but it detects a voltage of 5v1 +
0.7v = 5v8.
The detecting circuit is made up of the 5v1 zener and
base-emitter junction of the transistor.
These two components create a high impedance until a voltage of
5v8 because the zener takes no current until its "characteristic
voltage" has been reached.
Fig 71j shows
a Time Delay Circuit. The 100k is the time delay resistor. The
1M is the "sense resistor" and the the 330k is the voltage
The base of the Darlington transistor detects 1.4v and the
1M/330k produces a voltage divider that requires 3 x 1.4v = 4.2v
on the electrolytic. The 1M, 330k and transistor provide a
fairly high impedance detecting circuit that does not inhibit
the charging of the capacitor.
The circuit requires a supply of 12v.
Fig 71k shows two Time Delay Circuits as well as a latching circuit
(the 4k7 resistor), a buffer transistor (BD136) and a high frequency filter (the 15n
When the circuit is turned ON, the relay is not energised. The
signal on the base of the first transistor has any high
frequency component removed by the 15n capacitor (see below for
the effect of a filter on a signal).
The lower 47u is fully charged via the 1k5 a very short time
after the circuit is turned on and the output of the first
transistor discharges this electrolytic very quickly when it
receives a signal.
This turns ON the BD136 transistor via the 1k resistor
and the relay is energised.
The output of the relay is connected to a 4k7 resistor and this
resistor takes over from the effect of the first transistor to
keep the relay activated.
If the input signal continues, the top 47u starts to charge and
after about 2 seconds, the BC557 transistor turns ON and
removes the emitter-base voltage on the BD136. This turns the
In some circuits
using a relay, you will find a diode has been placed across the coil.
When the relay is turned OFF, it produces a voltage in the opposite
direction that can be much higher than the voltage of the supply. This
means the voltage appearing on the collector will be higher than some
transistors can withstand and they will either zener and absorb the
energy or be damaged due to the excess voltage. The diode across the
coil is connected so the voltage flows through it and the
transistor is protected.
This voltage is called BACK EMF and only occurs when the relay is
turned off suddenly when full current (or near full current) is flowing.
The size of the back EMF is due to the number of turns on the coil and
the metal in the (magnetic) core. It can be 10 times or even more than the
supply voltage and the diode will reduce this to about 0.7v.
Figs 71h,j and k above show a diode across a relay to remove the back
EMF and protect the transistor.
Figs 71m shows a relay connected in the emitter of a
transistor. This configuration is called an emitter-follower.
When the transistor turns off, the relay is de-energised and a
back-voltage is produced.
The voltage on the top of the relay becomes less than 0v and
this pulls the emitter DOWN. This has the effect of turning ON
the transistor and for a tiny fraction of a second, the effect
of the relay is cancelled by a flow of current through the
transistor. This prevents a high back-voltage being produced and
thus a diode is not needed.
One point about emitter-follower designs: The voltage on the relay is less than 12v due to the 0.7v
between the base and emitter and the base will be lower than 12v
by as much as 1v. Compare this with the common-emitter driver
where the collector-emitter drop will be as low as 0.4v.
Back EMF is also produced by motors and is known as "commutation
noise." This "noise" can also be suppressed via a capacitor and/or small
inductors in the leads. The size of the voltage must be measured when
the circuit is operating as it is a "spike" and this spike will puncture
a semiconductor (such as a transistor).
Back EMF is also produced by coils, called INDUCTORS. An inductor is
also called a choke.
When a piezo is placed across an inductor, and a signal is delivered to
the parallel-pair, the piezo will detect the high-voltage (Back EMF) and
produce a very load output. The inductor produces the high voltage when
the signal is turned off sharply. The magnetic flux collapses and
produces a very high reverse voltage. A typical circuit that takes
advantage of this high voltage is the:
HIGH FREQUENCY "NOISE"
we move on to the next phase of this discussion, there is one
interesting point that needs covering.
When a circuit has a number of amplifying stages, there is always a
possibility of noise being generated in one of the transistors in the
"front-end" (the first or second stage in the amplifier) and this is
amplified by the stages that follow. This is the case with the Hearing
Aid Amplifier in Fig 69.
The 330p between the
base and collector of the BC557 removes high-frequency noise. If
the 330p is removed a 1MHz waveform is generated in the
front-end and amplified by the stages that follow. This noise
cannot be heard but is visible on a CRO (Cathode Ray
Oscilloscope) and causes the circuit to take extra current. The
330p capacitor provides NEGATIVE FEEDBACK to remove the
We have studied circuits
that use components to produce NEGATIVE FEEDBACK. The first circuit we
studied was the self-biased common-emitter stage. The base-bias resistor
provided negative feedback to set the voltage on the collector.
Any component (resistor or capacitor) connected between the output and
input of a stage produces NEGATIVE FEEDBACK.
A resistor connected between the output and input produces about the
same amount of feedback no matter what frequency is being process by the
But a capacitor provides more feedback as the frequency increases.
That's because the effective "resistance" of the capacitor decreases as
the frequency increases.
This feature can be used to "kill" the amplitude of high frequencies and
thus only allow low frequencies to be amplified.
It can also be used to only allow high frequencies to be
amplified. When it is used to couple two stages, a low-value capacitor
will only allow high frequencies to pass from one stage to the next.
By using a resistor in series with a capacitor, the effect of the
capacitor can be controlled.
Using these facts, we can design circuits that will amplify low
frequencies or high frequencies. This type of circuit is called a
A Filter can be given a number of names. Here are a few: Active Filter contains a transistor or op-amp in the circuit High Pass Filter suppresses or rejects the low frequencies
Only the high frequencies appear on the output Low Pass Filter suppresses or rejects the high frequencies
Only the low frequencies appear on the output Notch Filter: A Filter that rejects or suppresses a narrow
band of frequencies.
To understand how a filter works, you need to know "HOW A CAPACITOR
Fig 72a shows a capacitor with a low-frequency signal entering
the left terminal.
The output amplitude from the capacitor in diag
will be small because the capacitor is able to charge and discharge as the
signal rises and falls.
As the frequency of the signal increases, the output increase in
amplitude because the capacitor does not have enough time to charge and
discharge and thus it does not "absorb" the amplitude of the
Fig 72b shows a capacitor connected between the "signal line" and
0v rail. When a low-frequency signal is on the "line," the capacitor has
little effect on attenuating (reducing) the amplitude, as shown in diag
because the capacitor charges and
discharges just like pushing a "shock absorber" up and down slowly.
As the frequency of the signal increases, it is reduced in amplitude
because the signal is trying to charge and discharge the capacitor very
quickly and it takes energy to do this and the energy is coming
from the signal.
Fig 72c Fig a shows a capacitor and resistor connected in series on the
"signal line." With a low-frequency signal, the capacitor reduces
the amplitude because most of the signal is absorbed by the capacitor
charging and discharging.
As the frequency increases (fig b), the output will be reduced by a smaller
amount because the capacitor has less time to charge and discharge and
less time to "absorb" the signal.
As the frequency is increased further
(fig c), the resistor starts to have an
effect on reducing the amplitude because these two components are
connected to other components in a circuit and a higher frequency has a
higher energy and more of this energy gets lost in the resistor - thus
reducing the amplitude slightly.
In addition, the capacitor is already charging and discharging as
quickly as possible and it is transferring as much of the signal as
possible. It is only the resistor that is creating the attenuation at
It does not matter if the capacitor or resistor is placed first or last, the
attenuation is the same.
Fig 72d Fig a shows a capacitor and resistor connected in series
between the "signal line" and 0v rail. With a low-frequency signal the
capacitor can charge and discharge and the voltage across it will rise
and fall so the effect on the amplitude of the signal is minimal.
The resistor has very little effect on reducing the amplitude.
The top plate of the capacitor rises and falls with the signal and the
bottom plate rises and falls very little.
As the frequency increases, the capacitor cannot charge and discharge
fast enough and more of the energy of the signal goes into charging and
discharging it. The top plate of the capacitor is
rising and falling very quickly and this is making the lower plate rise
and fall a small amount.
This puts a small current though the resistor and this has an effect
on reducing the amplitude.
amplitude of the output is reduced as shown in Fig
As the frequency is increased further as shown in dia
c, the top plate of the capacitor is
rising and falling as fast as it can and the lower plate is rising and
falling too. This puts most of the amplitude-loss in the resistor
but the signal is not reduced any more.
It does not matter if the capacitor is above or below the resistor, the
attenuation is the same.
Once you have a concept of the way a capacitor reacts to a high and low
frequency, you can see how a circuit will pass or prevent (attenuate) a
There are many different types of filters and they are all designed to
improve the output of a poor signal, such as removing background "hiss"
or "rumble" in audio recordings.
The following two circuits show the effect of adding capacitors and
resistors between the output and input:
Fig 72e is a low-pass
filter that provides unity voltage gain to all frequencies below
10KHz, but it rejects all frequencies above 10KHz at 12dB per
octave. It is used to remove high frequency noise from
Fig 72f is a high-pass filter
that provides unity voltage gain for all frequencies greater
than 50Hz. However, it provides 12 dB per octave rejection to
all frequencies below 50Hz. It is used to remove low frequency
noise from audio recordings.
The transistor is configured as an emitter-follower biased at
about half the supply value by the low-impedance junction formed
by the top 10k resistor and the lower 10k in parallel with the
Negative feedback applied through the filter network of the 33k
and 220n and the 10k and 220n creates an active filter
SQUARE-WAVE TO SINEWAVE
Square-wave to Sinewave
Converting a square wave to a sinewave
can be done with a
capacitor and resistor in series.
The capacitor takes time to charge and discharge, (according to
the value of the resistor) and the output of the circuit is
where the two components join.
The circuit is actually a TIME DELAY
circuit, but this time we are monitoring the rise and fall of voltage on
the capacitor, rather than just detecting the value after a
period of time.
The Square-wave to Sinewave circuit is drawn slightly different
to the Time Delay circuit so you can see what is happening.
The resistor is drawn at an angle of 90 degrees to the capacitor
to show the signal is attenuated (reduced) in the resistor and
appears across the capacitor as an amplitude.
The value of the resistor and capacitor are important.
When they are the correct values, the signal is a very-good
sinewave with maximum amplitude as shown in diagram A.
When the values are too high or too low, the result is shown in
figures B and C.
STAGE - or Digital State
also called the DIGITAL CIRCUIT or DIGITAL TRANSISTOR
There is no such thing as a DIGITAL TRANSISTOR or
an AUDIO TRANSISTOR.
All transistors are just "TRANSISTORS" and the surrounding
components make the transistor operate in DIGITAL MODE or ANALOGUE
It's a bit like saying money is: "food money" or "petrol
But we have some transistors that have inbuilt resistors to make it suitable for connecting to a digital circuit without the
need for a base resistor.
Here is the datasheet for an NPN transistor
BCR135w and PNP datasheet for
These transistors are called "Digital Transistors" because
the "base lead" can be connected directly to the output of a digital
stage. This "lead" or "pin" is not really the base of the transistor
but a 4k7 (or 10k) resistor connected to the base allows the
transistor to be connected to the rest of a digital circuit.
You cannot actually get to the base. The resistor(s) are built
into the chip and the transistor is converted into a
"Digital Transistor" because it will accept 5v on the "b" lead.
The 47k is not really needed but it makes sure the transistor is
fully turned OFF if the signal on the "b" lead is removed (in other
words - if the input signal is converted to a high-impedance signal
- see tri-state output from microcontrollers for a full
This transistor is deigned to be placed in a circuit where the input
changes from low to high and high to low and does not stop mid-way.
This is called a DIGITAL SIGNAL and that is one reason why the
transistor is called a digital transistor. (However you could stop half-way but the transistor may heat up and
get too hot).
Any transistor placed in a digital circuit can be called a "digital
transistor" but it is better to say it is operating in DIGITAL
All the circuits and stages
we have discussed have been amplifiers for audio signals.
However there is another signal that can be processed via an amplifier.
It is called a digital signal or "Computer" signal. It is a signal that
turns a transistor ON fully or OFF fully.
The simplest example of a digital circuit is a torch. The globe is
either ON or OFF. But a torch does not have any transistors. We can
simply add a transistor and the circuit becomes
2 STATES: ON
OFF. It is never half-ON or half-OFF.
The secret to turning a transistor ON fully is base current. If you
supply enough base current the transistor will turn ON FULLY.
The Digital Circuit is the basis of all computers. It produces an
outcome of "0" when not active or "1" when active.
This is called POSITIVE LOGIC.
Fig 72. A TORCH is
an ON-OFF circuit.
A Digital circuit is an ON-OFF circuit.
Fig 73. This is the
simplest DIGITAL CIRCUIT. The globe illuminates when the switch
Whenever a transistor is used in a stage that has a defined ON
state and OFF state, it is using the OFF and ON state of the
transistor and the transistor is called a DIGITAL TRANSISTOR.
It is simply an ordinary transistor used in a "DIGITAL WAY."
It is NOT a special type of transistor.
Two reasons why a Digital Circuit was invented:
1. It produces either "0" or "1" (LOW or HIGH) and these are
accurate values. By combining millions of "digital circuits" we can
produce counting and this is the basis of a computer.
2. When a circuit is OFF, it consumes no power. When a circuit is
fully ON the transistor also consumes the least power. This is because
the globe is illuminated brightly and the transistor
remains cool - as it has the lowest voltage across it.
The "ON" "OFF" states are called LOGIC STATES or
STATES and when two transistors are put together in a circuit with "cross-coupling"
they alternately flash one globe then the other.
Fig 74. This circuit
is called a FLIP FLOP or ASTABLE MULTIVIBRATOR.
(AY-STABLE - meaning not stable)
THE TRANSISTOR AS A SWITCH
Using a transistor as a switch is exactly the same as using
it in DIGITAL MODE or in a DIGITAL CIRCUIT or in a
LATCH CIRCUIT or any other circuit where the transistor changes from
OFF state to ON state VERY QUICKLY.
A transistor in this type of circuit is called a SWITCHING TRANSISTOR
and it may be an ordinary audio transistor but it is called a switching
transistor when used in a switching circuit.
The two Darlington transistors in Fig 74 are SWITCHING
TRANSISTORS and the circuit is an
One of the most common circuits is used to activate a relay. A relay
must be turned ON or OFF. It cannot be half-on or half-off. The
transistor changes from OFF to ON very quickly. It is called a switching
All transistors used in a DIGITAL CIRCUIT are switching
transistors. DIGITAL CIRCUITS or DIGITAL LINES are either
HIGH or LOW.
When a digital transistor is turned ON (saturated) the output is
LOW. When a digital transistor is OFF the output is HIGH.
The output is taken from the collector of a common-emitter stage.
This is called two MODES of operation. ON and OFF.
Any circuit that operates in TWO MODES is called a DIGITAL
DRIVING A RELAY
Any circuit that drives (powers) a relay is essentially a
DIGITAL CIRCUIT. Sometimes the driving circuit can gradually turn ON and
when the collector current is sufficient, the relay pulls-in.
When the collector current falls to a lower value, the relay
We like to think of the driver stage as a digital stage so that we
guarantee the relay will pull-in and drop-out.
Here's an important feature that has never been mentioned before:
A relay must pull in quickly and firmly so the contacts close with
as much pressure as possible. This prevents arcing when closing and
opening and ensures a long life for the relay.
That's why the driver circuit should be an ON-OFF or DIGITAL design.
The following circuits are NOT high-speed, but will activate a relay
Circuit A activates
the relay when light falls on the LDR. The level of illumination can be
adjusted by the 10k pot.
the relay when the illumination reduces. The level can be adjusted by
the 10k pot.
is an emitter
follower and although it works in a similar way to circuit B, the
voltage on the collector is less than 12v by about 1v and this creates
extra loss and added temperature-rise in the transistor.
A 12v relay might work on a 9v circuit, but as we said
above, the relay needs to close with as much pressure as possible to
prevent the contacts arcing.
A 6v relay will work on a 12v circuit if you add a 5v1 or 5v6 zener in
series with the coil.
A 12v relay will work on a 24v circuit if you add a 12v zener.
If the driver transistor turns on and OFF very fast, you will
need a diode across the coil to prevent the back voltage damaging the
transistor when the relay turns OFF.
The circuits above operate very slowly and a diode is not needed.
RELAY via CONSTANT CURRENT
A relay can be driven by a circuit called a CONSTANT CURRENT
CIRCUIT and this means the maximum current delivered to the
relay is fixed by the driving transistor. This means the supply
voltage can rise higher than the recommended voltage and the
relay will still be supplied with its rated coil current.
The circuit can also be called CURRENT LIMITING or
CURRENT CONTROLLING or MAXIMUM CURRENT.
For this circuit to work you must know the current required by
the coil for it to pull-in effectively.
This value can be obtained from the specification sheet for the
relay. It will specify the coil voltage and the coil resistance.
In our example the coil is 12v and the resistance is 360 ohms.
This identifies the relay as requiring 33mA to close the
Most relays will work on a slightly lower voltage and slightly
higher voltage, but if the supply ranges from 12v to 24v, you
need this type of circuit:
The BD 139 only
allows (a maximum of) 33mA to flow because the base sees 1.7v due to the
characteristic of the red LED.
The voltage between base and emitter is 0.7v and this
means 1v will be dropped across the 33R when the current rises
Suppose you fit a relay requiring 100mA. If it is a 12v relay,
it will have about 120R coil.
The transistor will turn ON and the current will increase. As
soon as the current reaches 33mA, the voltage across the 33R
will be 1v and since the transistor is only seeing 1.7v on the
base, if the current rises any further, the voltage across the
33R will rise and the voltage between base
and emitter will be less than 0.7v. This will cause the transistor
This means only 33mA will flow through the relay and it will not
be enough to close the contacts.
The relay will not work.
If you increase the supply from 12v to 24v, the current will not
rise above 33mA and the relay will still not work.
If you fit the 33mA relay and increase the voltage from 12v to
24v, only 33mA will flow though the relay and 12v will appear
across it. The other 11v will appear across the
collector-emitter terminals of the transistor and 1v will be
across the 33R.
THE UNIJUNCTION TRANSISTOR
The Unijunction Transistor (UJT) and
the Programmable Unijunction Transistor (PUT) and the ordinary
transistor ARE ALL THE SAME.
A voltage on the base or the "E" lead or the Gate, will turn them
all on and the resistance between all the
leads will reduce to a small value.
Actually a small voltage will develop between the leads and this is
called a characteristic voltage and cannot be altered.
BUT here are the differences. The voltage on the base of the
ordinary transistor is about 0.7v to turn it ON. The voltage on the
Unijunction Transistor can be up to 3.5v The voltage on the Gate for
the PUT is about 0.7v and can be slightly higher or lower.
When the voltage is removed from the base, "E", or Gate, the device
If you add a voltage divider to the base of the ordinary transistor,
you can turn it on at say 1v or 2.6v or any value. You can call this
"programming the transistor"
That's what a Programmable Unijunction Transistor is. It is an
ordinary Unijunction Transistor that can be placed in a circuit with
voltage divider resistors so the device turns on at a particular
Silicon Controlled Rectifier has a different feature. When the
voltage on the Gate turns the SCR ON, it "latches" and stays ON
until the supply voltage is removed and the current through the SCR
falls to zero.
The two diagrams opposite are NOT identical. The SCR remains "latched" via
the Gate. Making the Gate 0v, will not unlatch the SCR.
The UJT circuit shows a typical
arrangement with a low-impedance device, such as a speaker,
between B1 and 0v rail.
The way it works is this:
The capacitor charges via the 10k resistor. During this time the
resistance between B1 and B2 is infinite. The charging is a sawtooth
waveform because the charging is a rising gradient and the discharge
is very rapid - into the speaker. The emitter detects when the
voltage rises to about 3.5v to 4v (this is a characteristic of the
UJT device) and at this point the
transistor "turns on" and the resistance between the Emitter and B1
becomes very low and is effectively equivalent to a diode in
The voltage (the energy) in the capacitor is then passed to the
speaker and this produces a "click." This is the only way the
speaker gets its energy. The voltage across the capacitor falls very
rapidly and when it reaches less than 0.7v, the
transistor turns off and the cycle repeats.
Why can't you use an ordinary transistor in
the circuit above?
Because an ordinary transistor turns ON slowly when the voltage
is 0.55v and turned on more at 0.6v and turned ON fully at
about 0.7v. There is only a very small "gap" between 0.55v
and 0.7v and this would be the charging and discharging voltage
for the electro. But during this range the transistor is turning
ON too and effectively removing the charging capability of
the 10k to charge the electro and thus the circuit would never work.
You can't use an SCR because it "latches" and the cycle will not
One of the
main differences between Unijunction and PUT: A Unijunction Transistor needs about 2v to
3.5v to turn ON.
A Programmable Unijunction Transistor turns on at 0.2v to 1.6v
(normally 0.5v to 0.7v)
LATCH CIRCUIT -
an SCR made with transistors
Fig 75. Latch Circuit
Fig 75. Circuit B
is a LATCH. The two transistors instantly change from the OFF
state to the ON state.
This is also classified as a DIGITAL CIRCUIT. The circuit can
also be called an SCR made with transistors. Circuit A shows an
SCR in action. The top switch turns the SCR ON and it stays ON
when the button is released. To turn the SCR off, the lower
switch is pressed.
The SCR in circuit A produces a
The SCR can be replaced with two transistors as shown in circuit
Fig 75aa. Latch Circuit
Fig75aa is a LATCH and the PNP/NPN
transistors are "latched-on" by pressing S1. The circuit will
also turn on with a resistor as high as 15k across S1 as we only
need to put 0.6v on the base of the BC547 transistor. The 10k on
the base forms a voltage divider and this determines the
resistance of the "turn-on" resistor. The emitter of the BC547
transistor does not move when this voltage is applied and the
collector of the BC547 pulls the base of the BC557 down to turn
the PNP transistor ON. This action takes over from the 15k
resistor and the two transistor remain ON.
The base of the BC547 is pulled to nearly rail voltage and the
emitter is 0.6v lower. The 10u electrolytic charges to cater for
the voltage-difference between the collector of the first
transistor and the voltage on the emitter of the BC547.
When the first transistor turns on, the voltage on the collector
reduces and this pulls the positive lead of the 10u towards the
The negative lead of the 10u cannot fall as it is connected to
the emitter of the BC547.
This means the 10u discharges and when the first transistor
turns off, the positive lead rises and takes the negative lead
with it. This reduces the voltage on the emitter of the BC547
and the transistor turns OFF.
This is how the LED turns off.
Further blowing into the microphone will make the emitter lead
of the BC547 rise and fall and this will make the LED flicker,
just like trying to blow out a candle.
Fig 75a. Latch Circuit
Fig 75a. This circuit
is a LATCH. The two transistors instantly change from the OFF
state to the ON state when the input voltage rises above 0.6v
The 22k POSITIVE FEEDBACK resistor keeps the circuit
ON when the input voltage is removed.
The 6v supply must be removed to turn the LED off.
Fig 76. Touch Switch
Fig 76. This is a
circuit of a TOUCH SWITCH. Touching the "ON" pads turns
second and third transistors as they are a SUPER-ALPHA PAIR or
DARLINGTON arrangement and have a very high input impedance and
very high gain. The output of this pair goes to a PNP transistor
that amplifies the 5mA current from the Darlington to deliver
250mA to the globe.
A feedback line from output to input via a 4M7 keeps the circuit
ON when your finger is removed and provides a "Keep-ON" voltage
The first transistor removes this
"Keep-ON" voltage and current
when a finger is placed on the OFF pads. .
How can you tell a
1. Absence of capacitors.
capacitors in a
A switch or push-button will be activating the circuit.
The circuit will be driving a
ON - OFF
item such as a relay or globe.
The two states of a transistor in a DIGITAL CIRCUIT are: OFF -
called "CUT-OFF" and ON - called "'SATURATION."
To saturate a transistor the base current is simply increased until the
transistor cannot turn on any more. In this state the collector-emitter
voltage is very small and the transistor can pass the highest current
and the losses (in the transistor) are the lowest.
Fig 77. This circuit
has only two states. ON and OFF. The ON button turns off the
first transistor so the second transistor turns the
This is called a TOGGLE ACTION and the circuit is a BINARY
CIRCUIT or BISTABLE CIRCUIT called a BISTABLE SWITCH or a bistable
of the MULTIVIBRATOR family (BISTABLE MULTIVIBRATOR).
It can also be called a LATCH as it stores one bit of
information and is the basis of a COMPUTER.
ILLUMINATING A GLOBE (Lamp)
The base needs 12mA
"turn on" the globe
a globe is not easy. A globe (lamp) takes 6 times more current to
get it to start to glow because the filament is cold and its
resistance is very low.
For instance, if a lamp requires 100mA, it will take 600mA to get it
to start to glow.
This means the transistor must be capable of delivering 6 times more
current and when the filament is glowing, the current will reduce to
This means a transistor capable of delivering 800mA (BC337, BC338)
will be sufficient for the job, however the gain of the transistor
will have to rated at 50 because the current is getting near to the
The base current will have to be 600/50 = 12mA.
You need to supply the base with 12mA and the lamp will illuminate.
When the lamp is glowing, the transistor only needs about 1-2mA,
however it is difficult to reduce the current to the base and if the
12mA is supplied to the transistor, it simply means that 2mA
will be used by the transistor and 10mA will pass through the
base-emitter junction and be wasted.
Experiment: Shine a light on the LDR and the
globe will gradually get brighter.
You can change the 470R in the
diagram above for an LDR (Light Dependent Resistor) and watch
the globe illuminate.
The LDR chosen for the experiment has a resistance of 300k when
in the dark and about 100 ohms when a bright light is shone on
it. Do not shine a light too close to the LDR as its resistance
will be so low that a very high base current will flow.
If the circuit does not work, the globe requires more than 600mA
to get it to start to glow or 100mA when fully illuminated.
Fig 77b. This is
part of a counting circuit and since it takes many
transistors to create a circuit to count to "2" it is not
practical to make it using discrete components. That's why
INTEGRATED CIRCUITS were invented where dozens, then hundreds
then thousands then millions of transistors are connected to
produce counting chips and "bit-storing chips" and
Before we cover our next type of circuit, we will explain a 2-transistor
directly-coupled arrangement from Figs 52 and 66. It is interesting as
it can be used as a digital circuit or an analogue circuit.
Fig 78. Two facts to
1. Point "A" never rises above 0.6v as it is connected to the base of
the second transistor.
2. When the first transistor is turned ON, the collector-emitter
voltage is 0.3v and the second transistor is OFF - this is
because the base of the second transistor needs 0.6v to turn ON.
In other words, when one transistor is ON the other is OFF.
There is a very brief change-over point where the first
transistor turns ON a little more and the second transistor turns
OFF a very large amount. If you can find and maintain this change-over point, the two
transistors will work in analogue mode with high gain but if you
pass this point very quickly, the two transistors will
operate as a switch in DIGITAL MODE.
We can turn this circuit into a DIGITAL CIRCUIT.
The secret to doing this is FEEDBACK and the name of the circuit is a SCHMITT
THE SCHMITT TRIGGER
Fig 79a. Schmitt Trigger Circuit
Fig 79aaa. Basic Schmitt Trigger
Fig 79a. A Schmitt
Trigger takes a slowly rising or falling voltage and turns it
into a fast-acting ON-OFF signal. The secret is the feedback
line shown in red.
The circuit can also be called a
When the input is LOW the output is LOW.
It is a form of bi-stable multivibrator.
The distance between the lower voltage and the upper voltage (at
which the circuit changes state) is called the HYSTERESIS GAP. This
can be widened or narrowed via the 1k resistor (the 100k pot
needs to be re-adjusted when the 1k is changed).
The basic Schmitt Trigger only needs 3 resistors. This is
Fig 79aaa. When the first transistor is conducting (turned ON) and the
second transistor is OFF, a voltage develops across the 1k due
to the collector-emitter current of the first transistor.
When the first transistor turns OFF, about the same current
flows though the 10k collector-load, through the base of the
second transistor, plus current flows through the 4k7 load
resistor. This means about three times more current flows
through the 1k emitter resistor and thus the voltage across it
will increase about 3 times. This is the
voltage of the
Fig 79. Schmitt Trigger Circuit
Fig 79. This circuit takes a slowly rising or falling voltage and turns it
into a fast-acting ON-OFF signal to operate a LED or relay.
This is done via the positive feedback line shown in red. It is
called positive feedback because it ADDS to the change to speed
This circuit is fully explained in the:
Talking Electronics website CD.
Fig 79aa. A Schmitt Trigger
Fig 79aa is a Schmitt
Trigger made from NPN and PNP transistors.
As the voltage on the input rises, the first transistor is
turned on slightly and a small voltage is developed across the
100k emitter resistor that reduces the "turn-on" effect
slightly. This means the input voltage must rise more. As the
input voltage rises more, the second transistor starts to turn
on and the collector voltage rises. This voltage is passed to
the base of the first transistor to assist the input voltage and
because the collector voltage of the output transistor rises
considerably, it has a large effect on turning ON the first
transistor. They turn each other ON until they are both fully
The 2M2 has taken over from the 470k and made the base of the
input transistor slightly higher. The input voltage has to drop
a small amount before the pair will start to turn off.
The circuit has created a small gap between the low and high
input voltage (and between the HIGH and LOW input voltages)
where the circuit does not change from one state to the other.
This gap is called the HYSTERESIS GAP.
The output of the Schmitt Trigger in Fig 79aa is classified as
"high impedance" (due to the value of the 100k on the output)
and this must be connected to a stage with a high input
impedance so the voltage on the output of the Schmitt
Trigger is not affected.
Here is another Schmitt Trigger:
only problem with this circuit is the load is not pulled very
close to the 0v rail. The minimum voltage across the PNP
transistor is about 1.5v, however the circuit works well and
provides a wide voltage between ON and OFF. It has been
tested with a load of about 30mA across the 100R.
Changing the 470R to 1k increases the "gap."
Fig 79ab. The Monostable or
Before we leave the MULTIVIBRATOR family, the third type of
Multivibrator is the MONOSTABLE MULTIVIBRATOR.
It is only stable in ONE state. This is called the "rest" state.
The other state is "timed" via a capacitor.
The circuit is triggered and it changes to the other stage and a
TIMING CAPACITOR C charges via a resistor R (called a TIMING
CIRCUIT) and a multiplication of the two produces a value called
When it is charged, the circuit drops back to the
While the output is high, input pulses (trigger pulses) have no effect on the
circuit. Also, if the input is triggered and kept high longer
than the time constant of C and R, the output will NOT stay
high for longer than the time constant.
This circuit is also called a PULSE EXTENDER.
We have described the transistor as an amplifier and the fact that
POSITIVE FEEDBACK can turn a transistor ON more and more,
so it changes from: "not-turned-ON" to
in a very short period of time. When a transistor is operating
in this mode, it is said to be in DIGITAL MODE. We saw the
effects of DIGITAL MODE in Figs 74, 75, 76, 77 and 78. The
advantage of digital mode is the transistor dissipates the least
heat in either state.
The transistor can be put into a chip (IC - Integrated Circuit) and used
in Digital Mode. When this is done, the transistor is put into a circuit
called a GATE. A Gate is simply a BUILDING BLOCK in which
the output changes from LOW to HIGH or HIGH to
LOW very quickly. The simplest GATES are called AND, OR,
NAND, NOR and NOT. In general a GATE operates on a 5v supply (this
applies to gates in the TTL family. They cannot withstand a voltage
higher than 5.5v. CMOS gates operate to about 14v-16v and some are
up to 20v) and the
input has to change from LOW to HIGH or HIGH to
LOW very quickly and the output will change from LOW to
HIGH or HIGH to LOW very quickly. You may think the
gate is not achieving anything, but most gates have 2 or more inputs and
the output is "more powerful" than the input. The introduction of
GATES revolutionised the development of the computer and was the
beginning of the DIGITAL AGE.
shows AND, OR, NAND, NOR and NOT gates produced with
"n" indicates any number of inputs. ("n" is an
We have shown circuits with the load (such as a speaker or
LED) above the transistor or below (it cannot be in both places at the
same time). The position of the LOAD introduces two new terms: SINKING AND SOURCING
Fig 79b. When the
speaker (LOAD) is placed above the transistor, the circuit is said to
be SINKING the current. A BC547 does not have the
collector-current to adequately supply an 8R speaker. You really
need a BC338.
There is no advantage in one placement over the other. If the load is
connected to "chassis" such as a globe in a car, the circuit will need to source
Fig 79c. When the
speaker (LOAD) is below the transistor, the circuit is said to
be SOURCING the current.
The only difference between the two circuits is the voltage on
the base. Fig 79b will operate with a base-voltage less than 1v,
while Fig 79c is an emitter-follower design and will need a
voltage on the base from about 1v to full rail voltage.
Interfacing simply means:
When a circuit connects
a device (such as a microphone), to an amplifier, it is called
INTERFACING. The characteristics of the microphone are matched to
the input requirements of the amplifier. Or a relay may need to be
connected to the output of an amplifier (If the amplifier does not have enough current
to turn the relay ON).
In most cases, the output of a circuit or a "pick-up" device (sometimes called a
TRANSDUCER) does not have enough VOLTAGE or
VOLTAGE-SWING or AMPLITUDE to drive the next circuit or device and it
needs an amplifier.
That's why we have to add a circuit between.
The circuit we add has a number of names:
When it increases the CURRENT, we call it a BUFFER.
When it matches a high impedance to a low impedance or a low impedance
to a high impedance, we call it IMPEDANCE MATCHING.
Or when we need an increase in voltage, it is called an AMPLIFIER.
In ALL "stages" (common-base, common-collector and common-emitter) the
current is increased.
Interfacing can be as simple as adding a resistor or capacitor, but this
is usually called "connecting" or "coupling".
We have learnt that all devices and circuits have an ability to deliver
a "waveform" or "amplitude" or "voltage" and this can be weak or strong
according to the amount of current it can deliver.
We have also learnt that this current may be delivered from the load
resistor or from the device itself. It does not matter how the current
is delivered; the size of the current (the amount of current) is
We have also covered the fact that the input to a circuit (or "stage")
requires current and when these two are equal, the matching is ideal.
But this rarely happens.
If the input requires more current, the voltage (or
voltage-swing) from the previous circuit or device will be reduced. If the input requires less current, the
voltage-swing will be affected a very small amount. But in ALL
cases the voltage-swing will be reduced - because you ARE supplying SOME
energy to the stage that follows.
Interfacing is not easy.
You have to know the output voltage of the device and the
current it can supply.
The current it can supply is related to its OUTPUT IMPEDANCE.
OUTPUT IMPEDANCE basically means its output resistance. A
low resistance or LOW IMPEDANCE means it is capable of
delivering a HIGH CURRENT. A high-impedance device cannot
deliver very much current. A stage with a high output impedance
cannot deliver very much current.
All these terms are relative. When we say: "cannot deliver much
current" the value of current can be less than 1uA or 50mA.
It depends on the circuit we are discussing and if you are
working with low-current circuits or power circuits.
We have also learnt that the input impedance of a stage can be
high or low and the voltage-swing it will accept can be small or
large. (for instance, an emitter follower stage will
accept a large input voltage).
This gives us a wide range of values (parameters) that may need
to be joined together - INTERFACED.
In some cases the output voltage of a device or circuit will be
HIGH and by connecting a capacitor between the two stages, the
output voltage will be "absorbed" in the capacitor and the
energy from the output stage will be transferred. The "energy"
is a combination of the voltage-swing and the current.
But if the output voltage is very small, we may need to amplify
it to deliver a high voltage to a device.
This is the case in the following requirement.
A piezo diaphragm or electret microphone is required to be
interfaced to the input of a microcontroller.
The output of these devices is about 10mV and the input of a
microcontroller requires about 3.5v (3,500mV).
This involves an amplification (gain, amplification factor) of
10:3500 = 350 and requires two stages of amplification.
The output of a piezo and microphone are classified as high
impedance and the input of a microntroller is also high
This means the two amplifying stages can be low-current stages
(also called high-impedance stages) and the load resistors can
be high-value (about 22k - 100k).
The following two circuits have been designed for this application:
Fig 79d. In this
circuit the first transistor is self-biased and the 2M2 base
bias resistor turns the transistor ON and the voltage on the
collector is only about 1.8v.
This means the collector has to drop by only 1.2v for the second
transistor to turn off and the 100k will produce 5v on the input
to the microcontroller.
If the transistor has a gain of 100, the electret mic or piezo
has to produce a 12mV signal to activate the circuit.
When the load resistor is increased to 100k, the collector has
about 850mV on it, and it only has to drop 300mV for the signal to
enter the microcontroller. This makes the 100k load resistor
produce a more-sensitive circuit.
When no audio is being detected, the output of the second stage is
Fig 79e. This
circuit has been taken from Fig 71acc. It
is a bootstrap circuit and produces a very clever "switch."
The circuit sits with the first transistor turned ON and the
second turned OFF as can be seen in the first line at the top of
the output waveform - up to the red dot. When a
signal is picked up by the microphone (this is the
red dot on the waveform),
a negative-going signal of about 100mV will turn the transistor
off slightly and the second transistor will turn ON. The 4u7
will be "pulled down" and completely take over from the signal
from the microphone. It will turn the first transistor off more and
the second transistor will be turned ON more. This will continue until
both have completely changed states.
They will stay like this until the 4u7 is charged in the
opposite direction and the base of the first transistor sees
0.7v. This causes the second transistor to turn off and the 4u7
rises and turns the first transistor ON more. The 4u7 gets
slowly discharged and the circuit remains in this state.
The circuit produces a very clean output every time it detects
The duration of the low in the graph can be shortened by
reducing the value of the electrolytic.
Fig 79f interfaces a phase-shift
oscillator (see Fig 90) to a speaker. This is a very
difficult thing to do as the phase-shift oscillator has an output
that is very close to rail-to-rail and any loading of the output
will cause it to stop working.
In an attempt to interface the oscillator to a speaker we have added
an emitter follower transistor and a 1k separating resistor, plus a
100R in series with the speaker. This should give a loading of about
20k and the circuit should work. Otherwise the 10k will have to be
reduced or the 100R increased.
ANALOG TO DIGITAL
of the circuits we have described convert an ANALOG signal to a DIGITAL
These are called ANALOG TO DIGITAL CONVERTERS but we have not
given them this specific name because we have been concentrating
on other features.
We will now cover the concept of Analogue to Digital Conversion.
An ANALOGUE signal
rises and falls but doesn't have any defined amplitude or frequency.
This signal cannot be delivered reliably to a circuit that requires a DIGITAL
SIGNAL as the amplitude may not be large enough.
A DIGITAL CIRCUIT requires a digital signal and this type of
signal is either a constant HIGH or LOW and the amplitude must be very
close to rail voltage or almost 0v. And it must change from one state to
the other very quickly.
Delivering a high amplitude analogue signal may be recognised by
a digital circuit when it reaches a peak or goes to 0v, but this is not
guaranteed or reliable.
In addition we may want the signal to be a CONSTANT HIGH when the audio
This is what an ANALOG TO DIGITAL circuit will do. It will
produce a constant HIGH when audio is present and ZERO (LOW) when the
audio is not present. Or pulses that are nearly rail voltage and very
close the 0v.
To convert an analogue signal to a digital signal we need to deliver
ZERO OUTPUT (called a LOW output) when the signal has a small amplitude and a
HIGH output when the signal has a high amplitude.
To do this we use a common-emitter stage, as it has a high voltage-gain
and this is what we need.
There are many ways to convert an Analogue signal to a Digital signal
but the basic way is to amplify the signal by a large
amplification-factor so the resulting waveform will rise to the voltage
of the rail (or even higher). It cannot go higher than rail
voltage but you will see what we mean in a moment.
This is normally called "over-driving" the signal and if this is done in
an audio circuit, the result is distortion. But we are not going to
listen to the output, so we take advantage of this feature to produce a DIGITAL OUTPUT.
Fig 80a shows an analogue signal. It is made up of lots
of sine-waves and may be as high as 2v or only a few millivolts.
are generally expressed in mV, to make them instantly
recognisable and easy to talk about.
In general this type of signal will be too small to be detected by a
Digital Circuit. A Digital Circuit needs a signal greater than about 3,500mV so
the waveform appears on the input line as a HIGH, during the peak
of its excursion. It should be nearly 5,000mV for
80b. A Digital
Circuit will detect a waveform larger than 4.5v as a HIGH and
less than 0.5v as a LOW
Only the large excursion(s)
will be detected by a Digital circuit as the other parts
will not rise high enough to be detected. To increase the analogue signal to as
much as 5,000mV, an amplifier is needed.
The amplifier maybe one
or two stages, depending on the amplitude of the original signal.
Each stage of an amplifier will increase the size of the signal about 70
times. If you are very lucky, you may get an amplification of 100x (100
times). Thus a 5mV signal with one stage of amplification will produce a 350mV
to 500mV signal. This is not sufficient to be detected by
a Digital Stage. Another stage will easily produce a full 5,000mV signal.
The second stage only needs to amplify the signal about 10 to 12
times and a higher gain simply
drives the waveform into "bottoming"
and "cut-off" as shown in fig 80c.
This means the waveform
will be "clipped" at the top and bottom and converted to a fairly
Suppose you have a waveform that is higher than 5mV (say 30 - 50mV) and
want to know if it will trigger a Digital Circuit with a single stage of
Connect the components as shown in Fig 80d and write a program to illuminate a LED when the waveform is detected.
There is only one problem
with the circuit in Fig 80d.
At the end
of a whistle or speech, the LED may be illuminated or
extinguished. It all depends on the last cycle of the
waveform. The circuit sits with the output approx mid-rail
and the micro does not know if this is a high or low, and
takes the reading by the direction of the last cycle.
Some of the inputs of a micro are Schmitt Triggered. This means a HIGH
has to be 85% to 100% of rail voltage for it to be seen as a HIGH and
between 20% and 0% to be seen as a LOW.
The non-Schmitt Trigger inputs see a LOW as 20% to 0% and a HIGH
as above 2v for 5v operation.
If the last cycle went from zero to mid-rail the micro will
see the waveform as a low on Schmitt Trigger inputs and a
HIGH on the other inputs. This problem can be overcome by
adding a second stage that only produces a LOW when audio is
detected. It also increases the amplitude of the audio to
guarantee triggering of the Digital Circuit. This is shown
in Fig 80e.
transistor in Fig 80e is called
a DIGITAL STAGE. This simply
means a biasing resistor is not connected to the base of the second transistor
so it turns on fully when a signal greater than 650mV is
detected and is turned off at other times. This stage is ideal for
a micro or other Digital Stage as only two voltage levels are delivered. Either 0v or rail
voltage (5v).The other advantage is it does not take any
quiescent (idle) current.
This stage is only suitable if you are sure you have plenty of
"over-voltage" to drive the transistor into saturation. By this we mean
you must have at least 1v (1,000mV) drive signal so you can be sure the
transistor will turn on (saturate).
The fast rise and fall times means you have a "clean" HIGH and LOW.
Fig 80f couples a magnetic pick-up to the amplifying circuit
so the biasing of the first transistor can be determined
by the value of the base-bias resistor. The coil cannot be
connected directly to the transistor as the low impedance
(resistance) of the coil will upset the bias on the base.
this arrangement, the descending part of the input waveform of a
few millivolts will turn off the transistor, while the ascending
part of the waveform will not have any effect.
A coil of wire of any size will be suitable and to make it an
effective collector of magnetic flux, it should have a magnetic
core such as ferrite. No other impedance-matching is necessary.
shows an electret microphone connected directly to the base of a
two-transistor amplifier. This arrangement will work and
provides the best transfer of signal from the microphone to the
But biasing the first transistor is a very difficult thing to
do. The electret microphone needs a very small current to
operate and the series resistor allows this current to flow.
You will need to build the circuit, select values for the base
and collector then whistle into the microphone to see which
combination produces the highest gain.
If the resistor is a small value, the base current will be high
and the transistor will be turned on fairly hard. This is called
BOTTOMING and the collector voltage will be very low.
The electret microphone will produce a signal and this will
increase and decrease the current into the base. But the reduced
current will no turn the transistor off any appreciable amount
and the signal will not appear on the collector. If the base
resistor is very high, the electret microphone will not produce
a very large output signal and again, the waveform on the
collector will be very small.
There is no way anyone can predict the best values to use. It
depends on the type of microphone, the gain of the transistor
and the rail voltage. This is a very messy design and should be
avoided. It has been included because it has been seen in
circuits on the web.
LEVEL CONVERSION or LEVEL SHIFTING
3v3 to 5v
Suppose the output of a
device is 3v and you want to activate a device that requires 5v or
This circuit converts the signal that rises from about 0v to about
4v to a signal that rises from 0v to 5v or 12v. Note: The output is not inverted, it is in-phase with the
input due to two inversions within the circuit.
The resistor values and the type if transistor will depend on the
output current required.
3v3 to 5v:
When we talk about LEVEL CONVERSION we are only interested in
TWO STATES. The zero state or LOW LEVEL and 3v3 (or 5v) or
HIGH STATE (high level).
This circuit produces and output of 0v when the input is 0v and 5v
output when the input is 3v3.
The circuit is not linear throughout the whole transition but that
is not important for a DIGITAL to DIGITAL transfer. It is only the
HIGH or LOW condition that is important.
LEVEL CONVERSION with
LEVEL CONVERSION with
LEVEL CONVERSION with a Zener
3v3 to 5v with diodes:
Level conversion can be
done with 3 diodes to change 3.3v to about 5v, but you have to know
the current-capability of the 3.3v source and the current
requirement of the 5v section.
This arrangement is only suitable for very small current
The output is allowed to be taken HIGH by the input providing 3.3v
and the 3 diodes providing 2.1v. This allows the 4k7 to pull HIGH
with a very small current-capability.
The 10k resistor is only a safety resistor and can be removed when
the two sections are working correctly.
When the 5v line is 0v, the output is 0v. When the 5v line is 5v,
the output is about 3.2v.
5v to 3v3: A 5v signal
can be converted to 3v3 with two resistors called a VOLTAGE
This circuit will work with both analogue and digital signals.
5v to 3v3: A 5v signal
can be converted to 3v3 with a 3v3 zener.
There are over 20 different types of oscillators and many more variations.
We cannot cover them all
- so we will
concentrate on the most often-used and explain how they
Oscillators consist of one or two transistors. They start-up by one or
more components in the circuit producing "noise" or a spike
from the "mains" when the circuit is turned on. Some oscillators will
not start-up if the supply is increased gradually. When a spike or
noise is detected, the rest of the circuit amplifies it. In most cases the
noise comes from the circuit being turned ON but it can also come
from the noise generated within the junction of a transistor. This noise
is random and of little use, but it is fed to components such
as coils and capacitors as they have the ability to produce a waveform that rises and falls
smoothly and this is amplified to produce the output.
When coils, crystals, capacitors and resistor are combined with
transistors, many different effects and waveforms can be created and
this all comes under the heading of OSCILLATORS. And the circuits are
An amplifier can be
turned into an oscillator by providing POSITIVE FEEDBACK.
The purpose of providing NEGATIVE FEEDBACK is to prevent
The purpose of providing POSITIVE FEEDBACK is to create
Positive feedback is when you take a point that is rising a large
amount and pass it to a point that is also rising at the same time
but only a small amount.
In other words, the feedback line must be able to help or
assist the small-signal line. If it does not assist the
small-signal line, NO oscillation will occur.
Some oscillators have a name - either after their inventor, by the way
they are configured or by the shape of the wave. Some have 5 names.
Some have no particular name and are just called Feedback Oscillators
Fig 80. A Feedback Oscillator
Fig 80. The 10n
capacitor provides the positive feedback to keep the circuit
Fig 81. A feedback oscillator
Fig 81. The 10n
capacitor provides the positive feedback to keep the circuit
Fig 82. The positive feedback line creates the CALL tone
Fig 83. When the
third transistor is turning OFF, the collector voltage is rising
and this is passed to the base of the first transistor, to turn
This is how the circuit keeps "cycling" or oscillating.
Fig 83a. Globe flashes at 1Hz
Fig 83a. The
high-gain amplifier we studied in Fig 66, for example, has negative feedback
to prevent oscillation.
By using positive feedback we can turn the high-gain amplifier
into an oscillator.
This circuit is simply a high-gain amplifier with both
transistors turning ON via the 1k and 100k resistors. This makes
the voltage on the collector of the BC557 rise and the 22u and
4k7 passes this rise to the base of the BC547 to turn both
transistors ON more and more until they are fully turned
The 22u charges a little more and this reduces the current into
the base of the BC547 to turn it off a little. This effect is
passed to the collector of the BC557 and the two transistors
start to turn OFF. When they are fully turned off, the cycle
repeats by the transistors being turned on via the 1k and 100k.
Fig 83aa. Simple Tone Oscillator
The 2-transistor amplifier we studied in Fig 42 can be changed
slightly to drive a speaker. The two common-emitter transistors
turn on together and the 22u is "lifted" to turn on the NPN
Both transistors turn on until fully saturated and this puts
current though the speaker.
The 22u charges a little more and this reduces the current into
the base of the NPN transistors, turning it off a slight amount.
The PNP is turned off a small amount and they both keep turning
off until fully turned off.
The 10k and 50k start to charge the 22u to repeat the cycle. The 22u
produces positive feedback. It can be replaced by values from 100n to 22u
to change the frequency of the tone.
The two circuits above are examples of LOW
IMPEDANCE outputs. If the load (the globe or speaker) is
increased above about 47 ohms, the circuit will not work. They
simply "lock-ON." This is because the capacitor (electrolytic)
must be pulled down by the load at a very critical point in the
cycle. In addition, the 100k "turn-on" (or 50k and pot) resistor
must be a very high value. If it is too low, the circuits will
The critical point is this: When the circuit is fully turned-ON,
the right side of the capacitor is near rail
voltage and it is being charged via the bas-emitter junction of the
first transistor. As it becomes fully charged, the current into the
base of the first transistor reduces slightly and the transistor
turns off slightly. This effect is passed to the second
transistor and it turns off slightly too. The right lead of the
capacitor drops and this lowers the left lead to turn off the first
transistor slightly more. This is the beginning of the "turn-off
section" of the cycle.
If the second transistor did not have a very heavy load (low
resistance load), the slight turning-off of the two transistors
would not lower the capacitor and they would both remain ON.
You can see the importance
in a circuit. Some circuits will not work without feedback and some will
distort. Sometimes the feedback is POSITIVE and sometimes NEGATIVE. The trick to understanding a circuit is to locate the feedback
(component or "line")
and work out what it is doing.
Fig 83b. Positive feedback comes from the 22u electrolytic.
This is a very unusual circuit.
Normally the feedback is
Fig 83b. Here's an
oscillator circuit. We know it must have feedback to
operate, but where is the feedback?
In this circuit the 4 electrolytics are equivalent to miniature
When the circuit is turned on, they all get charged to a voltage
according to the surrounding components but the 22u is the
important component. The base of the BC557 sits at 4v and the
emitter must rise to 4.6v for the PNP transistor to turn on.
When it does, it turns on the BC547 and this transistor puts a
load of 220R across the circuit. This reduces the voltage across
the 470k/1M voltage divider and the base if the BC557 sees a
lower voltage. During this time the 22u is acting as a miniature
supply and maintaining the voltage of 4.6v on the emitter.
The BC547 turns ON more and more and even though the voltage on
the 22u drops, the circuit turns ON and this takes more current
from the 6v battery and produces a click in the speaker.
THE SQUARE-WAVE OSCILLATOR
transistors are cross-coupled as shown in Fig 84, you can
safely assume the circuit will oscillate. The frequency of
oscillation will depend on the value of the components but the oscillator is
known as a FREE-RUNNING OSCILLATOR or ASTABLE (ay-stable)
MULTIVIBRATOR and the
output is a square wave. It will have an equal-mark-space ratio
if the components are the same value.
This circuit is also called a FLIP-FLOP.
Fig 85. By
rearranging the components in Fig 84, we can draw the circuit as
one common-emitter stage driving another common-emitter stage
with a 100u providing positive feedback.
The circuit relies on the power being turned on quickly for it
to start up. Both
transistors will turn ON but one will turn on faster than the
other and prevent the other turning on.
The 100u connected to the turned-on transistor will start to
charge in the opposite direction and the second transistor will
start to turn ON. This will pull the 100u lower and the first
transistor will start to turn OFF. This will continue until both
transistors have changed states.
Here is the ASTABLE MULTIVIBRATOR with the LEDs in the emitters
instead of collectors (as is normal).
The frequency of oscillation is
approximately 1 second. The 330 ohm
resistors set the LED current to 12mA for a 6v supply.
The LEDs can also be connected as shown in this circuit. However
the circuit takes more current than the previous designs. In the
previous designs, one side of the circuit is taking current and
illuminating the LED while the other side is turned OFF
In Fig 86a, the "off" transistor is illuminating a LED while the
"on" transistor is drawing current though a 330R resistor. Both
sides are drawing current! This is a POOR DESIGN.
Fig 87. The
("ay" - meaning not-stable) MULTIVIBRATOR circuit is rich in harmonics
and is ideal for testing amplifier circuits. To find a fault in
an amplifier, connect the earth clip to the 0v rail and move
through each stage, starting at the speaker. An increase in
volume should be heard at each preceding stage. This Injector
will also go through the IF stages of radios and FM sound
sections in TV's.
Fig 88. The astable
multivibrator can be made with PNP transistors.
Fig 89. A circuit
can be made with one NPN and one PNP transistor. It ceases to be
a FLIP FLOP or Multivibrator as both transistor turn on at the
same time and the circuit becomes a Relaxation Oscillator.
THE SINE-WAVE OSCILLATOR -
also called the
A Sine-wave Oscillator can be made with a single transistor.
Fig 90. The Sinewave Oscillator
This circuit produces a sinewave
very nearly equal to rail voltage.
The important feature is the need for the emitter resistor and
10u bypass electrolytic. It is a most-important feature of the
circuit. It provides reliable start-up and guaranteed operation.
For 6v operation, the 100k is reduced to 47k.
The three 10n capacitors and two 10k resistors (actually 3)
determine the frequency of operation (700Hz).
The 100k and 10k base-bias resistors can be replaced with 2M2
between base and collector.
This type of circuit can be designed to operate from about 10Hz
to about 200kHz.
Fig 91. The phase-shift oscillator
has 3 "sections" made up of a 10n capacitor and 10k resistor.
This "section" is shown above and each
"section" produces a delay or "phase-shift" of about 60°
but the total must be 180°.
The base and collector of a common-emitter stage are 180° out-of-phase,
so the signal entering the base is 360° (IN-PHASE with the
output). This creates POSITIVE FEEDBACK.
This concept is very hard to understand so we need to explain it in simple
Points Y and Z are the ends of a long piece of rope and the three
resistors are weights tied to the rope.
You shake the rope up and down at Y and Z moves up and down at a later
time in the cycle. You know this because you can make a wave travel down
a rope. Exactly the same thing happens with a signal that enters at Y.
It takes time for the peak to reach Z.
Now consider the circuit at switch-on. The caps are uncharged and the 10k
collector resistor pulls the three capacitors high. Taking into account
the voltage-dividing effect of the three lower 10k resistors, the
collector is possibly at about 2v. The three 10k resistors start to
charge the three 10n caps and the voltage on the base falls. This makes
the collector voltage rise. This continues until the collector cannot
rise any further and the capacitors continue to charge and the voltage
on the base drops. The 100k base resistor takes over and starts to
discharge the 3rd capacitor and turn the circuit on. The collector
voltage drops and the energy in the three capacitors get passed into the
base to fully turn the transistor ON.
This all happens in a "sliding motion" that produces a sweeping output
called a SINEWAVE. It is a very "delicate" oscillator and any change to
the LOAD (10k) may stop its oscillation.
How to read the Graph: Get a ruler and hold it "up and down"
page (or on the screen) so you view the right-hand edge of the ruler
and can only see the word "phase" and "60° " Now slide
the ruler to the right and you will see the graph "A" gradually rising.
Keep moving the ruler to the right and you will see graph "B" gradually
This is how you "interpret" the graph and see how graph "B" lags (is
behind) graph "A." If you don't read the graph correctly, it looks
like graph "B" is in front of graph "A" - but this is not the case.
HOW THE CIRCUIT WORKS
power is applied to the circuit the transistor will immediately turn on
FULLY as the three 10n capacitors will acts as "short-circuits" and
current will pass to the base and the collector voltage will fall to a
very LOW level.
The capacitors will gradually charge via the 10k resistors and the
voltage across the second 10k resistor will drop (due to the charging
current into the middle 10n gradually reducing) and the voltage on the
left lead of the 3rd 10n will drop. This will reduce the voltage on the
right lead of the 10n and turn the transistor OFF. The first two 10n's
will start to charge but the voltage across the 2nd 10k will be very
small and the 3rd 10n will start to charge via the 100k resistor.
The charging of the first 10n is much faster than the charging of the
third 10n and the voltage on the output rises to almost rail voltage.
This allows the 3rd 10n to charge at a faster rate and just when the
collector voltage reaches a maximum, the base voltage reaches 0.6v and
the transistor starts to turn ON.
This lowers the collector voltage but the transistor still keeps turning
ON via the 100k and the three 10n start to charge. This action increases
the "turn-ON" of the transistor and it continues to turn-ON until the
voltage on the collector reaches a very LOW level.
The circuit works completely differently to anything described in any
Text Book. The transistor operates on a rising and falling
CURRENT (increasing and decreasing current) into the base. That's why
no-one has described it before.
The third 10n continues to charge and the "turn-ON" current for the
transistor reduces and it starts to turn OFF a slight amount. The
second 10k continues to charge the middle 10n and the voltage across the
10k reduces. This pulls the left lead of the third 10n towards the 0v
rail and the right lead follows. This action turns OFF the transistor
and the collector voltage rises.
Once you see how the circuit REALLY WORKS, it has nothing to do
with 60° phase shift for each section between X and Y.
No-one has actually sat down and worked out how the circuit works and
they just copy mistakes from others. The overall DELAY produces a signal
that ASSISTS in keeping the circuit oscillating, but the feedback signal
is not a simple sinewave.
It's possibly NOT important how the circuit works, but when you have
trouble seeing how a 10n and 10k can produce a
60° phase shift,
you have to look into the circuit and satisfy yourself.
You cannot possibly expect to fault-find a circuit if you don't have an
understanding of how it works.
Things you will learn in one circuit can be applied to another circuit
and this is how you progress.
That's why the operation of this circuit ahs been provided and is
completely different to anything that you would have considered.
THE TWIN-T OSCILLATOR
basic twin "T" filter is a passive notch filter composed of two
T-networks, with maximum attenuation occurring at
fnotch = 1/(2pRC)
One of these T networks has one resistor and two capacitors,
while the other has two resistors and one capacitor. At the notch frequency fnotch of the twin-T filter, the total phase
shift is zero, which
satisfies the requirement for oscillation. This is
the circuit generates a sine wave with a frequency equal to
Here's how the circuit
The transistor is turned ON because the base is connected to the
collector via the two 100k resistors. But any signal (such as a
voltage) coming from the collector will turn the transistor OFF.
So a rising signal has to be delayed to arrive at the correct
time to make the transistor oscillate.
There are two other factors that make the circuit work. The
first is the 4n7, 47k and 4n7. Any frequency on the collector
will pass through the two 4n7 capacitors and the 47k will be a
"load." As the frequency on the collector increases, more of the
signal will pass through the "T" network and appear on the base.
This means the top "T" network
allows the high frequencies to pass to the base.
The 100k, 10n, 100k allows the low frequencies to pass to the
base. This is because the "resistance" of the 10n becomes less
for a high frequency and it acts a "load" to attenuate the
We have two separate circuits and when the frequency at the
collector increases, the lower circuit reduces the signal. When
the frequency decreases, the top circuit does not pass the full
amplitude. There is a frequency where the amplitude of the
feedback-signal is the greatest and this is the frequency at
which the circuit operates.
Finally we have a situation where the "feedback signal" is
delayed by 180° by the capacitor-resistor networks and this
means it assists in making the transistor oscillate.
This circuit is not easy to design and is not a good performer.
It will stop oscillating if the output signal is passed to a
circuit that reduces the amplitude.
The Phase-Shift Oscillator is much more reliable.
The circuit below consists
of two "twin-T" oscillators set to a point below oscillation.
Touching a Touch Pad will set the circuit into oscillation.
Different effects are produced by touching the pads in different
ways and a whole range of effects are available. The two 25k
pots are adjusted to a point just before oscillation. A "drum
roll" can be produced by shifting a finger rapidly across
adjacent ground and drum pads
THE BLOCKING OSCILLATOR
Fig 92. The BLOCKING
OSCILLATOR circuit uses a transformer to produce POSITIVE
FEEDBACK to the base.
The circuit starts by Rbias charging Cbb to deliver voltage to
the base of the transistor via Rb (and also a small current). The transistor turns on and
produces expanding magnetic flux in the primary of the
transformer. This flux cuts the turns of the secondary (or
feedback) winding and increases the base voltage and CURRENT.
The voltage out of the top of the secondary winding is prevented
from "disappearing" by Cbb.
The transistor keeps turning ON until it cannot turn on any
more. At this point, the current through the primary is a maximum
but it is not expanding flux and its effect is not passed to the
secondary winding. The base-current reduces enormously to the
very small Rbias current
and the transistor turns off abruptly (it takes a high base
current to maintain a high collector current and the base
current is very small). The heavy current through
the primary is producing a very strong flux and it collapses,
producing a voltage in both windings of opposite polarity
and very high amplitude.
Fig 92a shows the base being "capacitor injected." This saves
one resistor and can produce a higher output. All the values
and the transformer needs adjusting for the performance
required. The start of each winding is shown with a dot. This
assumes the windings are wound in the same direction.
Figs 92b,c shows alternative ways to produce a blocking oscillator.
The difficulty with producing a Blocking Oscillator is getting a
Fig 93. A simple
BLOCKING OSCILLATOR circuit can be made with a 10mH inductor
and 80 turns of very fine wire wound on top.
The piezo diaphragm reacts to the very high "FLYBACK VOLTAGE"
produced by the primary when the transistor turns off. This type
of circuit is often used to produce very high voltages.
Fig 94. This LED
Torch circuit uses the "flyback" voltage of a BLOCKING
OSCILLATOR to illuminate a 3.6v super-bright LED from a 1.5v
Note: the 10n capacitor prevents the energy from the feedback
winding being lost. All the energy from the feedback goes into
the base of the transistor to turn it on hard.
Fig 95 shows a
Blocking Oscillator producing a regulated 5v from a 1.2v supply.
Fig 96. 2-transistors in
PUSH-PULL - as a Blocking Oscillator circuit
Fig 96. A simple
extension of the Blocking Oscillator in Fig 92c above, is shown in this
diagram. It consists of two BLOCKING OSCILLATOR transistors that
are turning each other off. The circuit starts by one transistor
being slightly faster than the other. It turns ON and produces
magnetic flux that cuts the turns of the other half of the
primary winding to increase the voltage from the battery and at
the same time it reduces the voltage to the base of the other
transistor - because the transistor allows only a very small
voltage to appear across the collector-emitter terminals when it
is turning ON. It keeps turning on until it is fully ON.
At this point the flux is no longer expanding and the generated
voltage in the winding that supplies the base voltage (and
current) ceases. This turns it off a small
amount and the magnetic flux starts to collapse and produce
voltages in the opposite direction. The voltage (and current) to the base is
less than before and this turns the transistor off more. The
voltage to the base of the other transistor starts to rise and
that transistor takes over. The two transistors operate in PUSH-PULL mode.
To reduce the wasted power in the 220R resistors,
Fig 96a uses Darlington transistors and 2k2 0.5watt resistors.
The circuit is used to drive a CFL lamp from a 12v battery.
The difficulty with producing a Blocking Oscillator is
getting a suitable transformer. A similar "flyback voltage" can be
obtained from an inductor. This will need an oscillator made up of two transistors
to drive the inductor.
This circuit is a "Buck Converter" meaning
the supply is greater than the voltage of the LED. It will drive one
high-power white LED from a 12v supply and is capable of delivering 48mA
when R = 5R6 or 90mA when R = 2R2.
The LED is much brighter when using this circuit, compared with a series
resistor delivering the same current.
But changing R from 5R6 to 2R2 does not double the brightness. It only
increases it a small amount.
The inductor consists of 60 turns of 0.25mm wire, on a 15mm length of
ferrite rod, 10mm diameter. Frequency of operation: approx 1MHz. This circuit draws the maximum
current for a BC 338.
When the circuit is turned on the 330p gets
charged. This turns on the BC547 and keeps the BC338 off. When the 330p
is charged the BC547 is not turned on as much and the 2k2 can start to
turn on the BC338. It pushes the charge on the 330p into the base of the
BC547 to keep it off. The 330p gets discharged by the 330R and the
voltage across the *R resistor turns on the BC547 to turn off the BC338.
The 1N4148 absorbs the high-voltage flyback pulse. The circuit is only
active for a very short period of time and off for a longer period of
time. This delivers a small amount of energy to the high powered LED and
allows the LED to be connected to a 12v supply (via the circuitry).
THE FLYBACK OSCILLATOR
A flyback oscillator is any oscillator where
the transistor turns off quickly and abruptly during part of the
cycle and allows the energy (the flux) in an inductor to
collapse suddenly to produce a high voltage IN THE OPPOSITE
DIRECTION. The circuit in Fig 97a consists of a
transformer with a feedback winding of 40 turns. It can be
constructed as a piece of test
equipment to test transistors, zeners and LEDs.
This circuit is a very good example of a flyback transformer in
The CFL needs a very high voltage to strike (start the
illumination) when the anodes (all projections into a glass tube
are called anodes) are not heated.
You cannot use the "turns- ratio" (50:500 in this case) to
determine the output voltage because the transformer is used in
its "collapsing mode."
This circuit will drive
a 5watt CFL tube from an old CFL lamp from 6v or 12v. It makes a
very handy emergency light.
The transformer is made by winding 500 turns for the
This consists of winding about 10 turns on top of each other before
advancing along the rod. The rod can be round or flat, from an old
AM radio. It is called a ferrite rod. The 500
turns have to be added before reaching the end and this means 100
turns has to take up 1/5th of the distance. This reduces the voltage
between the turns as the enamel will only withstand 100 volts.
Before you start winding, use at least 3 layers of “sticky-tape” to
prevent the high voltage shorting to the rod.
The size of the wire is not important and anything 0.25mm or thinner
will be suitable. After winding the secondary, the primary is 50
turns and the feedback is 10 turns.
The primary can be 0.5mm wire and the feedback 0.25mm.
Connect the transistor, components and tube and turn the circuit ON
very briefly. If the tube does not illuminate immediately, reverse
the wires to the feedback winding.
The transistor must be 2N 3055 (or the plastic version, TIP 3055).
It will get warm when illuminating the lamp and needs to be
heatsinked. The lamp must not be removed as the circuit will
overload and damage the transistor.
The circuit takes 250mA when driving a 5 watt CFL (or 18 watt
fluorescent tube) on 12v supply. The 1k base resistor can be reduced
to 820R and the brightness will increase slightly but the current
will increase to 500mA.
The circuit is more-efficient on 6v. The 1k base resistor is reduced
to 220R and the transistor remains cool.
THE BOOST CIRCUIT or BOOST CONVERTER
that converts a low voltage to a higher voltage is classified as
a BOOST CONVERTER or BOOST CIRCUIT.
Fig 97aa will
drive a super-bright white LED from a 1.5v cell.
The 60 turn inductor is wound on a small ferrite slug 2.6mm dia
and 6mm long with 0.25mm wire.
The main difference between this circuit and the two circuits
above is the use of a single winding and the feedback to produce
oscillation comes from a 1n capacitor driving a high gain
amplifier made up of two transistors.
The feedback is actually positive feedback via the 1n and this
turns on the two transistors more and more until finally they
are fully turned on and no more feedback signal is passed though
the 1n. At this point they start to turn off and the signal
through the 1n turns them off more and more until they are fully
The 33k turns on the BC557 to start the cycle again.
THE BUCK CONVERTER
that converts a high voltage to a lower voltage is classified as
a BUCK CONVERTER. Fig 97b will drive
a 1watt white LED from a 12v supply and is capable of delivering
300mA. The driver transistor is BD 327 and the inductor is 70
turns of 0.25mm wire wound on the core of a 10mH inductor.
The voltage across the LED is approx 3.3v - 3.5v
The 1R is used to measure the mV across it. 300mV equals 300mA
The diode MUST be high speed. A non-high-speed diode increases
This circuit is a very good design as it does not put peaks of
current though the LED.
The Armstrong, Clapp, Colpitts, Hartley, Wien Bridge and even
unknown oscillators like the one below all use capacitors, resistors and
coils to create a circuit called a RESONANT CIRCUIT and these two
components produce a sinewave when they receive a pulse of energy.
We are going to lump all these oscillators together as they are
variations on a similar design. There are two common oscillators that
can be recognised by the layout of the circuit.
The Colpitts oscillator has 2 capacitors across the coil with the signal taken
from the join or it may have a tuned circuit with the signal taken from
the active end. The Hartley Oscillator has a tapped coil
and these are difficult to obtain.
Fig 100. Colpitts Oscillator
101. Colpitts Oscillator
Fig 102. Hartley Oscillator
103. Hartley Oscillator
Fig 103a. Door Knob Alarm
This circuit can be used to detect when someone touches the
handle of a door. A loop of bare wire is connected to the point
"touch plate" and the project is hung on the door-knob. Anyone
touching the metal door-knob will kill the pulses going to the
second transistor and it will turn off. This will activate the
The circuit will also work as a "Touch Plate" as it does not
rely on mains hum, as many other circuits do.
The first transistor is a Colpitts Oscillator and the feedback
is via the 47p. Explaining the operation of this oscillator could take a
page of discussion. We are only going to explain one amazing feature -
how the oscillator creates the second half of its cycle. We know how the
stage turns on (via the base-bias resistor) - but how does it turn
OFF to create the other half of the waveform. Here's how:
We know that when a transistor turns ON, the collector voltage falls and
the emitter voltage rises.
Simply joining these two points with a resistor or capacitor will not
produce feedback as one is falling and the other is rising. We need to
join two points that are rising AT THE SAME TIME.
The secret comes from the inductor. The 16 turns of wire produces a
voltage in the opposite direction when the transistor is turned off.
In the first diagram of fig 103b we see the transistor turned ON and
current flows through the coil. The voltage at the bottom of the coil
will be slightly lower than the supply voltage. When the transistor is
turned off, it is effectively out of the circuit and the current flowing
through the coil produces magnetic flux that will collapse very quickly
and produce a voltage across the ends of the coil that will be OPPOSITE
to the applied voltage. This means the voltage at the bottom of the coil
will be HIGHER than rail voltage and we can think of the coil rising
above the power rail and producing a voltage 2, 5, 10 or even 100 times
higher than the power-rail voltage.
This is the amazing fact about a coil (inductor) and is the secret
behind the operation of this circuit.
In circuit 103b, this high voltage is produced at some
point in the cycle and it pulls the emitter UP a small amount via the
47p and this turns the transistor OFF. We are not going into what part
of the cycle produces the high voltage via the inductor but it DOES.
That's how the circuit produces the second part of its cycle. The inductor produces
a high voltage that starts to turn off the transistor and this allows
the inductor to produce a higher voltage and the transistor is turned
off even more. During this time the 47p feedback capacitor is charging
Most oscillators are described on the web and you
can decide which type you need for your particular application.
THE FEEDBACK CAPACITOR
The author was asked how to work out the value of the feedback capacitor
in an oscillator.
There are many different oscillator circuits and in some oscillators,
the feedback capacitor sets the frequency of operation for the circuit.
Rather than trying to work out the value mathematically, it is best to
refer to a circuit that is already operational and copy the value. You
will be able to increase or decrease the value 30% to get your required
frequency but any value less than 50% may not produce enough feedback to
keep the circuit oscillating.
You may need to increase the base-bias on the transistor to give the
transistor more gain.
In some oscillator circuits, the feedback capacitor does not set the
frequency. It is determined by a capacitor and coil in a TANK CIRCUIT.
The same suggestion applies. Refer to a circuit that already operates at
the frequency you require and use the same value capacitor.
You can halve the value and double the value and use a number of
different transistors to make sure the circuit still operates.
This proves the value you have selected is not at the extreme limit of
Look for a TUNED CIRCUIT and feedback line. It will be an oscillator.
Most have a high-impedance output and
must be connected to a circuit that will not "load" them (and reduce the
amplitude of the output) or prevent them oscillating. But some
oscillators have a very low output impedance and can drive a low-impedance device.
You have to be aware of these features.
HOW AN OSCILLATOR
STARTS All oscillators start-up due to noise.
In most cases this noise comes from a spike or peak of current
that occurs when the circuit is turned ON.
That's why some oscillators will not start if the voltage
gradually rises. Uncharged capacitors present a very small
resistance and allow a high current to flow (relatively
speaking) to other components and this starts the circuit
In addition we have the situation where a transistor produces a
small amount of noise in the base-emitter junction when a
current flows through the junction. This noise gets amplified by
the transistor and appears on the collector.
Other circuits rely on the transfer of the spike from the
collector to the base due to the capacitor being uncharged.
Some circuit rely on the fact that a tuned circuit starts to
produce a waveform when it receives a spike of energy and this
waveform is amplified by the transistor.
Other circuits rely on the waveform produced by a crystal when
it receives a spike of energy.
Some oscillators are very reliable as self-starting, while
others need the right voltage and very little loading on the
output to be reliable.
You can test an oscillator by gradually increasing the voltage
and see if it self-starts.
The starting waveforms can be so minute that you cannot detect
Every electronic component has a
value of resistance. You can measure this value with a
multimeter. But sometimes the value changes according to the
light it receives, the frequency it is operating-at, or the
voltage it is connected-to, or the sound it receives, or its
temperature or many other influences.
Sometimes the output from a circuit might be 2v, but if you put
a low-impedance device such as a speaker on the output, it
"kills" the sound.
Or you may have a nearly flat 9v battery. It measures 5v with a
multimeter but when you connect a 3v motor, it does not work.
These are both examples of poor IMPEDANCE MATCHING - yes, the
battery has a HIGH Impedance and that's why it cannot deliver
the current required by the motor.
What is IMPEDANCE MATCHING?
Impedance Matching is is connecting two items together so: "THEY
Some devices PRODUCE or DELIVER a signal or a voltage or a
Some devices ACCEPT a signal or voltage or current.
We need to connect these types of devices together.
Let's start with those that DELIVER:
An amplifier may be able to produce an output of 2v, but when a
low-impedance device (low resistance device) such as a speaker
is connected, it cannot deliver the CURRENT needed to drive the
speaker. The same with a flat 9v battery. It cannot deliver the
CURRENT needed to drive a 3v motor.
You cannot "test" or measure the output capability of a device.
You must connect it to the input of the project you are
designing and measure the waveform or voltage being delivered
If the voltage or waveform is considerably less than when it is
not connected, you have decide if the attenuation (reduction) is
acceptable. The ideal situation is NO attenuation - but in
nearly all cases you will get some attenuation.
There are no "rules to follow" and every case is different.
However when the output of a device is NOT reduced when it is
connected to a circuit, the two items are said to be IMPEDANCE
There are three ways to "Match Impedances:"
1. via a resistor. This is generally a poor way to match
impedances and is very inefficient. But it may be the only way
to connect two devices.
2. via a capacitor. This can be a very good way to match
3. via a transformer. Generally the most efficient way to match
impedances but requires a lot of calculation and expense in
getting the transformer designed and manufactured.
Impedance Matching can also be referred to as "MATCHING" and the
simplest example is connecting a 6v globe to a 12v battery. This
is called "Resistance Matching" or "Current Matching" or
"Voltage Matching" because the resistance, voltage and current
are known quantities.
[To connect a 6v globe to a 12v battery you can use a resistor
or put two 6v globes in series. Using a resistor will be very
difficult because a globe requires about 6 times normal current
to allow it to start illuminating and then it takes the "rated
But when when a device produces a signal; the voltage,
resistance and current changes during the production of the
signal and because these values are not constant, we use the
term IMPEDANCE MATCHING.
Impedance really means "resistance that changes during the
production of the waveform."
Impedance matching can be worked out mathematically, but you
need to know all the parameters of the device and the circuit
you are connecting together. This is rarely possible to obtain.
Rather than calculate the result, it is much easier and
more-accurate to connect the two items and view the result on a
CRO (Cathode ray Oscilloscope). But if you cannot do this,
simply connect them and listen or view the output from a
speaker, relay or LED etc.
We have already studied "Impedance Matching" in the circuits
above, but did not identify the concept.
We will now go over some of the circuits and show where
impedance matching took place:
In Fig 6, the
transistor matches the HIGH IMPEDANCE of your finger to
the LOW IMPEDANCE needed to turn on the LED.
The transistor converts the 50k resistance of your
finger to less than 0.5k (due to the gain or
amplification of the transistor of about 100 -200).
You can also say it matches the HIGH RESISTANCE of your
finger to the LOW RESISTANCE needed to turn on the LED.
In Fig 64, the
transistor matches the LOW IMPEDANCE of the speaker to
produce a HIGH IMPEDANCE output on the "out" terminals,
suitable for delivering to the input of an amplifier.
The transistor converts the 8 ohms of the speaker to
more than 800 ohms (possibly 1600 ohms) due to the gain
or amplification of the transistor (about 100-200) and
at the same time the 0.5mV produced by the speaker is
amplified to about 40mV to 80mV.
The 100n capacitor in Fig
71f matches the impedance of the electret
microphone to the input impedance of the transistor.
The impedance of the electret mic is about the same as
the input impedance of the transistor but the mic needs
about 0.5mA to operate and the voltage on the base of
the transistor needs to be very accurately set for "self
A capacitor separates these slightly different DC values
while passing the AC signal..
Matching is needed to separate or remove the DC
component of a signal. In
Fig 71e, the electrolytic matches the LOW
IMPEDANCE output of the amplifier to the LOW IMPEDANCE
of the speaker. The two impedances are almost identical
and you could connect the speaker directly to the output
of the amplifier, but the output has a voltage of approx
mid-rail and this would enter the speaker and shift the
cone when no audio is being reproduced. And the speaker
would only be able to amplify the negative part of the
To remove the DC component of the waveform, an
electrolytic is included.
SATURATING A TRANSISTOR
This is the last topic for discussion
because it is the last thing to attend to when designing a circuit.
We have explained the fact that a transistor turns ON when the base
voltage is above 0.7v and the current though the collector-emitter leads
is approximately 100 times the base current.
This means a transistor with a gain of 100 will deliver 100mA to a
collector LOAD when 1mA enters the base.
This is theoretically true and will occur in nearly all cases, but some
devices such as motors and globes need a lot more current to get them
started or to get them to turn ON because the cold resistance of
a globe is only about 1/5 its hot resistance. This means a 100mA globe
needs 500mA to get it to start to glow.
The same with a motor. The starting or "stalled current" is 5 times more
than the operating or "running current.
On top of this the transistor might not have a gain of 100 and the
voltage may be slightly higher than expected. All these things means the
transistor must be turned ON with more than 1mA into the base.
If you deliver 2mA, it does not mean the transistor will deliver 200mA
to a LOAD. If the load requires 100mA, delivering 2mA to the base will
simply turn the transistor ON harder and the collector-emitter
voltage will be slightly lower, but the load will still draw (or take)
But the devices we mentioned above require 500mA to get them started, so
the base current needs to be 5mA.
Now, here's the unfortunate part, 5mA base-current will not deliver
500mA collector current. The transistor needs more than 5mA base-current
to get it to deliver this HIGHER current. It needs about 7mA.
This process can be proven by experimentation.
Simply increase the base current until the device is turned ON,
then measure the base current. Add 1mA to 3mA to guarantee reliability
and the circuit is complete.
This process is called SATURATING A TRANSISTOR or GUARANTEEING
TURN-ON, or FULLY SATURATING the TRANSISTOR or FULLY
TURNING the TRANSISTOR ON.
(see also Schmitt
Trigger above) Hysteresis is a feature
of a circuit. It is when the circuit turns on at a particular voltage and
turns off when a higher or lower voltage is reached. The gap between
these two voltage-levels is called the HYSTERESIS GAP.
This is a very handy feature.
It prevents an effect called "hunting."
If a circuit turns on at say 6v, and turns off at 5.7v, any slight
variation in the supply voltage will cause the output to change state.
This may produce an undesirable effect of the circuit turning "on and
off" at the wrong time due to supply voltage fluctuations. By increasing the gap between
these two voltages, the circuit will remain in one state or the other -
until the input voltage (the controlling voltage) increases or
The Schmitt Trigger (Fig 79a) is an example of a circuit using
Any circuit with a positive feedback line, introduces the effect we
are talking about.
The feedback line has the effect of assisting the input voltage.
In other words, it widens the gap between an ON state and an OFF state.
This is called POSITIVE FEEDBACK because it ADDS to the effect of the
Even when the input voltage is falling, the feedback improves the ON or
OFF state by taking the
circuit past the point where the change takes place.
Rather than thinking of the feedback as "positive," consider it as
All HYSTERESIS feedback is AIDING or ASSISTING the
effect you are trying to produce.
This circuit uses Hysteresis. The main "assisting component" is the 22k.
This is how the circuit works:
When the circuit is turned on, the base of the second transistor
gradually develops 0.6v and the transistor turns ON.
The voltage between collector and emitter is about 0.2v and the
third transistor is OFF.
When the first transistor receives an AC signal, an increasing
voltage on the base causes the collector voltage to reduce and the charged 4u7
electrolytic moves towards the 0v rail. The negative lead of the
4u7 goes below the 0v rail by about 0.6v.
This allows the second diode to discharge the 10u electrolytic and
the 0.6v on the base of the second transistor is reduced. Let's say
it is reduced to 0.55v.
This causes the second transistor to turn off and the positive lead
of the 1u electrolytic rises toward the 12v rail. The negative lead
of the 1u rises too and this makes the transistor turn ON. In this
process the 1u starts to charge and it has the effect of slowing
down the "turning ON" of the second transistor.
But the pulses keep coming from the first transistor and 10u is kept
discharged. The 1u keeps charging and eventually it is fully charged
and now the pulses from the first transistor can finally turn off
the second transistor.
The third transistor is turned ON and the 22k connected to the
collector of the third transistor reduces the voltage on the base of
the second transistor by about 0.15v
This helps the pulses from the first transistor to keep or put a low
voltage on the base of the second transistor and even if these
pulses stop, the voltage on the base will
take time to rise via the 15k and this is called the HYSTERESIS
GAP. When the circuit changes state, the pulses from the first
transistor will discharge the 10u and this will be "fighting
against" the 15k and 22k resistors that will be trying to charge the
shows a capacitor between the base and collector. It provides
If we remove the capacitor, when the base
"moves down," the collector "moves up." In other words the
signal is inverted.
When the capacitor is fitted, we have to start with the collector because it has
more "power" and it is the lead that is driving the
action of the capacitor and then go to the base.
When the collector voltage "moves down" the right plate of the
capacitor moves down and it charges and tries to pull the left
plate down too.
This is the
opposite effect to the signal moving through the transistor.
This means the capacitor is working against
the action of the transistor.
The capacitor will have more effect on high
frequency signals while the low-frequency signals will be
Because the capacitor is working against the natural flow of
signal through the circuit, it is called NEGATIVE ACTION or
BIASING THE BASE
The base of a
transistor can be biased in many way. In other words,
the current to be supplied to the base can come from many different parts of
the circuit and delivering it is much more complex (complicated)
than you think.
Before we go into the section on SELF-BIAS, we will look at
the problem of adding a base bias resistor between the base and
In the following diagram a resistor is placed between the base
and 6v supply.
If we select the correct value for this resistor, the circuit
will produce the HIGHEST GAIN !!!!
This is the best gain you can get. But the circuit has a lot of
drawbacks and difficulties.
THIS CIRCUIT HAS THE HIGHEST GAIN !!!
If you select a
resistor for the LOAD and then select the base-bias resistor so
that the transistor turns ON and produces a voltage on the base
that is half-rail voltage, this circuit will produce the best
and highest amplification of ALL designs. If the transistor has
a gain of 250, the circuit will produce a gain of 250. If the
transistor has a gain of 450, the circuit will produce a gain of
BUT it will be very difficult to select the correct value. So we
will use a variable resistor.
If the surrounding temperate of the project increases, the
base-emitter current will increase and the collector voltage
will decrease. The same will happen if the rail voltage
increases. And the transistor will multiply the effect 250 to
This circuit is suitable for an individual design where the
base-bias resistor can be selected to suit the transistor but
for mass-production and reliability over a wide range of
transistors, the circuit needs individual attention to get the
biasing correct. But it is not suitable for mass-production as the gain of
the transistor is very critical.
The pot on the base is only used to set-up the circuit and is
replaced with the correct-value resistor.
In the SELF-BIASING stages discussed below, the voltage on the
collector is fairly stable and fixed due to the base-bias
resistor being connected between base and collector.
This feature is not provided on the circuit above and if you
want the collector voltage to sit at 0v (actually about 0.15v)
when the stage is turned ON, this is the circuit you must use.
Once you have selected this type of circuit, you need to select
the LOAD resistor. This is will be as high as possible so the
circuit takes the least current.
That's because the LOAD resistor will be effectively across the
rails for most of the time.
Now select the base-bias resistor so the transistor decreases
the collector voltage to zero.
Start with a very high base-bias resistor and monitor with
collector voltage with a digital multimeter so you don't put any
load on the circuit.
Decrease the value of the base resistor until the collector
voltage reaches 0v.
This circuit will operate differently to a self-biased stage.
It will only respond to a waveform or signal that LOWERS the
voltage on the base. This signal will TURN OFF the transistor
and the collector voltage will RISE.
The input and/or the output can be capacitor-coupled to other stages and
this arrangement will produce the maximum collector-voltage
Here is the main problem with the circuit.
If you adjust the stage so the collector voltage is at 0v
(0.15v) you have selected the correct value for the base
resistor. But if you decrease the value by 50%, the collector
voltage will not fall to a lower voltage but the transistor will
be turned on HARDER.
This means the transistor is turned on HARDER than necessary and
the input signal will have to be larger (higher amplitude) to
turn the stage OFF.
In other words, you will have decreased the gain of the stage by
using a low-value base resistor.
We say "decreased the gain of the stage" because it will require
a higher input signal. The actual gain of the transistor will be
250 or 450 but the "effectiveness" will be reduced.
If the base-bias resistor is TOO LOW, the stage will have an
"effectiveness" less than 250, and maybe less than 50. That's
why seeing the value of the base-bias resistor is so critical.
The transistor must be biased to the point where the voltage on
the collector is just at the point of reducing to less than 0.5v
To give a comparison, this circuit will produce a gain of about
250 with a transistor having a gain of 250, but if the base-bias
resistor is placed between the base and collector (as shown
below - to produce a self-biased stage) the stage will have a
gain of about 70 -100.
We will now cover some of the ways to bias a transistor using
the SELF-BIAS arrangement.
We will consider a transistor has a gain of about 100 in a
"self-biased" stage. (It can be as low as 10 or as high as 200).
This means the collector-emitter circuit can only deliver 100
times more current than is being supplied to the base.
But current supplied to the base is "wasted current" as it flows
ALL THE TIME and for battery operated circuits, this current
must be kept as low as possible.
Secondly, a transistor can only deliver current up to the
maximum rated value for the transistor. If the maximum current is 100mA, a current
of 1mA into the base will allow the transistor to deliver 100mA
via the collector-emitter circuit. That is: the load can take
100mA. But as the current reaches the maximum value, the
transistors gain decreases. This means a base current of 1mA
will only allow the transistor to deliver about 50mA. As the
output current requirement increases above about 50%, the base
current must be increased.
This is one of the hidden problems with a transistor. It
may take 2mA base-current to get to 70mA, 5mA to get to 80mA and
10mA to get to 100mA.
This is a big difference as the gain can drop from 100 to 10.
This is one of the factors you have to be aware of.
That's why the gain of a transistor is generally given for 10mA
Working out the value for the base current is a big problem and
we are going to cover only the small-signal stage.
The following 4 diagrams show how base current is delivered to a
small-signal stage in a "self-biasing" arrangement.
In this arrangement, the base-bias resistor is selected
so the voltage on the collector is half-rail voltage. In
this case it is 3v.
Circuits A and B have 3v on the collector.
But circuit A takes less current when "sitting around."
Circuit B takes 22 times more current. This is due to the
Why select circuit A or B?
We have already learnt the current delivered to the next stage
in a circuit depends on the current flowing through the
COLLECTOR LOAD resistor. Refer to Fig 13 and Fig 62. This means circuit B will deliver more current.
One other factor you have to remember, is this: Circuit A
requires a very small input current. Circuit B requires about 20
times more current. You can think of it this way: The input
energy (that is the input voltage-and-current) is "fighting
against" the base-bias resistor. It is much-easier to
fight-against a 2M2 resistor as it is delivering much less
current (to the base) than a 100k resistor.
Let's explain what we mean by "fight against."
For circuit A, the 2M2 is delivering current to the base. When
the input current increases, the transistor is turned on MORE
and the collector voltage falls. This means the current via
the 2M2 reduces. This means you have to supply more base current
to turn the transistor on.
Circuit B has a 100k base-bias resistor but the main
criteria in turning the transistor ON is the large amount of
current required by the base to get current-flow in the
collector-emitter circuit so that a voltage-drop is developed
across the 1k resistor. 22 times more current is required to get
the same voltage-drop produced in circuit A.
Circuits C and D show the wrong combination of
resistors, making the collector voltage too high or too low. If
it is too high or too low, the stage will not (equally) amplify the
positive and negative excursions of the input signal. However,
this may be what you want.
"Self Biased" stage
The second type of biasing is called a "Bridge" or "H-Bridge."
Circuits E and, F show two bridge circuits.
Circuit E is very similar to circuit A. It needs
about the same input current to circuit A and has about
the same output-current capability.
However circuit E has a gain of about 22. Circuit A
has a gain of about 70 to 100.
The gain of circuit E is defined as collector resistor
divided by emitter resistor 22k/1k = 22.
The gain of
circuit F is about 100.
Circuits A and F produce about the same gain and
the only difference is two extra resistors in circuit F.
The only problem with circuits E and F is the rail voltage must
be above 4v. If the rail voltage is less than 3v, the transistor
will not be turned ON as the base will see less than 0.6v.
The "self biased" stage will operate down
to about 1v rail-voltage.
type of circuit is biased just below the voltage needed to
turn the transistor ON.
This stage takes very little quiescent current but the supply
voltage cannot rise a large amount as this will turn the
transistor ON and change the characteristics of the stage.
transistor is biased "OFF"
point is to bias the first transistor so the voltage on the base
is just at the point of turning it ON.
This allows the 47k resistor to turn on the second transistor
and the diode does not see any voltage. This means the 1u does
not get charged and the input to the microcontroller sees a LOW.
This is called the QUIESCENT (standing, stand-by or idle)
The gain of the electret microphone is adjusted by the 10k pot
and when it receives a loud audio signal it produces an output
of about 20mV.
This signal is sufficient to turn ON the first transistor and
turn OFF the second transistor so that signal diode sees a HIGH
pulse via the 4k7.
This voltage is passed to the 1u and it gradually gets charged.
When the voltage on the 1u reaches about 4-5v, the
microcontroller sees a HIGH and the program in the micro
produces an output.
The main problem with this circuit is the 20mV required to turn
on the first transistor.
Different transistors have varying base voltages. You will need
to set the base voltage very accurately for the circuit to work.
What do we mean by:
Most silicon transistors start to turn ON when the base voltage
reaches 0.65v. But some transistors start at 0.55v. And
some transistors are fully turned on at 0.7v while others need
These different voltages are not important in most cases except
for the circuit above that keeps the transistor turned off until
We are now going to show how a transistor TURNS ON. In this discussion,
the base voltage is delivered from an adjustable power supply
An NPN transistor
is not turned on AT ALL when the base voltage is below 0.45v.
This is shown in diagram A:
As soon as the base voltage reaches 0.55v, the transistor starts
to turn ON. This is shown in diagram B.
A transistor is a current-controlled device and this is
how it works:
As the voltage from the power supply is increased from 0.50v to
will start to flow into the base and the transistor will start
to turn ON.
The transistor will turn on MORE when the base voltage rises
to about 0.65v.
The value of 0.65v shown in diagram C just lets you know the
transistor is turned on a certain amount. It's a characteristic
voltage produced by the transistor that we have no control over.
When we read this voltage we know the transistor is a point of
just being turned ON. Another way to look
at the situation is this: The transistor detects how much
current you are delivering from the power supply and it delivers
about 100 times this amount through the collector-emitter
Here's the highly technical part: The voltage from the power
supply must be CURRENT-LIMITED. This is done by adding the
resistor R. The voltage from the power supply flows
through resistor R and allows the transistor to develop a base
voltage called a CHARACTERISTIC VOLTAGE or BASE VOLTAGE. This is
a voltage developed by the transistor when a certain amount of
current is flowing and by measuring this voltage we know how much
current it is receiving. Adding the resistor allows the
transistor to take the amount of current it requires. If we
connect the power supply directly to the base, we will force
extra current into the base and over-ride the natural
requirements of the transistor.
As you increase the voltage from the power supply, more current
will enter the base and the voltage on the base will rise
slightly to indicate the new value of current. This is shown as
0.75v in diagram D. The transistor will deliver more current through
the collector-emitter circuit, but as the current increases to a
maximum value (every transistor has a maximum allowable
collector current), it may not be able to deliver 100 times more than
the base current. It may be only 50 times or 10 times.
Increasing the current (further) into the base will have no effect on the
base voltage. The base voltage is as high as it will go. The
transistor is SATURATED (turned ON as hard as possible) and no
further increase in collector-emitter current is possible.
Increasing the current into the base will simply overheat the
transistor and damage it.
LEAKAGE This topic covers small, unwanted, currents produced when two
or more transistors are connected together.
Headlight Extender Circuit
Headlight Extender Circuit - fixed
In the Faulty
Headlight Extender Circuit we see 3 transistors directly
When the 100u discharges, the BC547 turns OFF and this turns off
the BC557 and also the BD679.
But the relay remains activated.
This is due to the BC547 not turning off fully and a very small
current flows though the collector-emitter leads.
This current is amplified by the BC557 (about 200 times) and then by the
BD679 (about 20,000 times). The resulting current is sufficient to keep the relay
Two resistors are needed to turn the circuit off.
The 100k on the base of the first transistor discharges the 100u
and makes sure the voltage on the base is zero so the transistor
is fully turned off.
The 2k2 on the base of the BD679 does not remove the slight leakage
current though the BC557 but it flows through the 2k2 and this
produces a very small voltage-drop, that is too small to turn on
the BD679, and this makes sure the BD679 turns off.
This type of problem will occur whenever two or more transistors
are directly coupled together.
Even a leakage current of less than 1microamp will amplify
to many milliamps with the gain of two or three transistors.
Devices such as globes, motors, relays etc are called
LOADS. They can be placed above or below the driver transistor. When the
driver transistor is above the load, as shown in diagrams A and B the circuit is called HIGH-SIDE
SWITCHING. When the driver transistor is below the load, as shown in
diagram C, the circuit
is called LOW-SIDE SWITCHING.
The circuit for low-side switching is much easier to design and can be
less expensive however high-side switching is the only arrangement
available in cars and trucks as the load (globes and motors ) are
generally connected to chassis (to save wiring) and the control wire
(power wire) comes from the positive of the battery.
SWITCHING, the control line (the input to the base) can be active
LOW (as in the first diagram). In the second diagram, the base is taken
HIGH and the emitter follows. The voltage across the collector-emitter
terminals will be higher than the fist example and the transistor will
get hotter. This is because the base of the NPN transistor cannot rise
above the supply (in most cases such as cars and trucks) and the
voltage between collector and emitter will be about 0.65v.
The three diagrams above show the
voltage across the collector-emitter leads when the transistor
is fully conducting.
In diagram B the transistor is an emitter-follower and
the voltage is three times larger than diagrams A and
C. This means the heat generated by the transistor will be
three times larger than diagram A or C.
shows the problem trying to switch a load on a 12v supply, from
a 5v microcontroller. When the microcontroller is HIGH, the
voltage is not high enough to turn off the transistor. The
voltage on the base must be nearly 12v for the transistor to
This circuit will NOT WORK.
In diagram E, the voltage on the base will only rise to
and thus the globe will see only 4.4v and will not illuminate
fully. The transistor will get fairly hot.
The solution is to drive the LOAD via LOW SIDE SWITCHING as
shown in diagram F.
It is possible to switch a HIGH SIDE transistor from a
microcontroller by including a zener diode so the transistor is
turned off when the microcontroller is HIGH. The 10k resistor
makes sure the base sees 12v when the micro is HIGH. The voltage
of the zener is chosen so that when the micro is HIGH, the micro
rail voltage, plus the voltage of the zener is more than the
supply to the globe.
VOLTAGE TO CURRENT CONVERTER
This sounds very complex but it is very
The simplest voltage-to-current component is a resistor.
A resistor performs lots of different jobs, depending on the circuit.
One of its jobs limits the current to a LED. It is called a CURRENT
LIMITING RESISTOR. It can also be called a VOLTAGE TO CURREN2ONVERTER.
Here's how it works:
A resistor is a
VOLTAGE TO CURRENT CONVERTER
A red LED must be delivered a voltage of exactly 1.7v for it to work. In
other words it must be connected to a 1.7v supply.
But a 1.7v supply is very hard to obtain, so we use a 3v supply and a
The resistor converts the 3v to 1.7v.
This is easy to understand because the 3v supply is fixed at 3v and when
a voltage is delivered to the red LED it develops exactly 1.7v across
it. The resistor sits between the 3v and 1.7v
When the voltage of the supply is increased, The voltage across the LED
remains at 1.7v and the voltage across the resistor increases. This is
shown in the diagrams above.
When the voltage across a resistor increases, the current through it
increases. That's how we get 3mA, 7mA and 10mA. This is called VOLTAGE
TO CURRENT CONVERSION.
The VOLTAGE on the input goes up and down and the CURRENT through the LED
goes up and down.
The input CURRENT will also go up and down but we are only covering the
fact that the input VOLTAGE rises and falls and the output CURRENT RISES
Any circuit that produces this effect is called a VOLTAGE TO CURRENT
A transistor can also be connected to produce VOLTAGE TO CURRENT
The following circuit is an emitter-follower. It is also a VOLTAGE TO
CURRENT CONVERTER. A rising and falling voltage on the input
creates a rising and falling CURRENT on the output.
It also produces a rising and falling voltage on the output but we are
only concerned with the fact that the circuit produces a rising and
falling CURRENT on the output when the input VOLTAGE rises
emitter-follower is a VOLTAGE TO CURRENT CONVERTER
The circuit in Fig103e requires say 1mA input current. The output current will be
100mA. The circuit has the capability of increasing the current or
AMPLIFYING the current. The resistor circuit above does not AMPLIFY the
current. It is only a voltage-to-current converter.
The transistor performs a VOLTAGE TO CURRENT CONVERSION and also
produces CURRENT AMPLIFICATION.
A common-emitter stage also performs VOLTAGE TO CURRENT CONVERSION.
A common-emitter stage
is a VOLTAGE TO CURRENT CONVERTER
A slight increase in the voltage on the base of a common emitter
transistor will increase the current through the load by a large amount.
As you can see, there are lots of circuits that perform VOLTAGE TO
CURRENT CONVERSION but we usually identify them for other features and that's
why the term VOLTAGE TO CURRENT CONVERSION is rarely mentioned.
There are also special circuits (using op-amps) to perform precision
voltage-to-current conversion, but we are concentrating on transistor
ANOTHER VOLTAGE TO CURRENT CONVERTER
Here is another
VOLTAGE TO CURRENT CONVERTER:
A photo detector (A light dependent resistor - LDR) produces a very wide
change in resistance from dark to light conditions and when it is
connected to a LOAD RESISTOR, the voltage across this resistor will
But we cannot use this voltage-change to directly illuminate a LED or
drive a motor because it does not have enough current. We need to
convert this voltage-change to CURRENT-CHANGE.
The current through the resistor might be up to 1mA. A motor or LED or
relay needs 10mA to 300mA.
Here is what we have:
We have a wide (large) change in voltage and require this to be
converted to a wide change in current.
We say this voltage-change has very weak driving capability because it
will not illuminate a LED or anything else.
We want a POWERFUL DRIVING CAPABILITY and that's what a CURRENT DRIVER
is capable of doing.
In other words we want to convert a WEAK driver to a POWERFUL
Current to voltage converter
When light falls on the LDR, the voltage across the resistor increases
and this voltage-rise is detected by the base of the transistor and a
small current passes into the transistor. The transistor amplifies this
current and delivers up to 100 times more current to the circuit created
by the collector-emitter terminals. The LED is in this circuit and the
LED will illuminate brightly when 5 to 25mA flows. This is only a
demonstration circuit to show the effect of shining a light on the LDR.
It may cause the LED to shine too bright and be damaged as no
current-limiting resistor has been included to protect the LED.
TO VOLTAGE CONVERTER A resistor can be used
as a CURRENT TO VOLTAGE CONVERTER.
Fig 103g shows a resistor called a SENSE RESISTOR.
It is a low-value resistor in series with one line of a circuit and its
function is not to change the operation of the circuit in any
Measuring the "sense resistor"
Its function is to produce a very small voltage across it and this
voltage is detected by a circuit (basically a voltmeter (or milli
When a current flows though a resistor, a voltage is produced across the
resistor. You can also say a VOLTAGE DROP is produced across the
resistor. If the resistor is exactly 1 ohm, a voltage of 1v will be
produced across it when 1 amp is flowing or 1mV is produced for each 1mA
of current. Using a 1 ohm resistor produces an easy conversion.
If the circuit is 24v or 50v, a loss of 1 volt will not be noticed.
But if the circuit has a lower voltage, (say 5v) the resistor will be
need to be a lower value so the drop across the sense resistor does not
upset the operation of the circuit.
The actual value of the resistor is not important for this discussion,
It can be 1 ohm or 0.1 ohm.
The important point is to understand the function of a Sense Resistor.
In the circuit above, if the globe is replaced by a 20watt or 50watt,
globe, the current through the sense resistor will increase. We measure
the voltage (in millivolts) across the resistor and we convert the value
into CURRENT. This is a CURRENT to VOLTAGE CONVERSION.
A transistor can be used as a CURRENT TO VOLTAGE detector. Fig 103h
shows a 1 ohm sense resistor connected to a transistor. When the circuit
is turned ON, the charging current (the current flowing into the
battery) will be high and when the voltage across the sense resistor
reaches 0.65v, the transistor turns ON and the voltage on the collector
reduces. This turns on the red LED and reduces the voltage on the ADJ
terminal of the LM317T regulator and the regulator outputs a lower
current to the batter. This is how the circuit limits the charging
current. The resistor is converting the
current flowing through the circuit (and into the battery) into a
voltage, and the transistor detects the voltage. The transistor is not
detecting or measuring the current. It has absolutely no idea of the
amount of current flowing. It is detecting the voltage across the
resistor. The resistor is performing the CURRENT to VOLTAGE conversion.
The 1ohm Sense Resistor.
SQUEALING, BUZZING, OSCILLATING,
We have studied POSITIVE FEEDBACK
and the effect it produces. It turns an amplifier into an
The following circuit will not work:
The three stages of amplification will
produce so much gain that the circuit will self-oscillate. The output
will be a "buzzing-sound" and the fault will be impossible to find
because it comes from within the design of the circuit. The first
thing you must do is add "power-supply decoupling."
The unwanted sound produced by the circuit is called MOTOR-BOATING and is
generated in the "front-end" by very
small noises or "disturbances" and amplified by the stages that follow. Fig 104b shows where the noise starts. It can be produced by the
electret microphone or by the noise in the junctions of the first
transistor (due to current flowing in the collector-emitter circuit).
This waveform will be very small and almost impossible to detect via any
test-equipment, but it will start in the first stage and pass through the
coupling capacitor as shown in Fig 104c.
The next stage will amplify this "noise" and it will be amplified
further by the following stages.
There will be some slight cancellations from the various stages as the
signal will be "out-of-'phase" but the end result will be a
"putt-putt-putt" or squealing from the output.
The general term for this is called MOTOR-BOATING and is due to
the high gain of the circuit.
The noise will appear on the power rail and get passed to the front-end
where it will be amplified more.
Fig 104c. The positive feedback loop producing "Motor-boating"
This effect can be reduce and eliminated by a term called DECOUPLING.
Decoupling is achieved by adding capacitors [electrolytics] (and
resistors) across the power rails so that each stage is effectively
powered by a separate supply.
Adding an electrolytic can sometimes make a big difference and sometimes
it will make no difference.
It all depends where it is connected and the value.
Fig 104d shows an electrolytic connected across the power rails.
This is called DECOUPLING THE POWER RAILS and effectively
tightens up the power rails so that any noise on the positive rail is
But, as you can see, the power rails extend to the first
transistor and although the rails may be "tight" near the battery they
can "move" near the first stage.
This is due to the wiring between the stages or the tracks on the PC
board. That's why an electrolytic across the battery may have little
effect on removing our motor-boating problem.
Fig 104e shows an electrolytic connected across the supply that
electret microphone and 1k2 resistor to separate the supply we have just
created, from the main supply rail.
We have effectively created a separate power supply. It is fed by a 1k2
and kept "tight" by the 10u capacitor.
The electrolytic does not have to be a high value because the electret
mic takes very little current and the voltage-waveform (the AC signal)
produced by the microphone is very small (about 20mV).
These two items very effectively decouple the microphone from the supply
rails so the microphone has its own supply. The 1k2
resistor does most of the "separation." The voltage-drop across it will
be very small and it will not affect the operation of the circuit, but
the small voltage-drop will prevent any noise on the power rails being
fed to the microphone via the 10k resistor.
To remove any slight motor-boating problems (if they still exist); a
power-supply filter (called power-supply decoupling) made up of a 1k2
and 10u can be placed after the first amplifier stage as shown in
By selecting the value of capacitance and resistance, this arrangement
will remove almost all motor-boating problems. It is
a very-effective form of suppression.
Decoupling is most-effective on the pre-amplifier stages, however every
circuit is different and these two components only deal with the
low-frequency motor-boating type of instability. Some circuits also
produce high-frequency oscillations (about 1MHz) and these need removing
by a different value of capacitor-feedback.
BREAKDOWN and ZENER MODE There are two conditions or
states where a transistor can be instantly damaged. This is due to
voltage applied in the wrong direction or the application of voltage
that is higher than the rating of the transistor.
Voltage will kill a transistor faster than excess current.
A high voltage spike can damage a transistor instantly.
However if the excess voltage does not have enough current to damage the
transistor, it will recover and we can use this feature in a circuit.
Breakdown and zener mode are different.
In breakdown mode, suppose we have a transistor that has a
specification of 85v for the voltage it will withstand between the
collector and emitter as shown in Fig 104g:
It will "resist" a voltage of
85v and this voltage will appear
across the collector-emitter leads. When the voltage increases to 86v,
87v ... the transistor will suddenly breakdown and only a few volts will
appear across it. This fires the trigger transformer in the circuit
If the current is very small, the transistor will not be damaged and
when the voltage is removed and a lower voltage applied, it will operate
as an undamaged device.
In zener mode, the base-emitter junction is connected to a
voltage higher than 9v via a resistor. The junction will breakdown and a
voltage of about 7v will appear across the base-emitter leads and the
excess voltage will be dropped across the resistor.
The zener-effect or zener mode can be used to produce white noise or a
7v zener reference.
Fig 104h shows the first transistor with the base-emitter
junction reverse-biased to produce a "noisy zener" via the 1M feeder
resistor. The noise is picked off via the 100n and amplified by the
remainder of the circuit.
This circuit is basically a high gain
amplifier with feedback that causes the LED to flash at a rate
determined by the 10u and 330k resistor.
Remove one of the transistors and insert the unknown transistor.
When it is NPN with the pins as shown in the photo, the LED will
The circuit will also test PNP transistors. To turn the unit
off, remove one of the transistors.
ZENER TESTER The maximum voltage a transistor
can withstand is called the ZENER VOLTAGE of the transistor.
It is Vce - the voltage between (across) collector and emitter. It is also
the maximum supply voltage or circuit voltage or the voltage generated
by an inductor in the collector-circuit and can be tested via the
following circuit. This circuit will also test ZENER DIODES and LEDs.
TRANSISTOR and ZENER TESTER CIRCUIT
The circuit is a
flyback oscillator. This type of
oscillator energises an inductor then turns off very quickly and
the magnetic field (flux) produced by the inductor collapses and
produces a very high voltage in the opposite direction. The
maximum voltage produced by the circuit depends on the "maximum
voltage capability" of the transistor.
The voltage produced by the inductor is over 120v but the
transistor will zener at a voltage lower than this and thus the
output voltage will be determined by the characteristic of the
A diode on the output of the inductor passes this
high-voltage-spike to a 1u electrolytic, which stores the energy
and provides a high voltage output.
The circuit will test transistors up to 120v and zeners up to
the voltage produced by the transistor.
The project is built
on a strip of PC board cut into lands with a file or saw.
The following diagrams shows the parts placement and connecting
the 5 button cells to the board.
The project can be built in an evening and added to your TEST
When testing a transistor, fit it into the pins marked C B E. If
you have a LED connected to the LED terminals, it will glow.
If you remove the LED and measure the voltage across the 1u
electrolytic, it will provide the maximum working voltage for
TESTING A ZENER
When testing a zener, place it in the pins provided. If the
zener is around the wrong way, the voltage across it will be
less than 1v.
When it is placed correctly,
you can read the
zener voltage with a high impedance multimeter such as a digital
TESTING A LED
When testing a LED, fit it into the pins for the LED with the
cathode lead (the shorter lead) to the left. It will glow very
dim because the dropper resistor is very high and only allows 4
- 6mA to flow.
This will give you a good idea of the relative brightness of a
LED when compared to others in a batch.
THE TRANSISTOR & ZENER REGULATOR
A transistor can be used to amplify the characteristics of a zener. You
can also say the transistor is a BUFFER or EMITTER-FOLLOWER. It
is another example of the transistor as an AMPLIFIER - a DC
AMPLIFIER - indicating it amplifies the "steady-state" conditions
provided by a zener diode.
We will start with the simple Zener Regulator circuit, then add
the transistor amplifier. After that, we will remove the zener and add
another transistor to improve the smoothness of the output waveform.
A simple zener regulator circuit is very wasteful however it is the
basis for creating a stable output voltage from a voltage that may be
rising and falling a considerable amount.
The following circuit shows a simple zener regulator:
A Zener Regulator Circuit
A Zener Regulator Circuit consists
of a zener and a resistor. The resistor is called a Dropper Resistor and it
is designed to limit the CURRENT. It is not designed to limit the
VOLTAGE. The zener diode performs the task of limiting or SETTING the
voltage on the output.
The current through the Dropper Resistor will be shared between the
zener diode and the LOAD (on the output of the circuit). These two items
may or may not share the current equally, and the amount of share will depend on the
value of the LOAD. We can also say the Dropper Resistor is a CURRENT
LIMITER. If is is not included, a 12v zener connected to a 15v
supply would draw (or take) a very high current and "burn out."
Here's the important fact about the current-sharing
between the zener and load:
Suppose the SUPPLY VOLTAGE is fixed.
Here's an example of how the zener diode works:
Suppose we select a resistor so that 100mA flows through the zener when
no load is present. Fig (a)
When the load takes 50mA, the zener takes 50mA. Fig (b)
When the load takes 90mA, the zener takes 10mA. Fig (c)
When the load takes 100mA the zener takes 0mA. Fig (d)
Current-sharing between the zener
Up to this point the circuit works
perfectly. Even though the zener takes 0mA, the circuit is operating
perfectly and the output is smooth. If the load tries to take 101mA, the
output voltage will DROP.
This is point at which the circuit is said to FALL OUT OF REGULATION.
The load (the OUTPUT) can take more the 102mA and the output voltage
will drop further, but we are interested in the range where the output
voltage is STABLE (fixed).
In this example, the current though the Dropper Resistor is ALWAYS
100mA. The current is then split (or shared) between the zener diode and
This feature is always the case with a zener diode regulator.
100mA is always flowing though the Dropper Resistor and if the load is
taking only 10mA, this type of regulator is very inefficient.
When the supply rises, the current though the Dropper Resistor will
increase. When the Supply falls, the current through the Dropper
Resistor will decrease. During this time the output voltage of the
circuit will remain constant providing the current though the zener is
always at least a few mA and the maximum value does
not allow the zener to get too hot. If the zener gets too hot it may
The efficiency of the ZENER REGULATOR can be improved by adding a
transistor. The transistor is an amplifier. A CURRENT AMPLIFIER.
(also called a DC amplifier)
This type of circuit is sometimes called a SUPER ZENER or AMPLIFIED
ZENER. The transistor is connected as an emitter-follower as shown
in the following diagram:
An emitter-follower transistor
If the transistor has an
amplification-factor of 50, it will require 2mA (into the base) for each 100mA delivered
to the output.
This means our Zener Regulator only needs to deliver 2mA and the output
can deliver 100mA. The emitter-follower transistor must be a POWER
Here are some examples from 100mA to 2Amp:
The transistor has a gain of 50
In the circuits above, the output current
can range from 100mA to 2Amp. The zener will pass 48mA when the load is
100mA and drop to 10mA when the load is 2Amp.
If the output requirement is only from 500mA to 1Amp, the value of the
dropper resistor can be changed so the zener takes 20mA when 500mA
output current is required and 10mA when 1 amp is required.
When designing this type of circuit, the zener
is allowed to take 10mA when the maximum current is required. The 10mA
is about the minimum current for a 12v (300mW to 500mW) zener to keep it
in conduction. The actual minimum value depends on the wattage of the
zener and also its voltage. You will need to look at the specification
sheet for the zener you are using.
The term "keep it in conduction" means this: Suppose we have a 12v zener
and dropper resistor connected in series. As the voltage (the
SUPPLY VOLTAGE) on the
combination is reduced, the current through the zener reduces. If you
supply the combination with 11.5v, the zener will "fall out of conduction"
and it will appear like a very high value resistor or even an infinite
In the transistor / zener regulator circuit above, if the current taken
by the load increases above 1Amp, the current into the base increases
and when it reaches 30mA, the zener receives NO CURRENT.
Any further increase in current by the load causes more current to flow
through the Dropper Resistor and the voltage across this resistor will
increase. This will lower the voltage on the base and also lower the
voltage on the emitter. At this point the zener has dropped out of
If the transistor has a gain of 50, the maximum output current is
divided by 50 and this gives the base current of 20mA.
Add 20mA to 10mA to obtain the current through the Dropper
The value of resistance for the Dropper Resistor is obtained by the
Suppose the supply is 15v and the zener
is 12v . The value of the Dropper Resistor is:
The output voltage is 0.7v less than the voltage of the zener.
The following diagram shows an example of the voltages on a typical
The voltages on the regulator
A power transistor can be used to amplify the characteristics of a
zener. That's what the circuit above is doing.
The circuit is sometimes drawn as shown in the following diagram. It is
more difficult to see exactly how the circuit is operating, but this is
how it is drawn in many projects. By drawing the circuit as shown above,
you can see the voltages on each section of the circuit and you can't
make a mistake. One "circuit engineer" said the output was 1.2v above
the input voltage. But when you draw the circuit as suggested, you
can clearly see this is not possible.
That's why the layout of the circuit is MOST IMPORTANT.
The regulator circuit re-drawn
IMPROVING THE SMOOTHNESS OF THE OUTPUT
The quality of the output (meaning the smoothness of the output) of a
regulator - also called the smoothness of a POWER SUPPLY - can be
improved by adding a transistor that detects any increase or decrease in
the the output voltage and produces an opposing signal to counteract the
rise or fall. The end result is very smooth DC.
The action of this transistor is called NEGATIVE FEEDBACK.
In the regulator circuit above (and the circuit with the transistor
amplifier), the output is not being monitored and if the zener is noisy,
(in other words it breaks down in an irregular mode and creates ripple)
there is no feature to detect the changes, and reduce them.
The following circuit uses a transistor to detect the output voltage and
provide a feedback signal (feedback voltage) that will eliminate the
ripple. It is called a FEEDBACK SIGNAL or simply FEEDBACK.
The zener diode can be removed and two resistors used to monitor the
output voltage with the voltage at their join being passed to the
The base-emitter voltage of the transistor replaces the zener diode as a
"reference" and the transistor turns into a zener diode with the "zener
reference" appearing between the collector and emitter.
The following circuit shows the feedback transistor replaces the zener
diode in the circuit above and two VOLTAGE DIVIDER resistors on the
output are connected to the base of the feedback transistor.
When the circuit turns ON, the output voltage rises until the voltage at
the join of the resistors reaches 0.65v. The feedback transistor starts
to turn ON
and prevents the base of the emitter-follower transistor rising above
12v. This creates an output voltage of 11.3v.
Any reduction in the output voltage will turn off the feedback
transistor a very small amount and it will allow the voltage on the base
of the emitter-follower transistor to rise and this will increases the
The feedback transistor is also called an ELECTRONIC FILTER.
It has an effect equal to the gain of the transistor (approx 100) on
smoothing the output.
HOW DO YOU WORK OUT THE RESISTOR VALUES?
For the resistor values for the following circuit, we start with Ra and Rb.
The output current is 1amp and the transistor can handle more than 2
amps, so the gain at 1amp is 100. The feedback transistor also has a
gain of 100, but this is not important for these calculations.
Starting with Ra and Rb, we allow 10mA to flow though this voltage
divider so the stability of the circuit is very high.
The resistance of Rb = 0.65/.009 = 72 ohms.
The resistance of Ra = 10.65/0.01 = 1.065k
The resistance of the dropper resistor = 3/0.01 = 300 ohms
The circuit turns on via the 300 ohm dropper resistor pulling the base
UP. As the output voltage rises, a point is reached where the voltage
into the feedback transistor reaches 0.65v and the transistor turn ON.
It turns into a resistor and the join of this "resistor" and the
Dropper Resistor create a voltage of exactly 12v. At this point the
circuit becomes stable.
Ra and Rb can be any values providing the ratio is 72:1065. For
instance, the values can be 144:2130 or 108:1597 (where the values are
increased by 50%).
The value of the Dropper Resistor can be any value less than 300R
and although this will theoretically allow more current to enter the
base of the emitter-follower transistor, the transistor will not take
any more current than it requires to create the necessary
collector-emitter current (and thus the exact voltage of 12v at the
As the current through a transistor increases, the gain of the
transistor decreases. This means a transistor may have a gain of 100
when a small current is flowing thorough the collector-emitter circuit,
but as the current increases to say say 50% (of the maximum allowed for
the device), the gain may decrease to 50.
As the current increases to a maximum, the gain may decrease to 20.
All these values are variable and we cannot specify any exact values, so
you have to remember to takes these facts into account when designing a
That's why a transistor with a maximum collector current of 4 amps is
chosen for a circuit requiring 1 amp.
You are not over-stressing the transistor and it will provide a gain of
THE ELECTRONIC FILTER
Here is a simple circuit to reduce the ripple from a power supply by a
factor of about 100. This means a 20mV ripple will be 0.2uV and will not
be noticed. This is important when you are powering an FM bug from a
plug pack. The background hum is annoying and very difficult to remove
with electrolytics. This circuit is the answer. The 1k and 100u form a
filter that makes the 100u one hundred times more effective than if
placed directly on the supply-line. The transistor detects the voltage
on the base and also detects the very small ripple.
As current is taken by the load, about 100th of this current is required
by the base and if the load current is 100mA, the current into the base
will be 1mA and one volt will be dropped across the 1k resistor.
The circuit is suitable for up to
100mA. A power transistor can be used, but the 1k will have to be
reduced to 220R for 500mA output. The output of the circuit is about
2v less than the output of the plug pack.
By adding a zener across the electro, the output voltage will remain
much more constant (fixed).
If a zener is not added, the output voltage will drop as the current
increases due to a factor called REGULATION. This is the inability of
the small transformer to provide a constant voltage. The addition of
the 3 components only reduces the RIPPLE portion of the voltage - and
does not change the fact that the voltage will droop when current is
increased. It requires a zener to fix this problem.
An ELECTRONIC FILTER
THE ELECTROLYTIC AS A FILTER
The circuit above shows an electrolytic used to remove the ripple from a
power supply. 1. How does the electrolytic reduce the ripple? 2. Why do you need a larger capacity electrolytic for a higher
1. The electrolytic is just like a rechargeable battery.
When the voltage is higher than normal, the electrolytic gets charged
and this puts an additional load on the supply. Some of the extra
voltage (as well as the current being delivered by the supply-voltage)
is passed to the electrolytic as energy and the supply voltage is
When the supply-voltage drops, the voltage contained in the electrolytic
is slightly higher than the supply and it delivers energy. This prevents
the supply-voltage dipping too much.
You can see the electrolytic is receiving and delivering only a very
small amount of its stored energy and that's why its value must be very
large (about 1,000u for each amp delivered by the power supply). This is
because the electrolytic has to have a large ability to store a lot of
energy when the voltage rise and falls only a very small amount. 2. When the current is a large value (say 1 amp), the energy
contained in a few millivolts and a current of 1 amp, is a very large
and thus a high capacity electrolytic is needed.
When an electrolytic is placed across a power rail, it smoothes the
voltage by BRUTE FORCE. A large electrolytic will produce more
But when the electrolytic is connected to the supply via a resistor, the
electrolytic will take time to charge when the voltage rises and time to
discharge when the voltage fall.
This means the ripple on the electro will be much less than the ripple
on the power rail.
Depending on the value of the resistor and electro, it may be 1/100th of
the ripple on the supply.
The transistor detects this improved voltage and allows a high current
to pass to the load.
The zener improves the circuit enormously
The addition of a zener diode improves the output of the circuit
enormously. When the circuit delivers current, the voltage will "sag"
and sometimes the voltage will drop to 11v. If we include a zener, we
will only be delivering a voltage of 10v minus 0.7v = 9.3v and this
voltage will will NEVER have any ripple (in it). Thus this voltage will
never have any "hum."
THE TRANSISTOR AS A LOAD
This might seem an unusual topic but many circuits use a transistor as a LOAD or VARIABLE LOAD or partial load (in conjunction with a LOAD
RESISTOR) to dissipate (remove - take away) power - to prevent
another item (such as battery) being overcharged or a delicate device getting too
We are talking about wasting energy or losing energy in the form of heat
to prevent another item in a circuit getting too hot.
A power transistor such as 2N 3055 is ideal for this purpose however
there are smaller power transistors for smaller losses.
We use the gain (amplification factor) of the transistor to provide this
feature and by controlling the base current, the current though the
collector-emitter terminals can be adjusted. In most cases the
transistor is in series with a LOAD RESISTOR and the two items can be
adjusted to remove unwanted energy.
In addition, the percentage dissipated by the transistor compared to the
load resistor depends on the base current of the transistor.
This is quite a complex topic as the losses can be adjusted to any
percentage, irrespective of the supply voltage.
This is sometimes called an ELECTRONIC LOAD or ACTIVE LOAD
because the effectiveness in dissipating heat can be controlled by
current entering the base of the transistor.
A POWER RESISTOR (by itself) is called A DUMMY LOAD or STATIC LOAD. It's
dissipation is fixed (providing the voltage of the supply is fixed).
Our discussion introduces a variable dissipation, controlled by the base
current of a transistor.
This is another example of a transistor being used as an amplifier. The
current into the base is amplified by the transistor to produce a
current through the collector-emitter leads.
This current also flows through a LOAD RESISTOR and the resistor
increases in temperature.
The loss in the transistor and resistor is calculated in terms of watts
and when this is extended over a period of time, the result is energy -
watt-hours. This energy is given off as heat instead of raising the
temperature of a critical component in a circuit.
The following diagram shows the heat dissipated in the transistor
when maximum, medium and low current flows into the base of
When maximum base current is supplied to the transistor, it is turned
fully and only about 10% of the total wattage is lost in the transistor.
This means the total wattage of the load can be very high.
As the current is reduced, the wattage dissipated in the transistor
increases to about 50%, then drops off.
The following diagram shows a large wattage will be dissipated in the
resistor (and very little via the transistor) when maximum current is supplied to the
base of the transistor, but as the
base current is reduced, the size of the load must also be reduced
because more of the load is dissipated in the transistor (and the
transistor is the
It is impossible to work out the "load sharing" between the transistor
and resistor for any given base current because transistors from
different batches have considerably different characteristics.
The diagrams we have provided show percentages but not the base
current required to create the load sharing.
Even simulation software will produce false data as the actual
characteristics of the transistor you are using will be unknown.
Rather than spending time on trying to work out the probable
results via a software package, it is much easier to build the circuit
and apply current to the base.
As you apply current into the base, you can monitor the current through
the load via an ammeter and provided the transistor is correctly
heatsinked, it will not overheat.
A 2N3055 will dissipate 115 watts using a very large heatsink. This
gives a starting-point for the maximum wattage for the system.
When the transistor is turned on so it dissipates the same wattage as
the resistor, the total losses for the system can be as high as 230
watts, but when the transistor is fully turned on, the system can handle
about 1,0000 watts.
However the transistor must change very quickly from a state where it is
not turned to a fully turned-on-state. (If not, the transistor will be
damaged very quickly if it becomes partially turned on.)
In the fully-turned-ON state, the transistor is fully saturated and is
dissipating only about 10% of the total load and the resistor is
dissipating about 90%.
These are all points you need to know, when designing an ACTIVE LOAD.
THE TRANSISTOR AS AN INVERTER A transistor can be configured as an inverter - to change a
signal (that moves from LOW to HIGH) into a signal that changes from HIGH
This type of circuit can be used to transform a 0-5v signal into
a 0-9v signal. This is called VOLTAGE SHIFTING or LEVEL
SHIFTING. In this case, a LOW to HIGH (0-5v) signal is
converted to a "HIGH to LOW" (0-9v) signal. The
output changes when the base sees a voltage between 0.55v and 0.7v. The
remaining input voltage is dropped across the base resistor.
The output voltage will be initially HIGH and go LOW as soon as the
input voltage reaches about 0.7v.
5v to 15v Inverter
A non-inversion circuit is
shown in the following diagrams:
5v to 15v Non-Inverter
The following circuit does not work because the second transistor is
never turned off. Both transistors MUST be connected to the high voltage
This circuit does not work
The following circuit converts a signal that starts as 5v HIGH and
goes LOW. During this signal transition, the output start with a
LOW value and goes HIGH.
5v to 0v Inverter that
produces 0v to 5v Output
The following circuit is a Push-Pull Inverter:
All the circuits above convert an analogue signal or A DIGITAL SIGNAL into a digital signal.
This is due to the gain of the
transistor. In other words the output does not respond in a linear
manner, (to the input voltage). The output changes when the input moves from a
voltage of about 0.55v to about 0.7v. Input voltages below 0.55v
have no effect and voltage above 0.7v do not affect the circuit as the
circuit has already changed state.
One other characteristic of the circuit is this: It speeds up the waveform and removes noise from a noisy
TRANSFORMER One of the most complex electrical/electronic components is the TRANSFORMER. It is the simplest component and yet it
produces the most complex effects. A transformer is simply a coil of
wire placed near another coil of wire. The results and effects will amaze you.
There are so many different effects, we could write an eBook.
In fact we will write a chapter on the subject, but firstly we will
cover 5 things:
1. A single coil of wire is not a transformer but an inductor.
Without going into any complex mathematics, here is a fact you should
When an inductor is connected to a battery, the current does not flow
though the turns immediately, but a few microseconds or milliseconds
later. This is called CURRENT LAG. Don't ask why, it's just a fact.
But the most amazing thing is this: When the voltage is removed, the
inductor produces a HIGHER output voltage IN THE REVERSE DIRECTION.
If the coil is wound on a cardboard former, the core (the centre of the
coil) is air and the voltage (the reverse voltage) produced, may be twice the supply voltage. But if the core is steel or
other magnetic material such as iron (stalloy) or ferrite (a special type of iron)
the reverse voltage may be 100 times HIGHER or even 1,000 times HIGHER.
That's why a simple coil of wire is one of the most amazing things.
We can control the magnitude of this reverse voltage by adjusting the
frequency and/or the speed at which we turn the voltage OFF.
Even though the inductor is not called a transformer, we are
"transforming" a low voltage into a high voltage.
We are not getting "something for nothing." Conservation of
energy still applies. We are transforming a low voltage at high
current into a high voltage at low current. The watts IN equals the
When we add another winding, the two coils become a TRANSFORMER.
The first winding is called the PRIMARY and the second winding is called
We drive a transformer in a slightly different way to an inductor.
We deliver a rising and falling voltage to it slowly. This is
called an AC delivery and although the letters "AC" mean
Alternating Current, we really mean Alternating Voltage.
When we deliver a slowly rising and falling voltage, the primary does
not produce the high reverse voltage discussed above but it does produce
a reverse voltage that can be as high as 99.99% of the applied voltage.
But this is getting away from the point we want to cover.
The secondary winding produces an exact copy of the voltage flowing into
the primary and if you measure it on a piece of test equipment, it will
follow the primary exactly (but slightly delayed). If you reverse the leads to the test
equipment, the results will be a "mirror image." That's how we get a
reverse voltage out of the transformer.
Here's the next valuable fact: The voltage from the secondary will be
higher if the secondary has more turns or lower if the secondary has
fewer turns than the primary.
If it has more turns, the transformer is called a STEP-UP
transformer and if it has less turns the transformer is called a
Here's our last amazing fact:
If the secondary has less turns, the current from the secondary can be
higher than the current in the primary. But if the secondary has more
turns, the current from the secondary will be less than the current in
With all these facts and capabilities, we can do incredible things with
We forgot to mention one of the most beneficial uses for a transformer.
The voltage on the primary is totally isolated from the secondary. In
other words, the primary may have 110v or 230v on it and the secondary
may have 12v. You can touch either lead of the 12v winding and any
metal pipe and not get a shock. The transformer provides total
isolation. But if you touch either end of the primary winding and a
metal pipe you will be killed instantly.
That's one of the main uses for a transformer - to provide isolation
from the "mains." The energy passes from the primary winding to the
secondary via magnetic flux and the two windings are ISOLATED and
INSULATED from each other.
A transformer can be smaller that a grain of rice to the size
of a house and there are millions of different types. That's why
they are so complex.
As soon as the eBook article is written, it will be included HERE.
A transformer is a complex item. It takes up a lot of space on a PC
board and is expensive to make.
It is not added without a reason and a lot of thought.
Here are 8 reasons why a transformer is included in a project 1. To produce a voltage higher than the supply, 2. To produce a very low voltage, 3. To produce a high current, 4. To produce a sinewave wave, 5. To mix two different signals or frequencies, 6. To produce a feedback signal, 7. To produce a number of different, isolated voltages (and/or
current), 8. To produce isolation. And many other
Driving a transformer is
not like delivering current to a resistive load.
The primary winding of a transformer has a very small resistance but
when it is delivered an increasing voltage, the magnetic flux (produced
by the voltage) creates a voltage in the opposite direction that cuts
the turns of the winding and this voltage opposes the incoming voltage.
This effectively makes the winding appear to be a higher resistance.
When a transformer is delivering energy via the secondary winding, the
"back-voltage" produced by the magnetic flux will be less and the input
current (via the primary) will be higher.
A transformer is designed to receive an increasing and decreasing
voltage. During this time it can deliver energy to the secondary.
But when the voltage rises and remains HIGH, the opposing voltage
produced by the expanding magnetic flux ceases and input current
TRANSFORMER Designing a transformer is very difficult and complex. The easy
approach is to buy a product that contains a circuit similar to your
requirement and use the transformer.
It is very difficult to take a transformer apart as the laminations or
the ferrite core is dipped or glued or sealed so the windings do not
In some cases you can buy laminations or ferrite cores (called pot
cores) but there are many different types of materials and unless you
know the composition of the material, the resulting transformer can be
as low as 10% successful.
The other problem with taking a transformer apart is this:
Many transformers have an air-gap in the magnetic circuit to "remove" or
"use-up" the magnetic flux created by the DC component of the input
If this air-gap is not maintained in its exact thickness, the new
transformer will not be identical in performance to the original.
A transformer without an air gap must have "lapped surfaces" so the two
halves of the core touch each other.
All these technicalities will be covered in the eBook.
THE POTENTIOMETER A potentiometer is simply a resistor with the
resistance-material exposed to a wiper.
The resistance-material is called a TRACK and it can be straight or
curved. When the track is curved, we generally call it a "pot" (abbreviation
for potentiometer) and the pot is rotated to increase or decrease the
resistance. When the track is straight we call it a 10-turn pot. And a
screw is available on the end of the pot. Straight tracks are also
available in pots called SLIDERS. All pot have the same symbol.
In most cases a pot is connected to a circuit with a resistor on one end
or on the centre terminal (the "wiper") as shown in the following diagrams:
STOP RESISTOR and SAFETY RESISTOR
A resistor added to the top or bottom of the pot is called a stop
resistor. It stops the pot reaching full rail voltage or 0v. A
safety resistor is added to the wiper so the pot is not damaged when
turned fully clockwise and the resistance of the output is low. Fig A
above shows a pot with no external resistors. The voltage on the wiper
can be as high as rail voltage or as low as 0v. Fig
shows a pot with a resistor to positive rail and one to 0v rail. The voltage on the wiper
will not be as high as rail voltage or as low as 0v. By selecting a pot
with a particular value and resistors for the top and bottom, maximum
and minimum voltages can be set. Fig C
shows a pot with a top resistor. This sets a maximum voltage, while the
minimum will be 0v. Fig D
shows a pot with a bottom resistor. This sets a minimum voltage, while the
maximum will be rail voltage. Fig E
shows a pot with a resistor on the wiper. The allows the voltage on the
wiper to be as high as rail voltage or as low as 0v. The resistor is
called a "safety resistor." It prevents the pot being damaged if the
output becomes shorted as shown in the last diagram.
If the pot in the last diagram is turned fully clockwise, the wiper will
each rail voltage. If the wiper is connected to a low resistance, a high
current will flow and damage the pot.
A "safety resistor" will reduce the high current.
There are three reasons why a pot is included in a circuit.
1. To "pick off" a voltage.
2. To deliver a current
3. To "pick off" an amplitude.
"PICKING OFF" A VOLTAGE
The following diagrams show a pot "picking
off" a voltage. The pot values have not been shown because we are
dealing with the concept of picking off a voltage.
In actual fact the pot will be delivering a current (via the wiper) to
the circuit connected to the wiper, but to separate the functions of a
pot, we have identified this function as PICKING OFF A VOLTAGE.
The main difference between Picking Off A Voltage and
Delivering A Current, is the value of the pot.
The resistance of a pot for
Picking Off A Voltage
is generally a high value. The term "HIGH VALUE" is relative to the
In Figs F and G you can see the SAFETY RESISTOR and STOP RESISTORS.
The wiper in figure F picks off a voltage from the pot. The pot is the
load resistor for the MEL12 Photo Darlington Transistor and although it
is delivering a small current to the base of the transistor, this
current is very low and that's why we refer to the pot as "picking off a
The safety resistor in Fig F could be replaced with a stop resistor
above the pot.
This change can be done in some circuits and you have to build the
circuit to determine if the change can be made.
POT RESISTANCE The
resistance of a pot is selected from one of the following values:
100R, 500R, 1k, 5k, 10k, 50k 100k, 1M and 2M.
In most cases you will copy a circuit and use the same value for the
Working out the value is quite a complex task.
Here are three different circuits. The voltage on the top and bottom of
the pot is the same, but the value of the resistances is different.
The first circuit is classified as LOW IMPEDANCE. The second
is MEDIUM IMPEDANCE and the third is HIGH IMPEDANCE.
The output from each pot will range from 3v to 6v. So,
why different values of resistors?
The reason is to keep the current through the pot as low as possible.
The current through the resistors is WASTED CURRENT. If a project is
battery operated, wasted current is a problem.
The resistance of the load on the wiper also determines the value of the
Let's look at a 10k load connected to the wiper:
The voltage on the top and bottom of the pot changes when a load is
In circuit A, the
voltage reduces a small amount as the 10k load has little effect on the
low- value resistance of the pot and resistors.
In circuit B, the 10k load has a larger effect on the voltages.
In circuit C the 10k load has a major effect on the voltages.
This means it is necessary to choose values that are acceptable for
minimum current through the pot as well as creating the required voltage
on the top of the pot.
TRIM POT A trim pot is simply a pot without a
shaft. It usually has a screw-driver slot and is adjusted once in the
life of a circuit. It is usually small in size and can be any resistance
value to suit the circuit.
It can be connected as the only pot in a circuit or used in conjunction
with an ordinary pot to set a particular value or "setting."
It is identified in a circuit as follows:
TRIM POTS can be used to trim the value
of a POT
THE VOX - Voice Operated Switch
Basically, a VOX circuit is a very high gain amplifier that
detects faint sounds and turns on a relay.
Here are a number of voice-operated (sound operated) circuits
that turn on a relay or activate a device.
In general, a VOX circuit keeps the relay activated for a short
time between sounds so the device remains constantly illuminated or
The first circuit is a CLAP SWITCH. The LED illuminates for 15
seconds after the sound of a clap. For full details of the
circuit see Fig 71acd.
CLAP SWITCH USING PIEZO DIAPHRAGM PICK-UP
The circuit above takes about 20uA when "sitting around." That's
because the piezo diaphragm does not require any current.
The same circuit can use an electret microphone for the input
but the idle current rises to 200uA.
CLAP SWITCH USING ELECTRET MICROPHONE
Both circuits detect a clap but neither will detect faint
noises or talking.
The circuits do not keep the LED illuminated constantly but only
illuminate for 10 - 15 seconds and turn off for 10 - 15 seconds.
This circuit toggles the LEDs each time it detects a clap or tap
or short whistle.
CLAP SWITCH TOGGLES THE 2 LEDS
The second 10u is charged via the 5k6 and 33k and when a sound
is detected, the negative excursion of the waveform takes the
positive end of the 10u towards the 0v rail. The negative end of
the 10u will actually go below 0v and this will pull the two
1N4148 diodes so the anode ends will have near to zero volts on
As the voltage drops, the transistor in the bi-stable circuit
that is turned on, will have 0.6v on the base while the
transistor that is turned off, will have zero volts on the base.
As the anodes of the two signal diode are brought lower, the
transistor that is turned on, will begin to turn off and the
other transistor will begin to turn on via its 100u and 47k. As
it begins to turn on, the transistor that was originally turned
on will get less "turn-on" from its 100u and 47k and thus the
two switch over very quickly. The collector of the third
transistor can be taken to a buffer transistor to operate a
relay or other device.
The next VOX circuit activates a relay when audio is detected by
the microphone. The relay is kept activated for 5 seconds
after a silent period, by the 22u, to keep the relay fully activated during
normal speech. The circuit takes 0.5mA when "sitting around."
The circuit above is the best design as it uses the least number
of components and drives a relay.
The next circuit comes from Engineers Garage website. It uses
fewer components but takes more current (about 6mA) in the
quiescent mode and does not have any delay to hold the relay ON:
6v VOX CIRCUIT - no delay
(Not a good design - circuit takes 6mA)
The following two circuits detect audio and keep the LED
illuminated for about 5 seconds. The
delay is proved by the 100u capacitor on the output. The output
is normally HIGH and goes LOW when audio is detected.
The LED shows the condition of the output. It is removed when
you add the circuit to a project.
This circuit is the 12v version. Quiescent current (idle
current) is 0.5mA.
The addition of the diode in the 3v circuit is
needed to discharge the 22u so that it produces its "full
effect" to saturate the output transistor when required. It is
not needed in the 12v circuit as the base-emitter junction of
the output transistor "zeners" at about 5v and this helps to
partially discharge the 22u. But when only 3v supply is present,
the 22u has a maximum of only a few volt on it and none of its
voltage will be removed. The output transistor is turned on when
the middle transistor turns off. The 27k pulls the 22u high and
if it is discharged, it pulls the base of the third transistor
"up" and turns on the LED. During this time it gets charged
slightly and this charging current flows via the base of the
third transistor to turn it on.
When the second transistor turns on, the 22u effectively "drops
down" and the voltage across it (say 2v) will take the negative
lead of the electro BELOW the 0v rail of the circuit. As soon as
the negative lead is 0.7v below the 0v rail, the diode comes
As far as the diode is concerned, it sees a voltage of +0.7v on
the anode lead with respect to the cathode lead and current will
flow through it to discharge the electro. If the diode is
removed, it would take a voltage of about -5v on the electro
before it is discharged via the base-emitter junction of the
The 22u is discharged via the
The next circuit is
designed by electroschematic.com. It is not a good design.
The circuit takes 14mA when sitting around and the 470u electro
only needs to charge by about 0.5v before the circuit changes
state. This uses only a fraction of the possible time delay for
a 470u capacitor if the circuit is designed to charge it to a
higher voltage before changing state.
12v VOX CIRCUIT
(Not a good design - circuit takes
Here is the circuit re-designed to take less quiescent current
(0.5mA) and provide a longer delay with 100u electrolytic (20
The next circuit can be Voice
Operated or activated by a Video signal.
The circuit activates a relay when an audio or composite video signal is delivered to
the input. This allows you to use the tuner built into your VCR to turn on and
off older TVs that are not equipped with a remote. It can also be used to
activate surround-sound equipment, turn off room lights, turn on video game
When power is applied, the first transistor is not turned on and the second
transistor gets turned on via the 10k resistor. This prevents the third
transistor turning ON and the relay is not energised.
When an audio or video signal is delivered to the input, the first transistor
turns ON and this turns OFF the second transistor. The third transistor gets
turned ON via the 1k and diode after the 1u gets charged a small amount.
When the input signal ceases, the first transistor turns OFF and this turns ON
the second transistor. The third transistor no longer gets base current via the
diode but the 1u holds a small amount of energy and this is delivered to the
base to keep the relay active for a short period of time. After this the
transistor turns OFF and the relay is de-energised.
The next circuit
is a little over-complex and could be improved.
Here are some suggestions:
1. The 10u from the microphone can be as low as 100n without any decrease in
2. The 10k to the base of the first transistor should be a higher value to
increase the input impedance of the first stage.
3. The 100u on the emitter of the first transistor can be replaced with a link.
4. The third transistor has a gain of 10. This can be increased by reducing the
5. The 22k and the two diodes can be removed and the circuit re-designed as shown
6. The 4u7 on the base of the 4th transistor is only charging to 0.7v. The
delay section needs to be on the third transistor as shown above in the 12v VOX
Circuit and the fourth transistor should be a driver transistor.
This circuit can be improved
This clever circuit turns on a motor with a short whistle and turns the motor
off with a long whistle. It's a toggle arrangement.
VOX TOGGLE CIRCUIT
Short tone = ON Long tone = OFF
The circuit allows a whistle to turn an appliance ON
and OFF by sending a short whistle to turn a circuit ON and a long whistle to
turn a circuit OFF.
This is handy when you cannot see the result of your operation. A simple toggle
operation is not suitable as you do not know the state of the output at the
start of the operation.
By sending a long whistle,
you definitely know the output will be OFF and you can then control the output
A short whistle
is less than 0.25 sec and a long
can be any length longer than 1 second.
These times can be adjusted by changing the value of the components.
When a short whistle
is received, the lower 47u discharges and pulls the base of the BD136 towards
the 0v rail and turns the transistor ON. This activates the relay and the
contacts take the 4k7 to the 0v rail to keep the transistor ON.
During this time the top 47u charges via the 100k but not enough voltage appears
across it to turn on the BC557 transistor.
If the whistle
appears for a long period of time, the top 47u charges and turns on the BC557
and the voltage between the emitter/collector terminals is less than 0.3v. This
voltage is too low for the BD136 to remain on and it turns off.
When the whistle
stops, the BC557 remains on for 1 second and then turns off.
The circuit is then ready to be activated again.
VOICE OPERATED LATCH
The following circuit latches a LED
ON when sound is detected. It can be used to confirm a certain level of
sound has been reached or exceeded during an event.
Sound makes LED stay ON
The electret mic and first transistor are active when the circuit is "waiting
for a sound" and the 3rd and fourth transistors are biased OFF due to the 1M and
100k voltage-dividing resistors putting a voltage of between 0.27v and 0.54v on
the base of the second transistor. This voltage is not high enough to turn
the transistor ON. But the voltage helps to turn the circuit ON when audio is
detected and makes it very sensitive.
You can see a poorly designed VOX latch circuit (Called Puff to OFF LED) in our
Mistake eBook. These poorly-designed circuits show you how NOT to design a
circuit and are just as informative as a good design.
DIGITAL mode Now that we have covered more than
100 different circuits, you can see each transistor
in a circuit is operating in either analogue or digital mode.
Sometimes it is easy to see the mode of operation.
If a transistor is not taking any current, then gets turned on
(hard or fairly hard) it is operating in DIGITAL MODE.
If it is turned on with the collector at or about mid-rail
voltage, it is in ANALOGUE MODE.
Understanding these two modes is very important because a
transistor in digital mode wastes the least energy. However it
cannot amplify a signal that has an amplitude less than 0.6v. It
can only amplify a signal that is greater than about 0.7v.
That's why some circuits need both types of stages.
A well-designed circuit takes the least current in quiescent
We have also shown how one stage transfers energy to the next
stage via a capacitor. But a capacitor creates losses.
Direct-coupling transfers more energy and has no loss.
When designing a circuit it is best to refer to the circuits
covered in this eBook, to prevent designing something that may
not work correctly.
We have exposed many poorly-designed circuits in our "Spot
The Mistake" eBook, as explained above.
CLIPPING AND DISTORTION Most Analogue circuits
require a stage to reproduce a signal as accurately as possible.
After all, we don't want an amplifier to be distorted.
However some analogue circuits are designed to distort a signal. Therese
can be classified as "EFFECTS" circuits and the most common is a guitar
effect called FUZZ. A Fuzz circuit clips a signal so the full
amplitude is not delivered to the output.
There are many ways to distort a signal (or process a signal) so
a desired effect can be achieved and there are dozens of names
for these circuits.
There is no "electronics rationale" behind the design of these
circuits and many of them come from experimenting and placing
components in unusual places to create positive or negative
feedback or overdrive a stage or even under-drive the active
component (usually a transistor or op-amp).
There are hundreds of circuits to create these "EFFECTS" and
here are some:
An OVERDRIVE Circuit
This produces a ROCK PEDAL with very good performance.
This circuit takes the small signal from the magnetic pick-up
and amplifies it 10 to100 times and removes some of the noise
via the 47p capacitor.
Insert this circuit before Overdrive / Distortion / Metal /
Chorus / Delay and the result is like a very expensive high
Integration and Differentiation We are
going to show how two components, (a resistor and capacitor - in
series) produces different results according to the frequency of
a voltage (signal, waveform) delivered to them. We also have
different names for the two "series components," depending on
the actual circuit using the components.
already covered the TIME DELAY circuit and
shown it consists of a resistor and capacitor in series. The
join of the two components is detected by a transistor or
Integrated Circuit and when a particular value of voltage is
reached, the circuit produces results as shown below. When the
circuit is first turned ON, the
voltage gradually RISES when the capacitor is below the resistor
or gradually FALLS when the capacitor is above the resistor.
But if we connect the same two components to a
rising and falling voltage, a completely different result is
If the input is a sine-wave, the following results are produced when a
low-frequency, medium-frequency or high-frequency is supplied to
The circuit is a LOW PASS FILTER
You can see the output waveform almost disappears
when a high-frequency is delivered to the input. (It disappears
in amplitude, but a voltage appears across the capacitor that is
approximately the average of the high and low values.)
This means the circuit is capable of removing high-frequency
portions of a waveform. If the waveform consists of
a mixture of low-frequencies and high-frequencies, only the low-frequencies
will appear on the output.
In other words the circuit is a FILTER and it only passes the
In other words it is a LOW PASS FILTER.
We now look at the circuit above when ANY frequency is
delivered to the input. This time we will eliminate the effect
of the waveform when it is falling, by adding a diode to the
You will notice a voltage builds-up on
the capacitor. This is because the input voltage is charging the
capacitor a little-bit more on each cycle. The circuit becomes an INTEGRATOR. It does not matter if the
waveform is a sinewave or square-wave - the capacitor
gradually becomes charged.
The circuit is an INTEGRATOR
- it gradually charges the capacitor
If we deliver the waveform to the circuit with
the capacitor above the resistor, the output is not integrated:
The circuit is NOT an INTEGRATOR
So, what is the purpose of the circuit with the
capacitor above the resistor?
The following results
are produced when a low-frequency, medium-frequency or
high-frequency is supplied to the input:
We see the low-frequency waveform is attenuated (reduced) while
the high-frequency waveform passes through the circuit.
This produces the name "HIGH PASS FILTER."
It is also given the name "DIFFERENTIATOR."
This means only the high-frequency portions of the input
waveform will be delivered to the output.
If we have a waveform consisting of low-frequency and
high-frequency components, only the high frequency parts will be
delivered to the output.
The simplest type of "pull-up" and "pull-down" resistors
in the following diagrams:
The 47k is a "Pull-Up" resistor. The designer of the
circuit wants the base of the transistor to be at a known
voltage when the circuit is sitting around, waiting for a
signal. The 47k forms a voltage-divider with the 100k and BC547
and it makes sure the BC557 is turned off when the circuit is
waiting for a signal. It is pulling the base "UP" so the BC557
is not turned on.
The 470k is a "Pull-Down" resistor. It prevents the
BC547 generating a static voltage on the base and turning the
The 330R can be classified as a "pull-down" resistor.
It is also part of the load for the output stage.
CLASS-A CLASS-B CLASS-C
Each stage in a circuit can be given a name according to its
efficiency. Normally the output stage is the only stage that is classified
as "A" "B" or "C" because this
is where most of the efficiency or losses occur. However the
same criteria applies to the other stages in a circuit and this
can give you some indication of the performance of each stage.
The most inefficient stage is classified as CLASS-A. It
has an efficiency of 25% to 50%.
However it is the best stage for amplifying audio signals as it
produces perfect reproduction and amplification. High Fidelity
amplifiers are class-A throughout. The stage is biased so
the collector is at half-rail voltage. This allows the stage to
amplify both the positive and negative portions of the signal.
CLASS-B is basicallyan EMITTER-FOLLOWER stage.
It is also called PUSH-PULL when two emitter-follower
transistors are connected together. Push-pull has an efficiency
of 78% max. The stage only amplifies the positive portions of
the signal. When two transistors are connected,
one amplifies the positive portion of the signal
and the other amplifies the negative portion of the signal.
does not have base-bias. It consumes no current when
"sitting around" and efficiencies up to 90% are possible.
The energy to turn the transistor ON (and drive the base) must
from the input signal. The stage only amplifies the positive
portions of the signal and has high distortion, but it can be
used in certain applications with very good results.
THE DRIVER STAGE Any transistor that drives a LOAD is said to be a DRIVER
in a DRIVER
STAGE or OUTPUT STAGE. A LOAD is generally a relay, motor, globe, LED or other
device that requires a CURRENT.
The following diagrams show a DRIVER (or OUTPUT) TRANSISTOR driving a LOAD:
The transistor driving a LOAD
In most circuits, achieving a CURRENT to drive a load is the
most difficult thing to do.
The voltage for the LOAD is easy to get. It is the voltage of
the supply. The supply (RAIL VOLTAGE) can be set to match
the requirement of the LOAD. If you have a 6v motor, use a 6v
supply. If you have a 12v relay, use a 12v supply.
Achieving (getting, supplying) the current for the LOAD is
the requirement of the transistor - the DRIVER transistor or
commonly called the OUTPUT transistor.
The first thing you have to know is the amount of current for the
LOAD. This is generally expressed in milliamp (mA) or Amp
(A). 1,000mA = 1Amp.
some cases this is easy. It is on the specification-sheet - such
as a LED requires 25mA and a relay requires 90mA. But if a relay
states "12volt 100R", you will have to work
out its current. It's a simple Ohm's Law equation:
save mathematics, we have provided the current requirement for a
number of relays:
5v relay 100 ohms 50mA
5v relay 240 ohms 20mA
9v relay 100 ohms 90mA
9v relay 240 ohms 40mA
12v relay 100 ohms 120mA
12v relay 240 ohms 50mA
for some loads is unknown. For example, a 6v or
12v motor will take more current when starting or when heavily
loaded. The starting
current can be 6 times the running current and this is what the
driver transistor must be able to provide. In fact the running
current is not known until you try the motor. It can be 50mA,
100mA, 250mA or even 1amp. The starting current can be 6 times
the running current.
This is what you have to do:
Connect the motor to the supply (without a transistor driver) and
add an ammeter in series with one lead. Turn the motor ON and
hold the shaft. Quickly read the current. Release the shaft
and read the running current.
Suppose a motor requires 600mA to start and 100mA to run. In
other words, the motor will take 100mA when lightly loaded and
600mA when heavily loaded.
The driver transistor will be required to supply 600mA to start
the motor (and when it is under load) and when it is running
under almost no-load, the current will drop to
For a transistor to supply 100mA to 600mA to the motor, we must
deliver current to the base. The transistor simply amplifies the
current we supply to the base. The amplification factor is
called the CURRENT AMPLIFICATION or CURRENT GAIN and is normally
100, but this only applies when the transistor is lightly
transistor has a CURRENT AMPLIFICATION of 100, 1mA into the base will allow 100mA
to flow in the collector-emitter leads (collector-emitter
But if a transistor is designed to handle 100mA, it will only deliver about 50mA when 1mA enters the base.
It may take 2mA base-current to get to 70mA, 5mA to get to 80mA
and 10mA to get to 100mA.
To deliver 100mA, you will need a transistor capable of
delivering about 300mA. That's because the larger transistor has
a larger junction and is capable of handing the current.
For example, you will need a 300mA transistor to handle 100mA
and a 1 or 2 amp transistor to handle 600mA. This is a
fact that is rarely explained.
When you choose the right transistor, it will remain fully
saturated when 100mA is flowing and the voltage across the collector-emitter will be
less than 0.5v. In other words, the transistor will remain FULLY SATURATED
or FULLY CONDUCTING.
When the current exceeds the maximum rating, the transistor falls out of
conduction. In other words the voltage across the
collector-emitter terminals will increase above 0.5v and will gradually
rise to 1v, 2v, 3v, or more as the current increases. When this happens, the
wattage dissipated by the transistor increases and it gets very hot. The transistor may
be able to deliver the higher current but the voltage across the load will be
reduced as some of the voltage is lost across the transistor.
Let's take this in more detail:
When a transistor is fully saturated and passing 600mA, the
collector-emitter voltage will be as low as 0.2v to 0.5v. The
wattage dissipation will be: P=VI = 0.2 x.6 = 0.12watts or
P=VI = 0.5 x.6 = 0.3watts
If the transistor comes out of conduction and produces 3v across
the collector-emitter terminals, the wattage dissipation
increases to: P=VI = 3 x.6 = 1.8watts This is an
enormous increase in heat produced by the transistor. That's why
you don't want a transistor to come out of conduction.
How do you know if a transistor has a gain of 10 or 100? You don't. There is no way to know if a transistor has a
gain of 10 or 100.
However if you follow our suggestions, you will be able to
achieve the maximum gain:
Select a transistor capable of delivering 2, 3 or 5 times more
current than is needed for the project you are designing. This
will allow it to operate in its HIGH GAIN region because it will
not be over-loaded.
Feel the temperature-rise of the transistor and use a heatsink
to prevent it over-heating.
A transistor capable of supplying 1amp will be operating near
its maximum when supplying 600mA and
the base will need 60mA.
In the diagram below, the transistor is not controlling the current. It simply
supplies the current demanded by the motor. If the motor
requires 100mA, the required current will flow though the
collector-emitter leads of the transistor. When the motor "asks for" 600mA, the transistor will
deliver the current.
For this current to be available to the motor, the transistor
must be turned ON and fully saturated.
The 60mA base current flows ALL THE TIME so that when 600mA is required
by the motor, the transistor will deliver the current.
The following diagram shows the current required by the motor.
Here's the important point. The circuit must be designed so that
it can deliver 600mA AT ANY TIME. This means the
transistor must receive a base current of 60mA AT ALL TIMES.
If you use a high-power transistor, it may have a gain of
50 or more when delivering 600mA and the base current will be
lower, but we will take the case of the transistor having a gain
of 10 when 600mA flows.
Base current comes from the previous stage in a circuit.
There are two ways to deliver this current:
1. via a load resistor
2. via a transistor.
We have already seen the inefficiency of delivering a current
via a load resistor, as shown in the following circuit:
Base current supplied via a LOAD resistor
In the circuit above, current is always flowing through the 100R
LOAD Resistor. The BC547 is merely diverting the current from
the base of the output transistor to the 0v rail. We will not be
describing this circuit arrangement.
It is much more efficient to deliver a current via the following
type of circuit:
In the circuit above, current is only flowing through the 100R
LOAD resistor when the PNP transistor is turned ON. The 100R
CURRENT LIMITING resistor determines the base current for the
We have now designed the best circuit for driving a LOAD:
The approximate current in each leg of the circuit for 1Amp
LOAD is as follows:
This means 1mA will control (deliver) 1amp (1,000mA) into a
Here is the PNP-output circuit:
In Summary: When designing a circuit, we allow a
current-gain of 100 when a transistor is lightly loaded and a
gain of 10 when fully loaded. The driver transistor will have a gain of between
100 and may drop to 10 when full current flows. This means it
will need 1mA base current for 100mA it delivers to the load
when the gain is 100, but will need 10mA for 100mA, when it has a
gain of 10.
This also applies to a globe (lamp) as the load. A globe
requires 6 times more current to begin illumination because the
filament is cold and its resistance is less than when it is hot.
To make sure the circuit will illuminate the lamp, the driver
transistor must be able to deliver up to 6 times the operating
current for a very short period of time.
A transistor driving a speaker requires the same driver and
output arrangement as driving a motor as it is a very low
The situation is different with a LED. The current required for
a LED is small and a small-signal transistor can supply the
current and have a gain of 100. The current for a LED will be
about 25mA and the base current will be less than 0.25mA
The current for a relay can also
be provided by a small-signal transistor and the gain can be
The base current must come from a previous stage
For low-current requirements, use a BC547 for
the driver/output transistor. A BC547 will handle up to 100mA.
Similar transistors include: 2N2222A, BC107, BC108, BC109,
BC142,BC182L, BC337, and many others.
How to Design a DRIVER STAGE
For a HIGH
CURRENT requirement, the NPN driver transistor needs to handle 1 amp or more.
For 1 amp use: BC142, BC337, BD131, BD139,
For more than 1 amp use: 2N3054, BD131, TIP31A and many others.
Use a heatsink.
To make sure the transistor stays IN CONDUCTION, measure the
collector-emitter voltage when under FULL LOAD. The
collector-emitter voltage must
not be above 1v.
An 8R speaker puts a very heavy demand on any supply rail and
this can cause glitches that may affect the operation of other
sections of the circuit.
Fig A shows a speaker connected to a driver transistor.
The circuit will work but the first 0.5v of the signal will not
be reproduced as the transistor does not turn on until the base
sees 0.6v. The circuit takes no current in quiescent mode as the
transistor is not tuned on. The output will be distorted. Fig B shows a base bias resistor connected to the
positive rail. This resistor will have to be a low value (about
2k2 to 10k) to turn the transistor ON so that the collector will
be at about half rail voltage. This will allow both the positive
and negative excursions of the signal to be reproduced. Fig C shows the transistor in self-bias mode. Figs B and
C are very similar. The resistance of the base bias resistor
will be slightly lower in fig C and selecting the right value
will provide about half rail voltage on the collector.
But the problem with all these circuits above is the high
current taken by the stage.
As you can see, the load is only 8R and if the transistor is
partly turned ON, the idle current will be very high.
This can can be reduced by using the following
circuits. The output volume will be reduced, but the current
will be reduced considerably.
In the circuits above, the output impedance is 8R. In the
circuits below, the output impedance has been increased to 30R
The secret to the good performance of stages D and E is due to
the 100u on the emitter. This electrolytic turns the stage into a
common-emitter amplifier, just like circuits A, B and C, as far
as the audio signal is concerned, but turns the LOAD into 30R or
158R as far as the DC current is concerned.
We get the best of both conditions at the same time.
DESIGNING AN OUTPUT STAGE
The circuit at the left shows an output stage
as described in the video above, by the "Professor."
He tried to explain the base-biasing but failed.
In fact he did not know what he was talking about.
That's because the operation of the circuit is much more complex
than you think.
The circuit has two features. The H-bridge design allows a known
operating-point to be produced, even though the gain of the
transistor may change from one device to another. Suppose the
transistor is replaced with another having a higher gain.
The collector current will increase and more current will flow
though the emitter resistor. This will produce a higher voltage
across the emitter resistor and the difference between the base
voltage and the emitter
will turn the transistor OFF slightly and maintain the original
The second feature of the circuit is the electrolytic in series
with the speaker. This allows only the AC portion of the
waveform to enter the speaker and the cone is not "pulled" due
to DC flowing though the voice coil.
However, the operation of the circuit is very complex, so we
will explain it this way:
The design of the circuit starts by selecting the value of
base bias resistors A and B. The resistors are chosen so that a
voltage of about 0.7v is produced on the emitter. This allows
the stage to be turned ON with about the lowest quiescent
current. (This will produce a voltage of about 0.7 x 47 = 3.3v
across the load resistor.)
To get 0.7v on the emitter, we need 1.4v on the base. This gives
us the ratio of base bias resistor A and base bias resistor B.
The next thing we consider is the maximum current we want to
flow through the load resistor. When the transistor is
fully turned ON, this current can be as high as 230mA.
Let us allow the base current from the divider to be 100mA. The
additional current will be created when the signal arrives and
turns the transistor ON more.
Suppose the transistor has a gain of 100. This means the base
current must be 1mA.
This will cause 100mA to flow through the 10R emitter resistor
and produce a voltage of 1v.
The base voltage will be 1.7v.
The voltage across base bias resistor A will be 10.3v
The next design-feature to remember is this:
The current flowing through the base voltage-divider (resistors
A and B ) should be 10 times the base current to provide stable
This means the current through base bias resistor A is 10mA.
The value of base bias A is: 10.3/0.01 = 10.3k
The value of base bias B is: 1.7/0.009 = 1.8k
Forget about the actual amplitude of the incoming signal.
What happens with the incoming signal is this:
The signal-peak gets converted by the 10u electrolytic to
produce either a 2mA flow into the base or it cancels the 1mA
flow to produce 0mA into the base.
The 100u electrolytic on the emitter prevents the emitter rising
or falling and the stage operates just like the emitter is
connected directly to the 0v rail.
When the incoming signal cancels the 1mA into the base, the
transistor is turned off and the 47R charges the 100u connected
to the speaker. This current flows through the speaker to shift
the cone. A voltage develops across the electrolytic and energy
is stored in it.
When the signal rises, up to 2mA flows into the base and the
transistor is fully turned ON.
The energy in the electrolytic flows through the speaker and
moves the cone in the opposite direction.
stage takes considerable current when "sitting around"
this is one of the disadvantages of a "Class-A" stage. If you
want low quiescent current - use a class-B output stage.
The stage has a low input impedance (about 1k8) however it is
driving an 8R load (the speaker) and this is a ratio of more than 100:1 and the
stage is achieving a considerable "conversion."
You can call this circuit an INTERFACE, BUFFER, AMPLIFIER,
DRIVER, POWER-AMPLIFIER, or MATCHER. It
does all of these things.
The circuit must consume about "half-current" when "sitting
around" so the incoming signal can increase the current to
full-current or reduce it to almost zero, so the output signal
is not distorted.
The "design-feature" involving the two resistors on the base can
be changed to only allowing 2 or 3mA as "bleed current" so the input impedance
This will make the 1k8 resistor as high as 1.8 x 3 = 5.4k and
the incoming signal will not be attenuated as much. The
attenuation applies to both the rise and fall of the signal as
this 1k8 (5k4) is directly across the signal.
It's all a matter of building the circuit and see how it
performs as the transistor is at its limit of heat dissipation
in this amplifier.
A simpler circuit is shown here.
The base-bias resistor is chosen so that half-voltage appears on
the collector and the transistor is self-biased so transistors
with a slightly different gain will operate correctly.
This operation is commonly called "Mid-point operation" or
"Q-point" operation. For maximum output the load resistor will
need to be 8R, but this will create a current-flow that is more
than the transistor will handle. That's why we use "class-B"
(Push-Pull) output stages.
If you want to see the complex mathematics
behind the circuit above, click
for a .pdf from a
subscriber and teacher, Tom Wheeler
has provided a mathematical assessment to show the power
delivered to the speaker can be as low as 8% and in most cases
it is not much higher than this value.
This is one more proof that "class-A" configurations are very
Mr Wheeler is a lecturer at a tertiary college in Kansas City
If you are interested in taking a course in electronics, you
should investigate the courses available in a college near to
It is not easy to get through the jumble of course-names and
terminology provided by colleges, technical colleges,
institutes, Universities etc, but start by looking on the web and
refer the to course-structure to see what topics are covered.
Mr Wheeler's college is Metropolitan Community College in
Kansas City, MOwww.mcckc.edu
Here is a brief listing from the colleges introduction:
The Metropolitan Community College Business &
Technology Campus (BTC) is one of five MCC campuses that serve
the greater Kansas City area. BTC offers both degree and
certificate programs in Computer Aided Design and Drafting (CADD),
Cisco Networking (CCNA/CCNP), Engineering Technologies
Health & Safety (EHSS), Electric Utility Lineman, Heating
Ventilation and Air Conditioning (HVAC), Industrial Technologies
(Instrumentation & Controls/Industrial Maintenance/Energy
Efficiency), Solar/Photovoltaic Energy, Precision Machining,
Welding, as well as degree completion programs designed for US
military veterans and skilled trade apprenticeship completers.
We are located at 1775 Universal, Kansas City MO, 64120; please
visit our web site at
www.mcckc.edu/btc, or call 816-604-5200 to arrange a campus
THE TRANSISTOR AS A BUFFER We have already described the transistor as a
BUFFER but previously we used different wording, such as
DRIVER, EMITTER-FOLLOWER or SINKING TRANSISTOR
or SOURCING TRANSISTOR.
In all these descriptions the transistor is doing the same
thing. It is receiving a small current and/or a small voltage
and delivering a larger current and/or a larger voltage.
All the terms are interchangeable and BUFFER describes a
transistor that is placed between a LOAD that requires a
high current and a BUILDING BLOCK that is only capable of
delivering a small current.
A BUFFER or DRIVER TRANSISTOR is BUFFERING or JOINING the low
impedance of the LOAD to the high impedance of a previous stage.
If you remove the buffer transistor and join the previous stage
to the load, the result will be either very little volume from a
speaker, or a solenoid that does not operate or a globe that
does not illuminate.
These devices do not work because the CURRENT is insufficient.
Many devices ONLY work when sufficient CURRENT flows.
That's what the DRIVER TRANSISTOR does. It increases the
Sometimes a transistor will increase the voltage from one
section of a circuit to another.
This is called AMPLIFICATION and we do not use the word buffer.
BUFFER only refers to amplifying CURRENT.
BIASING A TRANSISTOR
We have seen many different ways to bias a
transistor and now we look at 3 and compare them.
Transistor NOT biased
Diode discharges capacitor
This transistor is
not biased AT ALL. However it will work if the incoming signal
The signal needs to provide all the energy to turn on the
transistor and it only amplifies the positive portions of the
The amplifier is classified as "Class-C."
To help discharge the coupling capacitor during the negative
portions of the waveform, a diode can be included as shown. It
comes into operation during the negative part of the cycle by
"flipping over" and conducting when the negative part of the
signal is less than -0.6v. All the waveform lower than -0.6v is
passed through the diode and the capacitor is discharged during
this portion of the signal.
amplifier takes zero current when "sitting around" (quiescent
current), and can be classified a very
efficient, however in most cases it will produce distortion.
This circuit uses a
base bias resistor between base and supply.
It is a "Class-A" amplifier.
This circuit will work successfully if the base bias resistor is
chosen so the collector voltage is mid-rail.
However if you use another transistor with a lower or higher
gain, the collector voltage will be higher or lower than for the
previous transistor. In addition, the gain of the transistor
will change slightly due to temperature and the collector
voltage will alter.
This is not a good design as the collector voltage is not
To make the
"Class-A" amplifier stable, the base-bias resistor is connected
If the gain of the transistor reduces due to temperature rise,
the collector voltage will increase. This will increase the
current through the base bias resistor and turn the transistor
ON more and lower the collector voltage. It will not reduce to
exactly the same voltage as before but the circuit will adjust
to a fair extent.
If the transistor is replaced by one with a higher gain, the
collector voltage will reduce. This is allow less current to
flow into the base and turn the transistor off slightly and the
collector voltage will rise. It will not rise to exactly the
same voltage as before but the circuit will adjust to a fairly
This is called a "self-biasing class-A amplifier."
To get a better
automatic biasing features as the circuit above, two additional
resistors are needed and the circuit forms a bridge. (The word
"bridge" comes from "Wheatstone Bridge" where 4 resistors are
placed in a square to form a balancing network).
Ra and Rb form a voltage-divider to produce a fixed voltage on
Re is a feedback resistor to create the stabilizing feature.
Suppose the transistor gets warm on a hot day and looses gain.
The collector voltage will rise and the current through the load
resistor will decrease. The current through Re will also
decrease and the voltage across it will decrease. This means
the voltage between the base and emitter will increase and the
transistor will turn ON more to counteract the voltage-rise on
The value of Ra and Rb are calculated so that 10 times the base
current "bleeds" through them. In other words, 10mA bleeds
through Ra and 9mA through Rb so that 1mA goes to the
This is very
wasteful, but let's look at why the bleed current needs to be
10x the base current.
There are two "design factors" in this type of circuit.
1. The bleed current through the voltage divider Ra/Rb is 10
times the current required by the base.
2. The values are chosen so that about 1v appears on the base.
This allows the transistor to be turned ON and gives about 0.3v
across the emitter resistor. This allows the emitter resistor to
stabilize the stage and allows the transistor to turn ON and
pull the output as low as possible when a large amplitude signal
Let's see why the bleed current is 10 times that
required by the base.
We will start with a bleed current of 2mA through Ra and 1mA
through Rb with 1mA going to the base.
Suppose the transistor has a gain of 100. The current through
the collector-emitter circuit will be 100mA and the value of Re
will be 3R. The voltage across the load resistor will be 5v and
the voltage on the collector will be 6v.
Suppose the transistor is replaced with one having a gain of
200. We already now the value of the load resistor (50R) and
emitter resistor (3R) and the base bias resistors have the
capability of delivering 1mA.
What will happen is this: The transistor will turn ON and
because 1mA base current is available, it will allow 200mA
flow through the collector-emitter circuit.
The voltage across the emitter resistor will be
0.2 x 3R = 0.6v and the voltage-drop
across the 50R will be 0.2 x 50 = 25v. We do not have 25v
available (only 11v) and the transistor will turn ON to a point
where it is fully saturated. This situation has occurred because
the base voltage was able to rise so the base-emitter voltage is
0.7v. Since the transistor is taking only 0.5mA, the voltage
across the 1k will theoretically be able to rise to 0.0015 x
1,000 = 1.5v
Now you can see why we want the base voltage to remain stable,
so a transistor with a gain of 200 is not fully turned ON and
the same collector voltage is produced for a wide range of
Up to now we have talked about the base-emitter voltage of a
transistor as being about 0.55v to slightly more than 0.7v
- depending on the transistor and how "hard" it is being turned
But there is also a situation where a transistor must be TURNED
OFF by providing a very low voltage on the base.
Sometimes you cannot provide 0v on the base due to the "control
voltage" coming from another chip or a set of voltage-dropping
Some transistors (including surface-mount types) have a
base-emitter voltage as low as 200mV, depending on the
temperature of the transistor. As the temperature increases, the
base-emitter voltage decreases and this will be quite a
The following two circuits are typical examples:
The voltage on Pin 3 of the 555
(when it is LOW) can be as small as 100mV (depending on the
SINKING CURRENT - the current flowing "into" pin 3 when a load
is placed between pin 3 and the positive rail).
As this current increases to 200mA, the voltage on Pin 3
increases to 2.5v - WHEN IT IS LOW !!
Since this voltage is unknown and widely-variable, it is a good
idea to add the extra 470R (shown in RED),
to reduce the voltage on the base to a "cut-off" value.
When the chip is driving into the base of a buffer transistor,
the current will only be very small when the output is low and
to make sure it is below "cut-off" the extra 470R HALVES
the base-emitter voltage - just tp make sure !!
In the second diagram, the voltage divider made up of the 22k
and 2k2 produces a voltage of 0.5v on the base, however the
inclusion of the 470R reduces this further and even though the
output of the micro may be as high as 200mV when sinking a
current such as 25mA, the base voltage will not go above 201mV.
(In the circuit above the LOW voltage on the output of the micro
will be almost zero - so, in this case, there is no problem).
Transistors with internal Resistors
Transistor with Resistors
transistors have internal resistors. This includes small-signal
transistors, power transistors, Darlington and surface mount
devices. You must know if the transistor you are testing has
internal resistors because the circuit will be lacking one or
two resistors and you will wonder how it has been
In most cases the values of the resistors will be the same as
what you would use with a transistor not having internal resistors however if you
are designing a new circuit, it is best not to include
transistors with internal resistors - you may want to change the
value and this is not possible with fixed values.
Some surface mount transistors and Darlington transistors have
different values to those shown in the diagram on the left and you need to use a datasheet to determine
the exact values. When testing these transistors they will
appear to be faulty due to the internal resistors.
The resistor-values are chosen for circuits using the
transistors with a medium to high collector current. They are
not suited for stages where a very low collector current is
required as the base resistor would be 47k to 100k.
Using a Transistor with a Higher Gain
There are 4 main features of a
1. Its power-handling capability,
2. Its voltage-rating,
3. Its maximum frequency,
4. Its gain.
What happens if you replace a transistor with one having a
The operating point (the "Q-point") will adjust in the
Self-Biasing circuit but if the new transistor has a much-higher
gain, the collector voltage will be lower.
The Bridge-Biasing arrangement will maintain a more-constant
collector voltage when the transistor is replaced with one
having a higher gain.
In other words, the
quiescent-point or operating-point (or mid-point
for the collector-voltage) of a Self-Biasing or Bridge-Biasing circuit
will self-adjust when the gain of the transistor alters. But a higher gain transistor will allow the
collector voltage to drop closer to 0v when the incoming signal
delivers the high portion of the waveform. The result will be a
slightly higher output because the output will go to a lower
value than before.
The opposite will occur with a transistor having a lower gain.
transistor and biasing
for details on how to design the two stages.
Using a Transistor as a CURRENT LIMITER
10 Second Delay
The circuit produces
a delay of a few seconds due to the TIME DELAY circuit
(made up of the 470k and 100u) taking time to charge the 100u.
The transistor is an EMITTER FOLLOWER and the emitter
rises at the same rate as the base but is about 0.7v less than
The circuit does not have a CURRENT LIMITING resistor in series
with the LED or in the collector circuit because the transistor can only allow a current
between the collector and emitter terminals that is 100 to 200
times more than the current entering the base and it effectively
becomes a resistor. If the transistor has a gain of 200, it
will be 470,000/200 = 2350 ohms. This will deliver 10/2350 =
Another way to explain the circuit is to work out the current entering the base.
across the 470k resistor will be about 10v max for 12v supply
and the current will be 0.02mA.
The emitter current will be 0.02 x 200 = 4mA max.
The transistor is acting as a CURRENT LIMITER.
You will notice the absence of a
current-limiting resistor on the white LEDs.
There are two reasons why this resistor can be eliminated.
The BC547 transistor can only pass about 100mA and three white
LED can easily accept 30mA each. The LEDs need to be tested to
make sure they all have the same "Characteristic Voltage" of
3.3v or 3.5v so they illuminate equally.
Secondly, they are only illuminated for a very short period of
time and your eye extends the effect by a feature called
PERSISTENCE OF VISION. This means they will never get
Even though the transistor is seeing about 15mA into the base,
it will still only allow 100mA collector current to flow and it is
acting as a CURRENT LIMITER for the LEDs.
TRANSISTOR REPLACES A RELAY
You need all the previous
discussions to understand how a transistor can be used to
replace a relay.
A replacement can only be used in some cases.
The main advantage of a relay is its ability to completely
isolate the driving circuit from the load.
That's the main reason why it is used. A transistor does not
provide this isolation.
However a relay has a limited life and if the circuit is
constantly switching on and off, the relay will soon fail. There
are a number of factors that need to be considered before a
replacement can be made.
The first is the way the load is connected. It can be connected "HIGH-SIDE" or "LOW-SIDE."
In other words, the load can be connected between the positive
and relay to create HIGH-SIDE. (the other terminal of the relay
is connected to 0v.)
Or it can be connected between the relay and 0v to create LOW-SIDE.
terminal of the relay is connected to positive rail.)
This will determine how the transistor circuit is designed. The
other important factor is the current-capability of the
transistor. It must be able to handle the current and it must be
driven into saturation so that the least voltage is lost
across the collector-emitter circuit.
Turning the transistor ON and OFF quickly will also reduce the
temperature of the driver transistor.
THE DIODE PUMP (CAPACITOR
Also called a CHARGE PUMP.
And the VOLTAGE
DOUBLER There are
two circuits in this discussion that look very similar but
produce different results. Circuit A is called a DIODE PUMP or CAPACITOR-INPUT PUMP
or CHARGE PUMP and it produces an output voltage that is never
higher than rail voltage. Circuit B is called a VOLTAGE DOUBLER. However it can be
called a CAPACITOR-DIODE CHARGE-PUMP or CHARGE PUMP. It produces
a voltage higher than rail voltage.
Circuit A is basically designed to activate a relay when
the voltage across the 22u rises above 0.7v. Only an AC signal
will pass through the 10u and when the BC547 turns off, the 4k7
pulls the positive lead of the 10u towards the positive
rail. This causes the negative lead to rise too and a voltage
appears at the join of the two diodes. The lower diode has no
effect at the moment but the upper diode passes this voltage to
the 22u and the electrolytic begins to charge. The 10u also
charges. The BC547 is designed to turn on after a short period
of time and it pulls the positive lead of the 10u towards the 0v
rail. The 10u has a small voltage across it at the moment and
the negative lead goes below the 0v rail. As soon as the lead
goes -0.7v, the lower diode starts to conduct and it discharges
the 10u. Only a small voltage of about 0.9v will be left in the
10u and the BC547 is now designed to turn off. The 4k7 pulls the
10u towards the positive rail and the 22u gets a further small
amount of charge. As soon as it reaches 0.7v, the BC337 turns on
and energises the relay. The 2k2 allows the 22u to charge to a
slightly higher voltage than 0.7v so that the relay remains
activated a short period of time after the signal from the BC547
This circuit coverts an AC signal into a DC voltage.
Circuit B is a VOLTAGE DOUBLER. The two circuits appear
to be very similar. The component values are the same. We have
reversed the 10u and placed the lower diode above the other
The operation of the circuit is completely different.
When the BC547 turns ON, the 10u charges via the top diode. When
the transistor turns OFF, the 4k7 pulls the left lead of the 10u
towards the positive rail. This pulls the negative lead up and
it rises above the 12v rail by about 11v. This puts 12v plus 11v
on the left lead of the second diode and it passes this voltage
to the 22u. The 22u charges to about 20v.
A practical version of this circuit is in our 50 - 555 Projects
A 12v battery can be used to charge another 12v battery by using
a voltage doubler circuit and although the full voltage is not
required, the circuit automatically adjusts and charges the
battery. When the battery is removed, the output voltage rises
to about 18-20v. Circuit C is a DIODE PUMP (VOLTAGE DOUBLER)
from an AC source (such as the "Mains").
The circuit takes a number of cycles to get up to full
output voltage and this is how it works:
The input voltage rises and because the 22u is uncharged, the
10u starts to charge as soon as the 0.6v across the top diode is
The 10u charges to about 10v and puts about 5v on the 22u.
When the AC reverses polarity, the top diode does not have any
effect but the lower diode becomes forward biased and it allows
the 10u to charge to about 15v.
When the AC reverses again so that the top input becomes
positive, the 10u already has 15v on it and the AC adds another
This means the positive lead of the 10u is 30v above the lower
rail and it charges the 22u to about 15v.
This happens a few more times and eventually the 22u gets
charged to 30v (minus 2 x 0.6v diode drops).
After 5 or more cycles, the 22u has about 30v across it and the
10u keeps "topping up" the voltage as follows:
Say the 10u has 14v across it, when the top input of the AC
becomes negative, the 10u immediately jumps to a position below
the 0v rail and the diode connected to the 10u allows it to be
charged to 15v, (the top diode effectively comes
"out-of-circuit" as shown in diagram D:
a summary of the features of the three basic ways a transistor
can be connected:
slightly less than 1
You can see the
input impedance of a COMMON BASE stage is equal to the
resistance of the emitter resistor
it is designed to interface (connect) a low impedance device to
the circuit. This is normally very hard to do as the speaker or
inductor may be only 8 to 50 ohms. Trying to connect an 8 ohm
device to the input of a 500R stage produces a lot of mis-match
and produces high losses.
The common base arrangement does this very well and that's why
it is so useful.
The COMMON EMITTER
arrangement needs more explanation as the summary above suggests
the input impedance is about 500R to 2k.
We have already explained (in Fig 11) that a transistor with a gain of 200
will produce a voltage amplification of about 70 in this type of
circuit. The reason is the 2M2 base-bias resistor.
It is acting as a
feedback resistor and is acting AGAINST the incoming signal.
For example, if the incoming signal is rising, the collector
voltage will drop and this will be passed through the base-bias
resistor to deliver less current to the base. This is opposing
the current being delivered via the signal and that's why it is
called NEGATIVE EFFECT or NEGATIVE FEEDBACK. Thus the transistor
cannot produce the output amplitude you are expecting. NOTE:
The value of 500R to 2k2 for the input impedance of a
common-emitter stage is very misleading.
You would think the input is like driving into a 2k2 resistor.
But this is not the case. Placing a 2M2 between the base and 5v
rail will turn the transistor ON fully. The 2uA via the 2M2 is
sufficient to turn the transistor ON. This is obviously nothing
like driving into a 2k2 and you should avoid thinking about the
input as a low value as it only applies when the transistor is
Transistors will respond to a fraction of a microamp and you
should think of them as having very sensitive inputs.
question at the moment is this: What effect does the input
impedance have on an incoming signal?
When a transistor is lightly loaded, as shown in the
circuit on the left, a very small current into the base will
produce an output waveform. It is too complex to talk about
input impedance of 500R to 2k. This value does not help us
explain the operation when lightly loaded.
The easiest way to talk about the input impedance is to discuss
the current entering the base when a signal is applied, by
looking at the "voltage waveform". The voltage (signal) on the
base can be viewed with a CRO (Cathode ray Oscilloscope) and
although we don't know the value of the associated current, the
voltage waveforms though the circuit produce an increase of
about 70 and this is adequate in most cases.
We see the waveform produced by the
electret mic is reduced when it passes through the 100n
This reduction is due to the impedance (resistance) of the 100n
capacitor and also the input impedance of the transistor. Rather
than trying to work out all these complicated values, the
easiest way to create the required amplitude on the output of
the stage is to reduce the 47k. When this value is reduced, the
output of the electret mic increases. Talking Electronics uses
only high quality electret mics and they require a load resistor
of 33k to 68k for 5v supply. Some junk electret mics need 10k or
as low as 2k2 and that's why this resistor can range from 2k2 to
The electret mic will produce an output of about 30mV and the
waveform on the base will be about 20mV. With a gain of 70, the
collector waveform will be 1400mV.
When a transistor is required to pass a high collector current,
the current entering the base is considerably higher than
discussed above. When the collector current is approaching the
maximum for the type of transistor, a transistor with a gain of
200 will not produce this high gain. The gain will be
considerably LOWER. It may be 100 or 70 or even 50. This is why
a high input current is needed and it will appear as though the
transistor has a low-to-medium input impedance.
COLLECTOR stage is also called the EMITTER FOLLOWER
The input is the base and the output is the emitter. The
collector is connected to the supply rail.
This stage is classified as having a HIGH INPUT IMPEDANCE
because the transistor allows you to deliver a high current to
the load by supplying a very small current into the base.
In other words the transistor amplifies your effort by about 100
This is due to the gain of the transistor.
When you are doing this, the load appears to be a much-higher
value of resistance as the transistor multiplies the resistance
of the load by a factor of about 100 times.
That's why the circuit is classified as having a HIGH INPUT
the Common-Collector circuit work?
Firstly we will assume you have a
voltage of 10v on the base and can deliver 3mA into the base.
What is the voltage on the emitter and how does appear on the
The transistor amplifies the 3mA by about 100 times and this
produces 300mA through the collector-emitter junction. This
current flows through the emitter resistor and say it produces
12v across the resistor.
This voltage is higher than the base voltage and this cannot
happen. What happens is the current increases and the voltage
increases on the emitter resistor until the voltage is 0.6v LESS
than the base voltage. At this point the transistor turns off a
small amount and the current reduces so the voltage reaches
EXACTLY 0.6v less than the base. That's how the voltage on the
emitter is 0.6v less than the base. If you raise the voltage on
the base, the emitter voltage will rise too. If the base voltage
is lowered, the emitter voltage will reduce.
The emitter voltage FOLLOWS the base voltage and that's why we
call the circuit EMITTER-FOLLOWER.
A "technical" person in a forum said: "Normally don't use
a bipolar transistor in common collector configuration for
switching loads." This needs some explaining.
If you have an input voltage that rises from 0v to full rail
voltage, you can use an emitter-follower circuit.
An emitter-follower is called an IMPEDANCE MATCHING CIRCUIT.
What this means is the transistor is increasing the resistance
of the load by 100 times and the previous stage thinks it is
delivering to a load that is 100 times higher in resistance.
You could use a common-emitter stage and the result would be the
same when you compare the maximum currents but the
common-emitter stage will turn ON when the voltage on the
previous stage is about 1v and this will cause distortion in an
If you are driving a motor, an emitter-follower stage will allow
you to regulate the speed since the input voltage rises from 0v
to rail voltage and the motor will respond to this.
If you are driving a relay, a common-emitter stage will speed up
the timing because it will react to the voltage when it has
risen about 1v. An emitter follower stage will turn on the
relay when the voltage has risen to about 80% of rail voltage.
COMMON-COLLECTOR stage has a number of problems. The main
one is the voltage dropped across the collector-emitter
terminals of the output transistor.
This transistor is also called an EMITTER-FOLLOWER and it
is always going to have a minimum voltage-drop of 0.7v when the base is at rail
voltage. But in most cases the base will be less than rail
voltage by at least 0.3v. This means the voltage across the
transistor will be 1v.
A transistor in a common-emitter configuration will be
about 0.2v. This means the common-collector arrangement will be
5 times hotter (approx) and the load will get 0.8v less voltage.
The situation gets
worse when a Darlington pair is connected as a common-collector
output. The voltage across the two transistors will be a
minimum of 1.35v and will normally be more than
Figs E and F show some solutions to driving a
load with the most-efficient design. Fig E is called a NPN/PNP pair and Fig F is a
Only the important voltage-drops have been shown and these
values will be higher for some transistors and will also
increase as the current increases.
This is just an example of the minimum values to expect.
THE TRANSISTOR AS A VARIABLE RESISTOR
Many circuits use
a transistor as a variable resistor but this fact is never
There are many ways to look at how a transistor is operating and
one of them is to see the transistor as a VARIABLE RESISTOR.
In fact, that is what the transistor is doing in 99% of
circuits. It is acting as a VARIABLE RESISTOR.
In a digital circuit, the transistor is turning OFF completely
then turning ON fully. This is equivalent to a very high-value
resistor in the first instance, then a very low-value resistor.
In an audio circuit, the resistance of the transistor is
reducing then increasing. It is working in series with a LOAD
resistor and the voltage at the join of these two is the OUTPUT
of the stage.
For a common-emitter stage, when the resistance of the
transistor decreases, the output is LOW (meaning the output
voltage is small) and when it increases,
the output is HIGH.
The transistor and the load resistor form a VOLTAGE DIVIDER and
the voltage at the join of these two components depends on the
value of the transistor.
When the resistance of the transistor is equal to the load
resistor, the voltage at the join will be 50% of the supply.
When the resistance of the transistor is twice, the
voltage will be 66% and when it is half, the voltage will be
The transistor as a
THE LIGHT DEPENDENT RESISTOR The Light Dependent resistor
(LDR) is a variable resistor. Its resistance varies according to the
amount of light it receives.
It's a very simple component and can be connected directly to a
transistor so the change-in-resistance of the LDR can be amplified about
Even though this arrangement is very simple, it takes a lot of
understanding to design a suitable circuit and see HOW IT WORKS.
An LDR has a resistance of about 300k when in total darkness and as low
as 200 ohms in bright light.
But it is very difficult to turn a light ON and OFF to get this extreme
range in resistance.
In most cases the change will be a LOT LESS.
The actual value of resistance will depend on the type of LDR and the
There are 4 ways to connect an LDR to a transistor:
How do you
connect an LDR? Circuit A
only works when the light drops to zero. Circuit B
produces a LOW on the collector when
light is detected.
produces a HIGH on the collector
when light is detected.
Circuit D detects a CHANGE in light
There are TWO WAYS to see how an LDR works.
1. It delivers a CURRENT according to its resistance.
2. It it part of a VOLTAGE DIVIDER when connected in series
with another resistor.
In the first circuit, Circuit A,
the LDR acts as a current-limiting device - or more-accurately A
CURRENT-DELIVERING DEVICE and the output will be HIGH when the
LDR does not receive any illumination, providing the LOAD
resistor is a small value.
This is the reason: The resistance of the LDR will be 300k and
the transistor will effective reduce this value to 1k and
produce a voltage divider of 1k:load resistor. If the load
resistor is 470R, the voltage on the output will be 66% of rail
When light falls on the LDR, it resistance decreases and if it
reduces to say 5k, the transistor will convert this to 5,000/200
= 25 ohms and the output of the circuit will go LOW.
If the LOAD resistor is 10k, the LDR in dark conditions will
provide a resistance of 300k and the transistor will convert
this to 1k. The ratio of 1k:10k will mean the output of the
stage is LOW for both dark and light conditions.
Thus the resistance of the LOAD resistor must be selected for
the light conditions experienced by the LDR and the gain of the
transistor. Circuit B effectively makes the stage less sensitive to
changes in illumination on the LDR.
The base resistor allows some of the current supplied by the LDR
to flow to the 0v rail. This means the transistor gets only some
of the current and more light has to shine on the LDR for the
circuit to change states.
The actual current "bled-away" by the base resistor is very
complex to determine and requires mathematics beyond the scope
of this discussion.
However we can say the base resistor is normally between 100k
and 1M and the LOAD resistor between 1k and 10k. In this
circuit, the output goes LOW when the LDR receives illumination.
In Circuit C, the output goes HIGH when the LDR detects
The base-bias resistor is normally between 2k2 and 10k and the
LOAD resistor between 1k and 10k.
This gives an enormous range of operating parameters (current
taken by the circuit) and we can simplify the discussion by
The base-bias resistor forms a voltage divider with the LDR and
when the voltage on the base is 0.55v, the transistor starts to
turn OFF. At 0.45v the transistor is OFF. Light must be directed
onto the surface of the LDR and it will reduce the effective
resistance on its terminals. The intensity of the light to turn
the transistor OFF will depend on the value of the base-bias
In Circuit D, the output will change when a hand is waved
over the LDR. This causes the resistance of the LDR to increase
The voltage at the junction of the LDR and the resistor to 0v
will fall the rise again.
The transistor can be biased ON slightly or OFF via the two
resistors on the base and when the voltage on the "output of the
LDR" changes, the change is passed through the capacitor to
change the state of the transistor.
HIGH IMPEDANCE If you want to go into
a more-complex discussion, here it is:
The LDR needs to feed into a HIGH IMPEDANCE stage because it is
classified as a high impedance device.
It is only capable of passing (delivering) a small current,
that's why it is called a HIGH IMPEDANCE component.
The transistor is called a BUFFER (or amplifier) as it
amplifies the current.
The transistor is a BUFFER because it "joins" the low-current
capability of the LED to the high-current requirement of a LED -
just like the buffer at the end of a carriage on a train or the
old-fashioned bumper on a car (containing springs).
A HIGH IMPEDANCE stage accepts very small currents and amplifies
If the LDR is connected to a low-impedance stage, the current it
is capable of delivering will enter the transistor and the
collector voltage will change VERY LITTLE. This is because the
transistor needs a higher input current. The LDR is not capable
of delivering this high current and the output of the circuit
will NOT CHANGE or only change a VERY SMALL amount. A
low-impedance stage has say 2k2 resistor on the base and 100R on
We use the word "IMPEDANCE" because if we measure the values
with a multimeter, the "resistance" of the junction of the
transistor is also measured and since this is a diode, its
"resistance" changes according to the current flowing and is not
a fixed value. Whenever a capacitor, diode or coil is included
in the circuit we are measuring, we call the value an impedance
to show that we are taking into account the inclusion of the
SHORT CIRCUIT CURRENT / shoot-through current
Here is one of the
major hidden problems when designing a circuit.
It's called a SHORT CIRCUIT current or "Shoot-through"
current and occurs when two
transistors are directly connected together as shown in the
output section of the circuit on the left.
The chip can deliver about 10mA to the base of the top
transistor and if the transistor has a gain of 100, the
collector-emitter current will be 1AMP !!
This current will
also flow through the lower transistor and is WASTED CURRENT. It
creates an almost SHORT-CIRCUIT across the supply rails.
The solution is to add a low value resistor such as 100R to
limit the current.
NO CURRENT . . .
No ON-OFF switch needed
In all the circuits we have
discussed, there is a small quiescent current flowing when the
circuit is "sitting around" doing nothing - called the
"Quiescent Current." This current is the
biasing current and is needed to turn the transistor(s) ON slightly
so it will amplify the smallest signal. If the transistor
is not turned on slightly, it will not amplify signals
smaller than 600mV as the base must see a signal more than 550mV
for the transistor to start to turn on.
The circuit above is a LOGIC PROBE and will detect a HIGH or LOW
on a digital circuit where the waveform is higher than 2.5v.
This circuit can be in a state of "cut-off" (not conducting)
and the input amplitude will turn the circuit ON.
The clever feature of this circuit is the lack of an ON-OFF
switch. The circuit takes no current when not detecting a
waveform because the voltage-drops across the semiconductor
junctions is greater than 3v (and the supply is 3v).
To turn on a transistor, the voltage between the base and
emitter must be greater than 550mV. But if the voltage is less
than 550mV, absolutely no current flows through the junction.
The same applies with the characteristic voltage of a LED. For a
red LED, the characteristic voltage is 1.7v but if the voltage
is less than about 1.5v, no current will flow. The same applies
to a green LED. Its operating characteristic voltage is 2.1v to
2.3v but when the voltage less than 1.9v, no current will flow.
The current-path for the circuit in "idle mode" is through the
green LED (1.9v), 100R, the emitter-base junction of the BC557
(0.55v), through two 10k resistors, the base-emitter junction of
the BC547 (0.55v), the red LED (1.5v)100R to the 0v rail.
When the minimum voltages are added we get 4.5v. But the supply
is only 3v and thus it is not high enough to meet all the
minimum junction voltages.
Thus no current will flow through the circuit when "sitting
with HIGH, LOW and PULSE
The same "cut-off" condition applies to the Logic Probe
circuit above. But this time an orange LED is added to the
The Pulse LED is connected to the HIGH section of the circuit
and its illumination is extended by the inclusion of the 2u2
electrolytic so a brief pulse can be
To extend the illumination of the orange LED, we need to charge
But an electrolytic needs a lot of current to charge it quickly
and the input of a Logic Probe is HIGH IMPEDANCE - in other
words it has no "charging capability."
This means the pulse from the input needs a lot of assistance to
charge the electrolytic.
To do this we have used 2 transistors. The first and third
transistors form a SUPER-ALPHA pair and this provides a lot of
"strength" (current) to charge the electro.
But we can only charge a very small-value electro, so we need
another transistor to ELECTRONICALLY increase the value of the
electro about 100 times. That's the purpose of transistor four.
We now have the circuit we need to extend the illumination of
the pulse LED.
But when all these components are connected, we have run out of
voltage to illuminate the LED with a HIGH of 3v.
To solve this problem we have added a single 1.5v cell as shown
in the diagram. The emitter of the 4th transistor will have
about 1v on it (with reference to the 0v rail) and the lower
lead of the 100R will have minus 1.5v (with reference tot he 0v
rail). This means the orange LED and 100R will see a voltage of
2.5v. This is sufficient to illuminate the LED.
negative voltage can be produced from a positive voltage with a
circuit called a DIODE PUMP or DIODE CHARGE PUMP. In its
simplest form it is called a VOLTAGE INVERTER.
To create the negative we use the circuit on the left, as shown
in fig A.
The transistor is fed a signal. It can be a square-wave,
sinewave or audio waveform. All it has to do is turn the
transistor ON and OFF and the voltage on the collector will
range from nearly 0v to nearly rail voltage.
This waveform is passed to the voltage-doubling circuit via a
capacitor. Let's see how the two capacitors and two diodes
produce a negative voltage.
The cycle starts with circuit B.
The transistor is OFF and capacitor C1 is connected to the
positive rail via the resistor R. It charges to about 10v
via the resistor and diode D1(actual voltage will be about 9v).
When the transistor turns ON, the left lead of the capacitor
will drop to the 0v rail and the right lead will drop 10v BELOW
the 0v rail.
The diode will "flip over" but it will not conduct when the
anode lead is negative. This is the most important part of the discussion.
The right lead of the capacitor produces the NEGATIVE VOLTAGE.
This is shown in circuit C.
In circuit D we connect D2
and the cathode of D2 is negative. This will make D2 conduct and
the anode of D2 will also be negative by about 8v to 9v and this
will pull the lower lead of C2 DOWN by a voltage of about 8v to
9v and create the negative output of the circuit.
We have shown the output to be 10v, but each diode will lose
(drop) about 0.6v and the capacitor will not charge fully so the
output will be about 8v.
When the transistor turns OFF, the left lead of C1 will rise to
about 10v and because the capacitor is fully charged, it will
not charge any more. It has done its job.
However, if the output of the supply delivers a current, the
output voltage will fall and the capacitor will not have 10v
across it. In this case is will charge a small amount and
restore the output to -10v.
circuits produce a negative voltage or negative spike at
some point in the
circuit. In other words the voltage will be LESS than
the 0v rail of the circuit. The circuit above is a good
example and this negative voltage is stored in the
But some circuits do not store the effect and and the
generation of the negative voltage is not realised.
That's why you have to know how a negative voltage can
be generated, so you can be aware of its generation.
It is due to the presence of a capacitor and the
animation below shows how a capacitor can produce a negative
When a charged capacitor is lowered
from a "high" position in a circuit, the positive lead may be
lowered by say 3v. This means the negative lead will be
lowered by 3v. (we are assuming the capacitor can be
lowered and is not directly connected to the 0v rail).
The result is a voltage on the negative lead that is
"less than" the 0v rail.
Generally, this "spike" or voltage is not obvious and
the voltage "stored in the capacitor" does not "go
negative." That's why this effect may be missed in a
You can see the electrolytic produces a NEGATIVE VOLTAGE
on the base of a transistor in the following animation, when the
transistors change state:
Many transistors don't like a negative voltage on the
base and they can be destroyed due to this effect if the
supply voltage is vey high. And you will wonder why the
transistors are being destroyed !!
We are not going into designing
a power supply but want to cover one of the mayor misconceptions
about a power supply:
A lot of readers to a Forum have asked: I want a 20mA power
supply, but all I can find is a 100mA supply. Will this BURN-OUT
Answer: If the project is correctly designed and requires
20mA when connected to a 12v power supply, it will take exactly
20mA and not burn out.
If the project requires 100mA, it can be connected to a 100mA
power supply and it will work correctly.
If the project requires 200mA and is connected to a 100mA power
supply, the power supply will deliver 100mA (and maybe a little
more) but the voltage will start to fall when the full 200mA is
required and the power supply will get very hot. The
project will not be damaged but it will not perform to its full
See 200 Transistor Circuits:
1-100101 - 200
for projects on Power Supplies.
Let's go over a number of
design-features we have mentioned in this discussion.
The three diagrams on the left show a self-biased transistor
with approx half-rail voltage on the collector.
What is the difference between circuits A,
We have shown that
passing the energy (the size of the signal) from one stage to
the next, depends on the value of the load resistor. A low-value
resistor will pass more energy. Thus circuit
A will pass more energy.
We have also shown the value of the coupling capacitor (this is
the same in all circuits above) and the value of the base-bias
resistor determines the amount of energy that will be
Circuit C will have the
least effect on attenuating the signal because the transistor is
turned on via the smallest amount of base current (the
highest-value base-bias resistor) and thus the input impedance
is the highest.
The third factor to consider (for battery operated devices) is
quiescent current. Circuit C
takes the least current.
The three circuits above will have the same gain but when they
are connected to a following stage, circuit
A will deliver more
signal-amplitude and the amplitude of circuit
C may drop to almost zero
because the input impedance of the following stage will be so
low that the energy from the 47k will produce a very small
waveform under "loaded conditions."
Now you know what causes the problem.
If an electrolytic is
connected across the emitter resistor, the gain of the stage can
be as high as 100 but it is best to keep each stage below 100
and use more stages. This will prevent circuit-instability such
as "motor-boating" and "squealing."
There are basically two ways to
bias a transistor. Circuit
A is called SELF-BIAS and circuit
is a BRIDGE.
We are assuming all resistors are correct to get half-rail
voltage on the collector.
Circuit A will maintain
mid-rail-voltage over a wide range of supply voltages without
having to change any of the resistor values. Changing the
transistor for one with a different gain will alter the
collector voltage by a small amount. It is a very reliable and
The only problem is the gain of the stage. It is about 70 for
all types of transistors because the gain is reduced by the
feedback effect of the base-bias resistor.
The collector voltage for circuit B
will change considerably when the supply voltage is changed and
if the transistor is changed for one with a different gain.
This is one of the most important topics when designing a
circuit. It has never been mentioned before in any text
When designing a circuit, each value of resistance and
capacitance has a "correct value." In other words, what me
mean is this: When a "qualified" electronics engineer sees a
circuit, he expects to see a certain value of resistance or
capacitance for every component and when this value matches
his understanding, he understands the circuit is
operating exactly as expected.
If the value is 50% higher or lower, it can still be within
the expected range, but when the value is 10%, he will look
into the reason for the variation.
Most circuits can tolerate a wide range of values, however
certain values are chosen to indicate the fact that the
value is non-critical.
We call these values "design values" and are the first
values you think of when designing a circuit.
If a circuit needs a slight adjustment, these values will be
increased or decreased slightly and and when an engineer see
this, he understands the values are fairly critical.
That's why you don't use 12k instead of 10k or 560R instead
of 470R when the "design values" will work perfectly.
You are simply introducing unnecessary complexities into the
circuit and reducing your status as a "design engineer."
We are still seeing engineers creating a circuit with a
parts list and when the circuit is displayed on Google, it
has no values!
Finally, remember to lay-out a circuit using symbols and not
line-diagrams of a chip with pin numbering 1,2,3,4 etc. A
schematic has nothing to do with parts-placement and a block
diagram of a chip tells you nothing about the function of
SPLIT POWER SUPPLY
need a voltage that is identified as 0v voltage at a
particular connection and a positive voltage and a negative
The reason for this is the output of the circuit has a
voltage that is half-way between the voltage on the top
rails and the bottom rail and it produces a signal that
rises almost as high at the top rail and almost as low as
the bottom rail.
This means the amplifying circuit, such as an op-amp or
AMPLIFIER, requires a voltage that is equally POSITIVE and
NEGATIVE, with the earth, or chassis or neutral classified
as having zero voltage. In other words it is the reference
point (0v) for all the other voltages.
As far as the
output is concerned, the two 470u electrolytics are
connected in PARALLEL and prevent the bottom lead of the
speaker moving, when a signal is delivered to the top lead.
The do not prevent the bottom lead moving like a fixed
connection but limit the movement or restrict the movement
just like a 2 ohm resistor.
A 100Hz, the two 470u electrolytics have a resistance
(impedance) of slightly less than 2 ohms and at a higher
frequency this value is a lot less. That means most of the
signal will appear across the speaker.
How A Capacitor Works we explain how you "see" the
Here's a new
It is INDUCTION
and it means a circuit or transistor is turned ON or OFF by
a voltage produced by a field. It means there is no direct
wiring between the two circuits and and they are linked by a
magnetic field or flux.
This is sometimes called TRANSFORMER ACTION.
The field from one winding INDUCES a field near
another winding and this field produces a voltage and
current to operate or drive of affect the other circuit.
This effect is identified by the fact that the two windings
DO NOT TOUCH. The only thing that couples them is the
In the circuit above the transistor oscillates and
produces a very high voltage at the the top of the
The centre of the coil is a magnetic material such
as ferrite. This is important
as it delivers the flux produced by the 3-turn
winding to the 275 turns of the transformer.
Removing the ferrite core will leave air to do the
transfer and air is not as good as ferrite as
transferring the magnetic flux. Ferrite is up to
1,000 times better.
The circuit is turned on a little bit by the 22k on
the base and this produces a small amount of flux in
the 3 turns.
This flux cuts the 275 turns and produces a voltage.
Because the 275 is only connected at one end, we
have to explain how it works.
The voltage produced by the 275 turns comes out both
ends of the winding and it will be a negative
voltage out one end and a positive voltage out the
The winding is connected so the positive voltage
comes out the bottom and even though the top of the
winding is not connected to anything, the voltage
out the bottom will deliver a small current to the
base of the transistor and turn it ON more.
This increases the magnetic flux and the transistor
turns on until it is fully
Up to this point in time the magnetic flux is called
EXPANDING FLUX and it is capable of cutting the 275
turns. But when the transistor is fully turned ON,
the flux is a maximum but it is not expanding. It
stops EXPANDING. It is STATIONARY FLUX and the
voltage produced in the 275 turns immediately STOPS.
The transistor turns OFF immediately because the
voltage (and current from the 275 turns stops being
delivered to the base) and the flux in the magnetic
core collapses. It collapses because the current
delivered by the transistor STOPS.
The collapsing of the magnetic flux produces a
voltage in the 275 IN THE OPPOSITE DIRECTION and the
voltage out of the bottom of the winding is
NEGATIVE. This turns the transistor off fully and a
very high voltage is produced in the winding because
the magnetic flux collapses very quickly.
This is the spark or CORONA or HIGH VOLTAGE produced
by the "Tesla Coil."
When the magnetic flux has full collapsed (and
converted to a very high voltage), the voltage ou of
the bottom of the 275 turns is zero and the
transistor gets turned on a small amount by the 22k
to start the next cycle.
Because the 275 turns is only connected at one end,
the energy into the base can only be very small
because the winding does not have at connection at
the top to act as a "fixed point" to deliver a high
It's like trying to push a car while standing on a
skateboard. The voltage has a "little bit of grip"
as it expands and this is just enough to get the
circuit to work.
In actual fact, the voltage out to bottom of the
winding is just as high as the top spark, but the
base of the transistor converts this energy to a
small voltage and small current to turn the
This gives the winding a "fixed point" to produce a
spark at the top.
The high voltage is INDUCED or PRODUCED in the 275
turns when magnetic flux in the core collapses very
Decoupling is done with capacitors.
To understand how a capacitor works, read this article:
Decoupling is soldering capacitors or
electrolytics to a circuit to prevent signals from one stage
entering another and creating a problem. And there are other
explanations too. There are quite a number of words and terms
that are used for capacitors in a certain location in a circuit.
The problem might be squealing, motor-boating or causing a stage
or an output to pulse or operate at the wrong time. Or another
10 problems. We will only cover a few.
The capacitor (electrolytic) absorbs the unwanted signal (by
charging and discharging) and this is how it is removed. You can
also say the capacitor prevents the voltage rising or falling
because the voltage is actually a very narrow pulse of high
The capacitor converts this waveform into a low voltage - by
simply absorbing the energy contained in the waveform. What the
capacitor is actually doing is absorbing the energy and
releasing it during another part of the cycle.
Whenever a device such as a relay or speaker is activated, it
takes additional current and the voltage of the supply drops a
This "dip" is transferred to the previous stages (or
other stages) and these
stages can be sensitive to very small changes in voltage.
In the circuit below, the first stage thinks the noise is due to the
signal from the radio station and it gets amplified along with the real signal the
The signal eventually appears in the speaker and is then
transferred to the supply rail as a "glitch" or "spike"
where it is again passed to the "front-end."
Very soon the signal is amplified sufficiently to produce a
squeal or "putt-putt-putt" noise called "motor-boating."
Audio amplifiers and radios are particularly difficult to design
because you need an enormous amount of amplification and this
allows feedback in the form of squealing to be produced.
The following circuit has this problem and the cure was to add a
22n to the output stage to prevent the high frequency signals
being amplified and cycled through the circuit.
The layers of aluminium foil between cardboard form a tuning
capacitor to select the stations.
The unwanted signal
starts at the output because a loud signal takes more current
and this causes the supply voltage to drop. It puts a pulse on
the top rail.
This pulse is passed to the first transistor where it is
amplified about 50 times.
The second transistor amplifies it another 50 times and it
cycles around the circuit again and again and again - each time
getting louder and louder.
The 22n absorbs the signal and prevents most of the squeal.
This is an unusual form of decoupling and it is classified as
DECOUPLING because the intention of the 22n is not to alter the
audio but remove the squeal.
simply means the decoupling component(s) are near the battery.
In the following circuit the 22n is across the wires that come
from the battery.
The circuit above is
called a high frequency circuit (90MHz) and the wires and leads
and tracks on a printed circuit board from the battery to the
circuit are 100 times more critical than an audio circuit.
Let me explain why.
The coil in the diagram is wound with wire but it can be created
by using a zig-zag type track on the PC board or a circular
track like a snail-shell. It only has to be 5 turns and if one
end is held stationary, the other end will have a signal equal
to about 6v on it.
This shows what can happen with a short length of track.
Where the coil meets the top rail in the circuit above, the join
is trying to move up and down at 90MHz and if the track between
this junction and the battery is long and thin, it will become
part of the coil and start to move.
Two things will happen. It will change the frequency of the
circuit and, because it does not have a 39p capacitor across it, the
energy delivered to it will be lost.
The 22n effectively tightens up the power rails and prevents
We also say the 22n reduces the impedance of the rails.
improved enormously by adding a resistor in series with the
In the hearing-aid circuit above, the second 100u across the
battery is designed to improve the performance of the battery
when it is getting weak and at the end of its life.
The 100u is providing the pulses of high current when the
battery is weak. It also reduces the amplitude of the pulses on
the power-rail, as these pulses will be sent back to the
front-end and cause motor-boating.
This type of amplifier has a very high gain and even the
smallest signal from the output will cause instability.
To reduce this a second decoupling arrangement is provided for
the microphone and first transistor.
It consists of a 47u and 330R resistor.
The resistor increases the effectiveness of the 47u by about 100
It works like this: The resistor and capacitor in series
is called a TIME-DELAY circuit. When a peak or pulse appears
at the right-hand lead of the 330R, it passes through the
resistor INSTANTLY and begins to charge the 47u. Suppose the
pulse is 10mV. After one-hundredth of a second the voltage on
the capacitor will be 3mV. This means 100Hz signals will
be reduced to 30% and any higher frequencies will be reduced to
10% or less. When the current is very small, as in this case,
the effect of the two components is to reduce the signals that
create a squeal by as much as 99%. They
are highly effective. The squeal is completely eliminated.
The two capacitors (electrolytics) and resistor
form a pi filter. This type of circuit is called a low-pass
filter, but we are using it to remove high frequencies
and we are considering it as a "high-frequency eliminator."
The 330R and 47u are enormously effective in doing this as the
calculation for the 330R and 47u show that signals above 10Hz
are reduced or eliminated completely. Feedback squeals are 3kHz
and above. These signals are completely eliminated.
DECOUPLING A 7805 REGULATOR
This circuit shows two things.
The 100n on the input and output have nothing to do with
filtering or smoothing the voltage.
They are designed to prevent high-frequency oscillations (approx
1MHz) being generated by the 3-terminal regulator.
The regulator is actually a high-gain amplifier and it will
produce oscillations if the 100n capacitors are not fitted.
Secondly, the diagram shows that the capacitors should be placed
very close to the common lead of the regulator by showing the
leads going diagonally to the 0v point. This must be
done so they have the greatest effect on eliminating the
These capacitors are called DECOUPLING CAPACITORS
because, in electronic terminology, when you isolate a component
from a source of interference, it is called decoupling.
Even though the 100u is a much-higher value, it has almost no
effect on reducing the self-oscillation.
We mentioned above, the effect of a PC board track on 90MHz
If the distance between the waveform and the capacitor is long,
a signal will be generated by the track.
This time the track is inside the electrolytic. The electrolytic
is made up of a spiral layer of foil and a high-frequency signal
on the positive lead will not be reduced.
Tantalum capacitors are constructed differently to aluminium
capacitors and WILL have an effect on reducing the oscillations.
Another form of Decoupling is BYPASSING. In the following
circuit the capacitor is allowing the signal to bypass the
emitter resistor and it does this by effectively making the top
of the electrolytic equal to the 0v rail. It is does this
because the electrolytic takes time to charge and discharge and
the emitter thinks it is connected to the 0v rail. It holds the
emitter lead tight and prevents it rising and falling.
The Bypass Capacitor is called and EMITTER BYPASS CAPACITOR
100n BYPASS CAPACITOR
The 100n capacitor connected to
the positive and 0v pins of the chip is called a BYPASS
It prevents noise on the power rail getting into the chip it is
Some projects may have 2, 10 or 50 capacitors that are designed
to prevent the rails "moving" and prevent spikes and glitches
passing from one section to another.
These capacitors are all part of "DECOUPLING" and it
is a very complex area to discuss as we are talking about very
small, high-speed, signals that can create an enormous number of faults.
This is just about the last topic we will be
covering, but it is one of the most important characteristics to
consider because it is a hidden problem.
It is hidden because a transistor has a number of different ways in which
it can fail due to excess voltage. This failure can be
CATASTROPHIC and the transistor will be damaged or the failure
can be temporary and the transistor will survive when the
voltage is removed.
This characteristic can be used to produce a circuit that
oscillates. But first we will explain BREAKDOWN:
There are two different
types of voltage breakdowns in semiconductor devices and
the first is called ZENER BREAKDOWN.
The other is called JUNCTION BREAKDOWN.
the diagram, the zener is a 25v zener and the
transistor is rated at 25v between the collector-emitter
terminals. It is classified as a "25v transistor."
As the supply voltage is increased, both devices will
"breakdown" at 25v. (or just above 25v), so to explain
the result we will increase the supply to 30v. Both
components have "broken down" and the zener will have
exactly 25v across it and 5v will be across the load
The transistor will "break down" and zero volts
will be across the collector-emitter leads and 30v will
be across the load resistor.
If we did not include a LOAD resistor, both devices
would be permanently damaged because the current-flow
through each device would be so large that they would
get hot and EXPLODE. With a load resistor
included, both devices will not be damaged if only a few
milliamps flows through the load resistor. (it is said:
they will RECOVER.) The LOAD resistor is also called a
Here is the zener diode used to
increase the point at which the transistor gets turned ON.
The voltage on the electro needs to rise to 5v1 plus 0.65v for
the base-emitter voltage and then the transistor will respond.
Here's how to view the action of the zener:
As the voltage on the electro rises from zero volts, the zener
can be considered to be removed from the circuit. Then, when the
voltage on the electro reaches 5.1volts, the zener immediately
sits in the circuit. Now, as the voltage rises above 5.1v, the
rise is immediately transferred to the other end of the zener
and appears on the base of the transistor.
This circuit shows the zener as a VOLTAGE LIMITING device. A
555 IC can only work to 18v and a 15v AC supply can produce
peaks as high as 23v.
The 15v zener will limit the peaks to 15v however the excess
voltage will appear across the 100R and we need to work out the
heat generated in this resistor.
The voltage across the resistor will be 23 - 15 = 8v.
The current will be V/R = 8/100 (amps) = 0.08 = 80mA. The heat
(wattage) generated in the resistor will be I2R. But
the voltage is only entering the
circuit on every half cycle, so the current will be 40mA. The wattage will be
0.04 x 0.04 x 100 (watts) = 160mW. A 0.25watt resistor
will be perfect.
The wattage lost in the 15v zener will be 40mA x 15v = 600mW.
A 1watt zener will be perfect. This circuit will deliver
transistor can be placed in a circuit "around the wrong way" and
the a junction inside the transistor will breakdown at a fairly
You can try all the different combinations and use almost any
type of NPN transistor to see the result.
The electrolytic will charge and when the voltage across the
junction of the transistor reaches a voltage between 7v and up
to about 18v, the junction will breakdown and pass a fairly high
current. This will allow the energy in the electrolytic to
illuminate the LED and when the voltage across the electro drops
to 7v or so, the junction will suddenly recover and cease to
allow current to flow.
This will allow the electro to charge again via the 1k resistor and
repeat the cycle of flashing the LED. There are some transistors
(Uni-Junction Transistors) that use this feature to produce
Here is a circuit using the breakdown of a transistor to produce
a frequency to drive the BOOST circuit.
The only transistor tried in the circuit was BC547, so try all
Some will not breakdown at less than 18v and obviously will not
work in this circuit.
base-emitter junction can also be used but the voltage at which
the junction "breaks-down" and the voltage at which it does not
conduct is very small.
The difference between these two values can be as small as a few
millivolts to about 200mV.
We can use this feature in the circuit on the left. It produces
voltage between emitter and base and this needs to be up to
above 7v for some transistors to start the process of "breaking
This process changes the voltage between emitter and base and
thus the flow of current changes. This effect is amplified by
the second transistor.
The result is NOISE and because this noise is random, we call it
See if you understand what we have covered above.
In the two diagrams on the left, the supply increases from 0v to
What happens to the brightness of the globe?
In the top diagram, the zener will start to pass current when
the supply is 12v and when it is 20v, the globe will have 8v
In the lower diagram, the transistor will break-down when the
supply is 12v and the globe will immediately glow very bright. It
will get even brighter as the voltage increases to 20v. It may
The zener diode has been fitted the "other way around." What
The zener diode will act just like a normal power diode and drop
The lamp will illuminate from zero to full brightness.
The zener diode will become a power diode with the reverse
voltage characteristic of 12v, whereas a normal power diode will
have a reverse voltage characteristic of 100v to 400v.
used these characteristics in a number different circuits.
breakdown of a transistor in a photo-flash circuit to trigger
The breakdown of a 1N4148 diode (130v) in a telephone ring
Using a transistor (with a high voltage) in a circuit that
showed problems due to breakdown when using a low voltage
4. Using 3 zeners in series to get 33v. A single 33v zener only
dissipates 250mW. But 3 zeners increases the dissipation to over
The wattage of a zener diode can be 300mW, 500mW or 1 watt and
these diodes are very small.
To make sure they do not overheat, the tracks and lands around
the diode MUST be as large as possible as this will remove a
large portion of the heat. You will be amazed at the
effectiveness of a PCB track.
A power supply came to us with two 330R resistors in series.
They were metal film and rated at 1 watt. One resistor was
getting hot and "burning out" while the other had no problems.
Since they were in series, they were taking exactly the current.
It took us a long time to find the fault.
The tracks and lands to the resistor were smaller than the
others. This slight difference overheated the resistor. The
tracks take away an enormous amount of heat. This was
corrected with the next run of PC boards.
CIRCUIT 1 The input to a microcontroller needs a HIGH when a
microphone picks up audio. This is the requirement from a
customer. The circuit in Fig 104 was designed to meet the
customers requirements. The 10mV audio waveform from a
microphone is converted to a 4v-5v CONSTANT HIGH. The following
circuit is the result:
The starting point is to bias the first
transistor so the voltage on the base is just at the point
of turning it ON.
This allows the 47k resistor to turn on the second transistor
and the diode does not see any voltage. This means the 1u does
not get charged and the input to the microcontroller sees a LOW.
This is called the QUIESCENT (standing, stand-by or idle) condition.
The gain of the electret microphone is adjusted by the 10k pot
and when it receives a loud audio signal it produces an output
of about 20mV.
This signal is sufficient to turn ON the first transistor and
turn OFF the second transistor so that signal diode sees a HIGH
pulse via the 4k7.
This voltage is passed to the 1u and it gradually gets charged.
When the voltage on the 1u reaches about 4-5v, the
microcontroller sees a HIGH and the program in the micro produces an output.
2 How does this amplifier get biased?:
One of the most difficult amplifiers to design
and service is a DC (Directly-Coupled) amplifier. The voltage
on the output is fed back to the input to create the idle
(quiescent) state and the biasing of the input is created from
the output. So, where do you start?
All the facts we have learnt in this discussion are needed to
understand how this circuit works.
The circuit has high gain and without the 22k feedback, we would
not be able to create an output "set-point." The first
transistor has no DC voltage gain as but it does have an AC
voltage gain of about 22. The BC557 provides a voltage gain of
about 70-100 and the output transistors only provide a current
gain. This gives the circuit a voltage gain of about 2,000. A
50mV input will produce an output of about 10v.
The aim is to get the output voltage near to mid-rail so it can
swing both positive and negative and create a relatively distortion-less waveform.
The starting point is the voltage divider made up of the 27k +
27k and 100k. This puts 10v on the base of the first transistor.
Now we come to the 470R resistor on the base of the BD140
transistor. This resistor is a very low value and is keeping the
BD140 turned ON and the emitter voltage will be very small.
Here's the interesting part: The collector of the BC557 can pull
the base of the BD140
UP without any difficulty to
less than the positive rail, due to the two 1N4148 diodes.
The Two Biasing Diodes
These two diodes prevent both output transistors turning ON at
the same time. If the transistors are both turned ON at any
point in the cycle, a very high current will flow and create a
How do the diodes work?
Let's remove the diodes and see what happens.
due to the base-emitter voltage-drops across the two output
transistors. But this only raises the collector
To be able to pull higher, the transistor must turn on harder and
since the bottom transistor is being pulled down by 470R, the
top transistor is also being pulled down via the two 1R
resistors. The BC557 sees the base of the BD139 as a 470R
resistor, plus the actual 470R resistor. This make it 220R.
To raise the voltage on the base of the BD140, requires current
through the 470R and the BC557 needs to be turned on a certain
amount to provide current through the 470R and into the base of
the BD139 AT THE SAME TIME.
At the moment the join of the two one-ohm resistors has a very
low voltage on it and the BC547 is also an emitter-follower and the emitter is
about 10v minus 0.7v.
This puts a current through the 22k resistor of less than 1mA
however this current also flows through the emitter-base
junction of the BC557 and if the transistor has a gain of 100,
the emitter-collector current can be as high as 100mA.
However the 220R (470R and 470R in parallel) resistor only needs
a flow of 22mA to create a voltage of 5v across it, so we have
plenty of gain to begin to turn on the circuit.
The BC557 creates a current-flow through the 470R and the BD140 starts to get
pulled UP. This puts less current though the BC547 and less
current through the base of the BC557, so the BC557 starts to
The actual settling-point has a lot to do with the 27k + 27k and
100k base-bias resistors as this puts 10v on the base and the
emitter 9.3v. Suppose the output settles at 7.5v.
This puts 1.8v across the 22k and creates a current-flow through
this resistor. Approximately the same current flows through the
emitter-base of the BC557 and about 100 times this current is
available to be divided between the 470R and base of the BD139.
This is how the output becomes
biased at very nearly half-rail voltage.
Select the best circuit
between Figs 106 and 107:
From the theory discussed above, can you see the
problem with driving the BC237 in Fig106.
It is being pulled HIGH via the 1k resistor. If the transistor
has a gain of 100, Q4 will effectively be equal to a 10 ohm
resistor. For 100mA current delivered to the output, 1v
will be dropped across this transistor and it will start to get
hot. This is wasted energy.
A BC237is only capable of delivering 100mA.
Fig 106 has been re-drawn as Fig 107 with improvements
The output transistor has been changed to a BC327. It will handle 800mA.
A 1N4001 is not a high-speed diode and using an Ultra Fast 4004
will deliver an extra 50mA to the output.
Transistor Circuits for details.
Fig107a shows a 560R
resistor to discharge the 47p coupling-capacitor.
The circuit is a 27MHz transmitter with buffer. The buffer is an
amplifying stage to increase the output.
You will notice two things: the buffer stage is not biased ON
and a low value resistor is connected between base and 0v rail.
This called a "Class-C" stage.
This resistor discharges the capacitor so it will transfer the
maximum amount of energy (on each cycle), from the oscillator
stage to the output stage.
The resistor is not needed when charging the capacitor but it is
very important to discharge the capacitor.
Remove the resistor and the output will be nearly ZERO!
Another point to note with a
All the energy to turn-on the Buffer stage comes from the
Fig107b shows a
transistor that is turned on via a diode on the
This is a BAD design. The transistor is said to be in a HIGH
IMPEDANCE STATE, when not turned ON.
This means the base is FLOATING when the anode end of the diode
is at 0v.
When the anode of the diode is LOW, it does not deliver any voltage to
the base and the effective resistance on the base is infinite
and any noise picked up by the base will
turn the transistor ON. To prevent this from happening, a 100k
resistor is connected between base and 0v rail.
The high-value collector load also gives the
transistor in the first circuit a high possibility to pick up
noise on the base and produce pulses on the collector.
Solar Night Light
Here is a poorly-designed circuit.
A 1 watt white LED has a characteristic voltage of 3.2v to 3.6v
and it takes 300mA. The voltage drop across 8R3 will be V=IR =.3
x 8.2 = 2.46v The base voltage will be 3.2 + 2.46 + 0.7 =
The LEDs will not turn on very brightly. A white LED will start
to turn on at a lower voltage but the full brightness is not
achieved until 300mA is flowing and this will produce a voltage
of about 3.2 to 3.6v across the LED.
There is another major fault with the circuit.
The transistor is only designed to pass 500mA. It is
over-stressed. The base current will be about 20mA to 40mA. This
current must be supplied by the 4k7 pot.
This current is too much for a pot.
Secondly, the current must flow through the LDR when it receives
illumination, so that the current is removed from the base of
the transistor. This current is too high for the LDR.
You can learn a lot from other designer's mistakes.
Anil wanted to increase the volume from his
mobile handset using a single transistor and a few components.
He has a choice of using an emitter-follower or common-emitter
amplifier, as shown in the two circuits.
The first circuit will only increase the current.
The second circuit will increase the current AND the voltage of
the waveform and is the best circuit to use.
This circuit is a
But it does not have an emitter resistor. The emitter resistor
allows the stage to self-adjust the current through the
collector-emitter of the transistor to produce an approximate
mid-rail voltage on the collector.
Without this resistor it is very difficult to produce mid-rail
voltage when the supply rail can vary from 12v to 18v and the
gain of the transistor can be anything from 100 to 300.
The solution is to change the biasing to SELF BIAS.
This involves a resistor from collector to emitter. The stage
will now have a voltage on the collector and by testing a number
of transistors, you can determine the correct value for the base
There is one other fault with the circuit. The load resistor
(3k) is too low.
The circuit is a pre-amplifier and if the collector resistor is
increased to 33k, the output signal will be increased 10 times.
It works like this: The incoming signal supplies a small current
and this is amplified by the transistor (about 100 times) to
produce a current in the collector-emitter circuit. This current
flows through the collector-load-resistor and produces a voltage
across it. If the resistor is a high value, the voltage produced
is high and thus the waveform is high and thus the stage
produces a HIGH GAIN.
This circuit is unusual in
appearance but it does NOT work.
The transistor is in a COMMON-BASE configuration and we have
seen this using an NPN
You need to turn everything up-side-down to work with a PNP
The problem lies in the value of the 220k resistor. The value is
If the transistor has a gain of 330, it will convert the 330k to
1k and become a 1k resistor.
1k in series with 220k means very little voltage will be dropped
across the 1k and the collector voltage will be about 9v.
This means the 470n will be fully charged.
Suppose the transistor turns OFF fully.
The 470n is discharged via the 220k.
When the transistor turns ON, it can only charge the 470n with
the energy that has been removed by the 220k.
This means that although the transistor can theoretically act as
a 1k resistor, in reality it is only as good as a 220k and it
will deliver very little energy to the 470n.
The 220k resistor needs to be reduced to 1k to deliver the
maximum energy to the 470n.
The transistor can only CHARGE the 470n. The collector resistor
DISCHARGES the 470n.
This circuit above is poorly designed and has a
number of fundamental faults.
You can learn a lot from other peoples mistakes and this is a
NEVER try to control the bias on a stage via the base
resistor. This applies when the resistor is connected
between base and positive rail. See more on this type of biasing
This resistor operates entirely differently to a resistor
between collector and base.
A resistor between collector and base operates as a NEGATIVE
FEEDBACK resistor and adjusts the voltage on the collector to
about mid-voltage when the correct load resistor is used and the
appropriate base resistor is selected. When this resistor is
changed, the voltage on the collector changes a SMALL AMOUNT.
But when the resistor in the circuit above is altered, the
collector voltage changes a VERY LARGE AMOUNT.
The arrangement above is NOT a self-biased stage and getting the
transistor to a point where it will produce the highest gain is
a very difficult thing to do.
The highest gain is when the voltage on the collector is
mid-rail, but we need the transistor to have a collector voltage
below 1.8v so the final stage is not tuned ON. The voltage
cannot be mid-rail.
This means we need to deliver extra current into the base to
turn the transistor ON more so the collector voltage is reduced.
This means, to activate this stage, we need to deliver more
energy from the 4n7 in the form of a negative pulse, to turn it
OFF and allow the 4k7 to deliver current to the stages that
This means the first transistor has to turn OFF more so the 4n7
can charge to a higher voltage so that when the first transistor
turn ON, the higher energy will be delivered to the base of the
second transistor to turn it OFF. (In other words, the energy
from the 4n7 reduces or removes the current delivered by the 2M2
Having the 2M2 as a manual adjustment will decrease the gain of
the two stages considerably.
If you turn the transistor ON too hard, the circuit will require
a lot of energy from the 4n7 to remove this turn-on current to
turn the transistor OFF.
This is the first major fault.
The next point to note is the sensitivity of the diode pump has
been reduced by the inclusion of the third transistor. This
transistor is an emitter-follower and does NOT assist in the
operation of the circuit. In fact it reduces the sensitivity of
the circuit. It adds an extra 0.6v to the requirement of the
The diode pump is perfectly capable of turning ON the output
transistor without the need for the current-amplification of the
The output transistor only requires a very small current into
the base to operate the piezo sounder (less than 0.5mA) and the 4k7 will deliver this
With the third transistor removed, the output transistor is
turned ON via the 4k7 on the collector of the second transistor.
The 1k on the base of the output transistor is far too low for a
transistor delivering 50mA collector current and should be 10k
to 47k.If we do this, we have already increased
the gain of the circuit by 10 times to 47 times.
If the second transistor is AC coupled, ALL of the signal from
the 40kHz transducer will be passed to the diode pump.
If the signal on this transistor is 1,500mV, the output will
respond if the third transistor is removed.
This requires 30mV into the base of the second transistor.
To get 30mV on the collector of the first transistor the
transducer needs to produce less than 1mV. This will be the maximum sensitivity for the circuit and it is
AUTOMATICALLY self-adjusting and AUTOMATICALLY delivering the
maximum overall gain without any need for adjustment.
This is how to work out the requirements of the circuit.
It is a VOLTAGE
not a CURRENT
Here is a description of the
circuit above, including an understanding of how the two "coils"
work in the first stage:
THE "FRONT-END" The secret
to getting a long-range 27MHz link is a powerful transmitter and
a sensitive "front-end" on the receiver.
A 27MHz transmitter of only a few milliwatts (10 to 30 mW) will
reach 100 metres providing the receiver has a very sensitive
The Front End is the first stage of a receiver.
In our case it is a very weak 27MHz oscillator and thus it is
actually a 27MHz transmitter (or more accurately - a 27MHz
radiator) as it fills the surroundings with a 27MHz signal.
When another 27MHz signal enters this field it upsets the
transmission of the receiver and increases the amplitude of the
oscillator. This has the effect of producing a cleaner signal
and the background noise or "hash" is reduced.
This is a very clever way of making the front-end very sensitive
as it takes a lot of energy to "excite" an oscillator that is
sitting in a dormant condition. It is also very difficult to get
an oscillator to sit in a condition that is just before the
point of oscillation. So we cause it to oscillate at a very low
level and an incoming signal will increase the amplitude.
The fact that the transistor in the front end is oscillating can
also be referred to as a REGENERATION circuit as the output of
the transistor - at the collector - is fed back into the circuit
via the 39p between the collector and emitter.
The signal delivered by the 39p is prevented from being lost to
the 0v rail by the 50 turn inductor. This is called "emitter
injection" as the transistor is configured as a common base
amplifier in which the base is held firm by the 22n and the
signal fed into it via the emitter.
The operation of the circuit is kept at a low amplitude and when
the antenna picks up a signal of exactly the same frequency, the
amplitude increases. This is called SUPER REGENERATION or
increase of the regeneration.
It is a very simple way of getting an enormous result from very
few components. Normally you would require 2 or more stages of
amplification to produce the same result.
The only problem with a SUPER REGENERATIVE circuit is the
background noise it produces. But since this project is designed
only to activate a load, this background "hash" is not a
problem. Nearly all the background noise is removed by the
feedback capacitors on each stage.
A capacitor placed between the collector and base of a
transistor has an enormous effect on reducing the high
That's why a small capacitor such as 2n2 can be used. The gain
of the transistor (about 70) multiplies the effective resistance
(impedance) of the capacitor by about 70 and the background
noise is removed because it mostly consists of high frequencies.
But there is still a lot of skill to get the front-end to
oscillate very lightly, while being sensitive to signals coming
from the antenna.
A high-value resistor in the collector only allows a small
current to flow and if the supply voltage is high, this produces
a circuit that will oscillate
with a small waveform and can be easily "upset" or "modified" by
an injection at the highest point of oscillation. If this
injection is timed accurately, it will increase the amplitude of
oscillation and the high voltage supply will allow this to
The output of the front end is taken from the collector of the
transistor - but not directly from the collector as this would
load the circuit and stop it oscillating. The top of the
oscillator coil will have a small percentage of the waveform
(that is generated on the collector) and we can amplify it via
three stages of amplification. The important thing is we can
only pick off a very small percentage of the energy from the
oscillator so the front end keeps oscillating.
The signal appearing at the "pick-off" point consists of: 1. - 27MHz frequency called the "carrier frequency,"
2. - a lot of noise and "hash" produced by the circuit
and also from the antenna picking up background noise from the
surroundings and 3. - a tone from the transmitter.
The tone is about 1kHz and its frequency is considerably
different to a 27MHz frequency so a simple PASS FILTER can be
used to remove the 27MHz and only allow the tone and noise to
pass to the next stage.
The component that does this is the 22n across the 4k7 in the
This capacitor effectively passes (shunts - removes) the 27MHz
to the positive rail where it is passed to the 0v rail via the
47u electrolytic. Thus it NEVER gets passed to the amplifying
We only want to VERY LIGHTLY load the front so we don't stop the
We do this by using a resistor and capacitor in series.
If we just use a capacitor, the "resistance" of the capacitor
will be quite small at some of the frequencies of the "hash" and
it will load the circuit and reduce the amplitude when a signal
is being received. Even though the top of the oscillator coil is
connected to "earth" via a 22n, another 22n "pick-off" will
reduce the signal slightly as it will have a "resistance of
about 5k to 7k because it is only passing audio frequencies. To
increase this resistance we add a resistor in series with the
The combination of the resistor and capacitor in series reduces
the LOADING effect.
We now have a very small signal that can be amplified by the
following 3 stages.
The second transistor amplifies this signal and removes a lot of
high frequencies (hash) via the 2n2 feedback capacitor.
The third transistor does exactly the same and we finish up with
a signal that is almost equal to rail voltage from the fourth
We need a very high amplitude signal for the PIC12F629
microcontroller. That's why there are three stages of
The microcontroller counts the number of cycles in 100mS and
determines if the tone is from button A or B. We could have a
count in 10mS because the two frequencies differ by 100%, and
this is something you can do when you start writing your own
programs for a microcontroller.
It also counts in increments of 100mS and if the signal is
present for 5 x 100mS, it is counted as a long pulse.
INDUCTOR The inductor on the
emitter of the transistors plays a very interesting part in the
operation of the circuit.
The transistor is configured as a COMMON BASE amplifier and the
base is held rigid by the action of the 22n on the base.
Amazingly, this capacitor has a "resistance" of less than 0.5
ohm at 27MHz and so you can see the base is held firmly.
The signal on the collector is passed to the emitter via a 39p
and its "resistance" is about 150 ohms at this frequency. So you
can see the signal on the collector has a fairly low resistance
path to the emitter and this capacitor will have an effect on
the amplitude of the signal on the collector.
However this capacitor does two things. It turns the transistor
ON more and it turns the transistor OFF.
Let's see how the capacitive reactance (the resistance at 27MHz)
has an effect on the amplitude of the signal.
We know the base is held rigid but the emitter is also held
rigid when the signal is not increasing or decreasing.
It is held rigid because the inductor has a very small
resistance (3 ohms) and the 2n2 capacitor between the inductor
and 0v needs time to charge and discharge and so it also holds
the voltage on the emitter at a rigid potential.
But let's look at the effect of the base-emitter junction. When
the 39p tries to LOWER the voltage on the emitter, it has the
effect of turning the transistor ON more and the base-emitter
junction acts like a very low resistance. This means that when
the voltage on the collector is reducing, the 39p acts like a
very low resistance to lower the voltage on the emitter and the
voltage on the emitter drops by only a few millivolts.
During this time what is the inductor doing?
To answer this we will explain how a capacitor and inductor
A capacitor is like smashing up against your opponent in a
football match. He wont let you pass him and he grabs on to you
so you cannot back away.
That's what the 2n2 capacitor does when it sees an increasing or
decreasing voltage. It RESISTS the increase or decrease and the
top of the capacitor is just like connecting to the 0v rail.
But an inductor is different.
It's like your team mate lifting you up to reach the ball when
you charge at him.
If you supply an increasing voltage to one end of an inductor
(the other end is at 0v), the inductor will produce a voltage of
the same amplitude and it will seem the INDUCTOR HAS DISAPPEARED
This is the most important concept you will learn about
inductors. Once you realise the inductor disappears, you can
see how it has no effect on reducing the signal.
That's exactly what happens in this circuit.
The inductor allows the signal through the 39p to have an effect
on controlling the voltage on the emitter without absorbing any
of the energy.
The inductor does not have any positive effect on the circuit,
is just does not have any loading effect.
In other words, the voltage developed across the inductor does
not add to the waveform. But the coil on the collector DOES have
an effect on the amplitude of the waveform.
INDUCTOR The top coil (inductor)
is actually part of a building block (circuit) called a TANK
It just needed a capacitor to be called a tank circuit. The
capacitor can be in series or parallel, but it is normally a
parallel circuit. This name came from the early days of radio
where they realised the coil and capacitor stored energy during
the first part of the cycle and released it during the second
half. It stored energy like a water tank.
This coil works in a completely different way because it has a
capacitor connected across it.
The bottom inductor acts like a HUGE opponent that NEVER lets
The top coil acts like a much smaller opponent that you can
reason-with and create a double "high-five." The transistor
delivers energy to the coil and it produces a voltage on its
lower lead so that the voltage across the coil is equal to about
The transistor now turns off and the magnetic flux created by
the coil collapses and produces a voltage in the opposite
direction and this voltage charges the capacitor across the
This reverse voltage can be equal to about rail voltage.
This means the amplitude of the positive voltage and the
amplitude of the negative voltage can be as high as twice rail
voltage and this signal is passed to the antenna. In our case
this amplitude is much smaller, but for a transmitter, the
amplitude can be 2 x rail voltage.
OSCILLATOR The first transistor is a
How does it oscillate?
It oscillates because a signal from the output is fed back to
The feedback signal must be POSITIVE FEEDBACK. Negative feedback
will not start an oscillator operating and any negative feedback
fed to an operating oscillator will reduce the output and may
even kill the oscillator.
So we need POSITIVE FEEDBACK. But this is not the complete
answer. The feedback signal must deliver slightly more energy
than the transistor requires, at each part of the wave. As the
transistor turns ON more and more, it requires more energy from
the feedback signal to do this. That's why the value of the 39p
feedback capacitor is larger than necessary.
Suppose we use a 10p capacitor, the transistor will turn on and
feed some of the energy from the collector to the emitter. But
as it turns ON more, the base-emitter junction requires
additional energy and the 10p cannot provide this. The result is
the transistor does not turn ON fully and it stops oscillating.
Alternatively, it may oscillate but at a lower amplitude.
The feedback capacitor will turn the transistor ON faster than
We want a 27MHz oscillation.
To get the transistor to turn ON and OFF at the required
frequency, we have two components on the collector.
These two components are a coil an capacitor.
When they are connected in parallel, the coil picks up signals
(called electromagnetic radiation) from the surroundings and a
very small voltage is produced by the turns.
This voltage is passed to the capacitor and it charges.
When it is fully charged, it sends its energy to the coil and
this takes a certain period of time because a capacitor cannot
charge and discharge instantly. This energy-flow back and forth
produces a microscopic amplitude and if a surrounding signal has
a frequency that exactly matches the natural frequency of the
two components, it will increase the amplitude of the signal and
continue this oscillatory effect.
The point to understand is this: The coil and capacitor has a
natural frequency at which they operate and if we connect them
to a transistor, they will output a waveform that is picked up
by the feedback capacitor to turn the transistor ON and OFF at
the same frequency.
The coil and capacitor form a circuit called a TUNED CIRCUIT and
they determine the frequency at which the circuit operates.
The transistor is simply tuned ON and OFF at exactly the time.
There is just one more interesting part to the circuit.
The top of the coil is fixed to the supply rail via the 22n and
when the transistor turns ON, the lower end of the coil becomes
slightly negative (slightly less than the voltage on the supply
The transistor keeps turning ON until it is fully turned ON. At
this point the feedback capacitor stops providing energy to the
emitter and the transistor turns OFF a small amount. This
reduces the current through the coil and the magnetic flux
changes from expanding flux to collapsing flux.
This has the effect of producing a reverse voltage from the coil
and now the end connected to the collector has a voltage HIGHER
than rail voltage. This voltage raises the voltage on the
emitter and has the effect of turning the transistor OFF.
In other words, the transistor is effectively removed from the
circuit and the coil/capacitor combination perform their energy
transfer as mentioned above.
The end result is a voltage on the antenna that can be twice
rail voltage. In our case the amplitude is kept very low but for
a transmitter, this voltage can reach twice rail voltage.
The 3 amplifying stages are called self-biasing common-emitter
The base bias resistor is chosen so the voltage on the collector
is about half-rail voltage. This allows both the positive and
negative excursions of the waveform to be amplified.
We are not concerned with the quality of the signal (the shape
of the wave - the waveform - as we only need to produce a
maximum amplitude to feed the microcontroller so it can count
the number of pulses - waves - per second.
The value for the components for each stage are chosen by
selecting the collector load resistor so the average current
will be about 0.5mA. We then select base-bias resistors so the
collector voltage sits at about 2.5v.
We then input the signal from the front-end and try capacitors
for the negative feedback.
Why waste time working out the values mathematically when it
only takes 30 seconds to physically fit a component?
After you generate a mathematical answer, you will want to try
different values to see the effect and so it's best to just
experiment and see the results on a CRO.
You don't know the gain of the transistor so it's pointless
using a mathematical equation to get a result.
The feedback capacitor has an enormous effect on removing
waveforms such as "hash" that are higher in frequency than the
tone signal, but it also reduces the amplification of each
stage. So you have to chose between reducing hash and reducing
overall gain of the stage.
Each transistor is called a STAGE.
A stage is a self-contained circuit with a capacitor at the
input and and output.
This makes the voltages on the stage entirely generated by the
components within the stage because the capacitor on the input
and output prevent DC voltages on adjacent stages having any
This makes diagnosis and testing very easy.
If a stage is not working, the first thing you do is check the
voltage on the collector of the transistor.
If it is too low or too high, the component may be the wrong
value or the transistor may have a very high gain or very low
gain. The feedback capacitor may be leaky or damaged or the
tracks on the PC board may have a splash of solder.
Audio circuits are very difficult to diagnose and our BENCH
AMPLIFIER project can be used to listen to the tone entering and
leaving the stage.
You can also use a CRO to view the waveform.
If you want to be really technical you can say the 3 transistors
form a "Frequency Selective Strip" - the frequencies being
selected are LOW FREQUENCIES.
Another question from
What is the effect of adding the 3v3 zener?
Circuit A turns on when the
resistance of the LDR decreases. Nothing happens until the
resistance of the LDR reduces to put a voltage of 0.7v on the
base of the transistor. At this point the LDR and 47k form a
voltage divider and no current flows into the base.
As the resistance of the LDR decreases, current will flow into
the base of the transistor and start to turn it ON.
Take note of the level of illumination for this to occur.
If a 3v3 zener is paced in the base-lead, the resistance of the
LDR will have to decrease until a voltage of 3v3 plus 0.7v is
developed across the 47k resistor. This means more light will
need to be detected by the LDR.
As soon as 4v is reached, current will flow through the zener
and into the base of the transistor.
Because the change in resistance of the LDR is not linear, the
extra amount of illumination is not known, but it will be more
than the circuit above.
Once the transistor in both circuits is turned ON, the same
amount of current into the base will have the same effect on the
LOAD but since the level of brightness on the LDR is different,
it may require additional illumination to produce the same
effect in the second circuit.
Instead of fitting the zener, the same effect can be obtained by
lowering the value of the 47k resistor.
BD139 is a 80v transistor. Can I use BD135 - a
The maximum voltage the
transistor will see on the collector is about 12volts, so any
transistor with a rating above 25v can be used.
There are no inductors (coils or transformers) in the circuit
and thus no spikes will be generated.
In many cases, when an inductor, coil or transformer is present
in a circuit, a high voltage spike will or may be produced when
the current is switched OFF and this can be up to 100 times
greater than the supply voltage.
These spikes can damage a transistor and need to be suppressed
(prevented ) by using a damper diode (a diode connected in
reverse across the coil).
If the spikes are say 70v, an 80v transistor can be used.
Another question from
Why is 3p for the base capacitor? Won't a 300pf be better.
The 3p is a very small value for this reason:
The frequency of operation of the circuit is 303MHz and you have
to have a coupling capacitor that will charge and discharge
almost fully during this time-period.
This is also the reason for the 3p between emitter and base.
Why is the 15uH inductor connect to the middle of the
The reason for the inductor connected to about the middle of the
coil is to produce POSITIVE FEEDBACK.
In other words we need feedback to keep the transistor turning
ON more and more until it cannot be turned ON any more. At this
point the feedback signal ceases and transistor starts to turn
OFF and the feedback signal makes it turn OFF more and more
until it is fully turned OFF. Then the cycle starts again.
For a transistor, when the voltage on the collector is falling,
the voltage on the base must be rising.
You cannot connect a capacitor or a resistor to the two terminal
to do this. It has to be a TRANSFORMER.
Only a transformer can deliver a "reverse waveform." or an
INVERTED WAVEFORM or what is actually and really a NEGATIVE
FEEDBACK SIGNAL. A signal that is opposite to the signal being
When the circuit is turned ON, the transistor is turned on via
the 12k and the collector voltage falls. This is transferred to
the base via the 9p and 3p so it does not turn on so quickly.
At the same time the 3p between the emitter and base transfers a
signal from the emitter to the base to turn the transistor ON
more. All these signal fight each other and they result in a
short period of time for the transistor to turn ON. This is what
produces the first part of the 303MHz waveform.
The 3p on the emitter "has the last say" and when it cannot hold
the transistor in a state of "being really turned ON" the
transistor loses some of the ability to be turned ON and starts
to turn OFF. This lowers the voltage on the emitter and the 3p
transfers this waveform to the base.
But this capacitor does not perform any "delaying action" as the
transistor can turn ON and OFF really fast.
It is the 9p capacitor that creates the time between the turning
ON and the turning OFF. It creates the FREQUENCY.
You can see the 3p on the end of the PCB track is much smaller
than the 9p but this does not matter as it can be discharged and
charged in the opposite direction during a cycle and and still
have a "transferring factor."
But what actually makes the transistor OSCILLATE?
It is the INVERSE VOLTAGE (INVERSE WAVEFORM) coming from the
lower part of the oscillator coil.
The voltage coming out of the end of the PCB track to the
base of the transistor is very small and it is also very
weak. It does not carry a lot of amps. In fact it carries
The base of the transistor is really a very
low impedance line and it takes a lot of current to raise
So, the component between the end of the PCB track and the
base must interface two different and widely differing
Because the frequency is so high, you have to have a
capacitor that charges and discharges very quickly or it
will slow down the circuit to 10MHz or less.
This means we must have a very small capacitor.
the capacitor does something amazing. It converts the few
hundred millivolts rise and fall on the end of the PCB track
(with very little current capability) to a much lower amplitude
(because the base causes the amplitude to decrease because it is
such a low impedance) and with a much higher current capability.
This is something that no text books have ever described before.
Now, here's where the feedback feature comes
When the transistor turns ON, the collector falls slightly and
it produces an expanding or increasing waveform. As this signal
gets larger and larger, it transfers some of its energy to the
other part of the coil via the "air".
This energy produces a waveform within the length of the track
and since the end connected to the inductor cannot move, the
waveform produces a wave that increases in amplitude at the end
going to the base.
This is how a downward movement on the collector produces an
upward voltage on the base.
The voltage of this waveform fights against the downward voltage
produced by the 9p. The 9p produces the "timing factor" and the
lower track produces the voltage of reverse polarity.
Here is a circuit from DIYODE Magazine - by
Bob Harper, from their Educational section.
circuit will work, it
uses incorrect values for a stage such as this.
This stage is called pre-amplifier stage and it should
take the least current possible because as the current increases
in all the components, they produce background noise -
especially the transistor.
The author starts by saying the electronics microphone has a
resistance of 2k2 and you need a load resistor of 2k2. He is
possibly using the reasoning of half-voltage across the load and
half-voltage across the component.
So, we put the microphone into a circuit and the voltage across
it is 8.4v and 0.6v across the 2k2. What does this tell us???
All it tells us is the "resistance" of the microphone is more
An electret microphone is an active device containing a FET
transistor and you cannot simply measure the "resistance" across
its leads as this value will change enormously when placed in a
circuit. And when you measure it with an analogue meter, the
"resistance" is 10k. The main reason is the voltage delivered by
a multimeter to read the "resistance" is different to the
voltage it will see when placed in a circuit.
Electret microphones are very sensitive and if you allow too
much current to pass through them, the result will be
overloading and they produce a "crackling sound." For a 9v
supply the load resistor should be between 47k and 100k.
This has simply been found by experimentation. NO specifications
sheets can be used in this determination - only trial and error.
The next major fault with the circuit is the biasing components
for the transistor.
All the values should be increased 10 times to 100 times.
The transistor in a pre-amplifier stage should pass very little
current to keep the background noise to a minimum.
Again, all the accepted values (from qualified engineers) have
been determined by experimentation.
Bob Harper shows absolutely no understanding of how to design
this type of circuit.
Even if we take his reasoning that the microphone is 2k2, he is
trying to connect a 2k2 impedance to the input impedance of the
"H-bridge Transistor stage" that has an input impedance of less
than 270 ohms. This immediately says the signal will be
dropped to 10% and it will be dropped further by the coupling
And it will be dropped even further because the impedance of the
microphone is really about 10k.
He has overcome some of this by driving the microphone 10 times
harder than is necessary.
In his article,
claims the BC547 transistor is a 500mA device!!!! He has
obviously NEVER used a BC547 in a circuit. You would never make
a mistake like if you had used them this as they start to fail
when trying to deliver 100mA.
When I commented about the faults in the design, my comments
were immediately removed.
How can someone, who has taught students at technical schools
for 30 years, be so devoid of basic understanding ??
That's why I have produced this article on The Transistor
Amplifier, so you don't make a fool of yourself.
Here's away to look at the design. Remember the days when
you had a crank-handle to start a car. Small cars were easy to
"crank over," while big cars were almost impossible to "crank."
A little old lady could not crank a BIG CAR.
The electret microphone is like a little old lady. It is very
weak and delicate and its output is very weak. The transistor
amplifier circuit above has very low impedance and it is like an
eight cylinder car. The output of the microphone will be almost
completely lost when trying to deliver its signal into the 47n.
In an attempt for the microphone to try and work, it has been
overloaded 10 to 20 times will extra current and this will
create all sorts of background noise. The whole circuit is
a disaster in design.
A POORLY DESIGNED CIRCUIT FROM
DIYODE MAGAZINE FEBRUARY 2019
Lab Electronics produces a "stand-alone" trainer
that covers the common-base, common-emitter and
109 shows the circuit for the trainer and how it
can be wired to produce all the stages we
have covered in this discussion. By feeding each stage with a
sinewave at the input, you can view the output on a CRO and see
how it works. This is only part of the picture to understanding the operation
of each stage as the input and output impedances are also
important and the third important thing is the effect of the
capacitor(s) and/or electrolytics that connect one stage to
another and/or those connected to the emitter to provide EMITTER
We have already explained the advantage of a common-base stage
(to connect a very low impedance device to an amplifying
circuit) and the advantage of a common-collector
(emitter-follower) circuit to drive a low-impedance load. e load.
A "trainer" only provides a fraction of the knowledge needed to
understand "circuit-design" - but it helps. You must build
"real-life" circuits to get a complete understanding.
The trainer above has lots of faults in its design. You cannot
get a full understand of the common-base stage with 1k in the
emitter. It should be 100R or less. The 10k feeding the 33u will
attenuate the sinewave and is not needed.
The common-emitter stage does not provide any self-biasing
option. The 56k base-bias is too low and the collector and
emitters resistors are the wrong values to get any appreciable
gain from the stage. When the 33u is put across the emitter
resistor, the gain will increase enormously.
It would be much better to work on the circuits we have presented above
and view the output on a CRO.
This trainer does not give you a full understanding of the
operation of the three stages. (33u and 15v is rarely used in
modern designs), I
would give it a MISS.
Fig 110 shows
another trainer. It covers the common-emitter stage.
When a common-emitter stage drives a transformer or speaker as a
load in the collector circuit, we want the sound to be free of
distortion and to do this this we must bias the stage so the
collector is at half-rail voltage when no audio is present.
This allows the transistor to turn ON and OFF to provide
the maximum voltage-swing. If the transistor is not sitting at
mid-rail, either the positive or the negative peaks of the
signal will hit either the positive or negative rail and produce
distortion - because the full excursion (height) will not be
But biasing the transistor at mid-rail means the current though
the speaker or transformer will be about half the peak current
and this is wasted as it flows at all times, even when audio is
not being processed.
That's why this type of stage is not efficient and it heats up
the output transistor considerably, even with no audio.
This type of circuit is called "CLASS-A" and the trainer above
has a "Bridge" circuit as a pre-amplifier and is
capacitor-coupled to a common-emitter stage as an output stage -
driving a transformer - as a class "A" amplifier. Since transformers are expensive,
difficult to purchase and add weight to a project, they have
generally been replaced by complementary-symmetry push-pull
class-B output stages.
All the features in this trainer have been covered in the
Which circuit is best? Fig 111 shows four different circuits driving
a speaker. Which circuit is best??
The 4 circuits in Fig 111 drive an 8 ohm speaker and are
called OUTPUT STAGES or DRIVER STAGES. They are
all different in performance and have different input voltage
requirements. Circuit A is really only a one transistor
emitter-follower amplifier as
the other transistor discharges the electrolytic.
However it is fully discharged and represents only a few ohms
resistance (impedance) in series with the speaker. The input voltage-swing must be as large
as possible (called rail-to-rail swing) to achieve the maximum
is a two-transistor amplifier (called a Darlington Pair)
and requires only a very small input current for the circuit to
work, but a rail-to-rail
voltage-swing. The speaker is AC coupled and only the audio
current enters the cone and the cone is not displaced by any DC
current. However the 100u is discharged via a 330R and the
electrolytic is equivalent to a 330R in series with the speaker.
The output from this circuit will be very low.
directly connected to a speaker. The input is very sensitive and
requires less than 1v swing for full output. However DC flows
through the speaker and will heat up the coil as well as shift
the cone and maybe reduce the output capabilities of the
speaker. The BC547 driver transistor will not be able to deliver
much current. A BC337 is a better choice.
is a high gain
and has a sensitive input and requires less than 1v for full output.
However the electrolytic is discharged via a 100R and this means
it is equivalent to a 100R in series with the speaker.
The best circuit is "A" but it needs a pre-driver
transistor to achieve the gain (amplification) of the other 3
Fig 111a is a "Class-A amplifier" with an
emitter resistor that is by-passed with a 100u capacitor.
The quiescent (idle) current taken by the stage will be low due to the 330R emitter
resistor but when a signal is delivered to the base, the
transistor will operate as if the emitter is connected directly
to the 0v rail. This means the stage will provide very good
amplification while the quiescent current is quite low.
Note: Fig111a needs to have a base-bias resistor and a
capacitor coupling the base to the previous stage, to qualify as
a class-A amplifier. If the base-bias resistor is removed, the
stage becomes "Class-C" where the stage uses the energy from the
previous stage (via the capacitor) to "turn it on."
Here is a discussion on the type of output stages and the
advantage and disadvantages of each stage. Output stages are
classified as "class-A, class-B" etc and explain how current is
"steered," or controlled, within a power amplifier before it is
delivered to a speaker load.
Class A is the simplest, most basic
topology. Reproduction of music requires speaker motion both in
and out. To do this, amplifiers must "source and sink" current.
In a Class A amp only one direction of current flow is used. It
relies on the movement of the code to return it to the neutral
position or movement in the reverse direction due to the inertia
of the cone.
To reproduce a sinewave (or similar waveform) the output stage
must be biased to half-rail voltage so the waveform can rise and
fall. This means the output stage is consuming half-power when
NO SIGNAL is being delivered. The heat losses are
high but the output is very high quality.
Class B uses two variable output
stages, one to source current and the other to sink current.
This topology overcomes the poor efficiency of pure Class A,
only delivering power as needed.
Both output stages turn completely off during the time when the
signal passes from one output to the other. Time delay and
low-level non-linearities cause some distortion, called
"crossover distortion," during transition from source to sink
output stages. This type of distortion is worst at low output
levels. Pure Class B is only used in the lowest-cost,
Class A/B is a combination of Class
A and Class B. Using two variable output stages like Class B but
keeping them from completely turning off, you get near Class B
efficiency with near Class A's low-distortion performance.
Class C combines active devices
with resonant magnetic components for high efficiency at radio
frequencies. This topology is not used in audio-frequency
designs. The output stage is not biased, but gets all its drive
capability from a previous stage.
Class D uses source and sink output
stages that consist of full-on or full-off switches. These
output stages toggle from full sink to full source at a rate
significantly higher than the highest audio frequency to be
reproduced. The ratio of time sinking to time sourcing controls
the audio output, with a 50% ratio delivering zero output.
Class D offers significantly higher efficiency than Class B.
Losses in Class D designs are limited to turn-on time of the
switching devices and resistive losses in these devices and
Class D amps require more complex circuit designs with extensive
shielding and filtering.
Class G & H are variations on Class
B that use multiple source and sink output stages. Low-level
signals are handled by one pair of output stages, while
higher-level signals are handled by other pairs. Each pair is
optimized for the power range it delivers.
More efficient amplifiers can deliver the same output power with
smaller transformers and less heat sink.
Circuit complexity increases, which adds cost. Switching
distortion similar to Class B's crossover distortion occurs at
each output level transition.
Bridge Mode takes advantage of the fact that speaker loads can
be driven differentially. Using separate amplifiers to drive
both the positive and negative speaker terminals with
opposite-polarity waveforms yields an effective doubling of the
voltage swing for 4 times the power.
Provides high power levels using lower-voltage components.
Increased circuit cost/complexity and inability to ground
reference either speaker lead.
Here is a text
book containing a series of questions and answers.: Self Teaching Guide
I have included it to show you that even electronics authors who
have been in electronics for decades, make mistakes.
See the question on base and collector current for a 24v globe.
The author has made no mention of the fact that the globe takes
6 times more current when turning ON. That's why you will find
the circuit he designed, will not work.
He also describes a two transistor circuit where one transistor
turns the other off. This is a very current-wasteful circuit.
He also describes the bleed current for the base
(voltage-divider) for a circuit that does not need a voltage
The following software allows you to design your own
single-stage Common-Emitter, Common-Base or Common-Collector
It has been created by Didaktik Software.
This is version 1.1.1 created 23-6-2012
Download: TransistorAmp (.zip
TransistorAmp unzips to
TransistorAmp.msi (620KB) and will install on your computer with
a desktop icon.
Or you can download: TransistorAmp (.exe)
Unzip.rar in a folder "TransistorAmp" and it will create
TransistorAmp.exe Click on the file and the
image above will appear.
How to use the software TransistorAmp
TransistorAmp is very easy to use.
You start every design with the menu item: "New Amplifier". In
the pull-down-menu you choose your desired circuit. You can
choose between common-base-circuit, common-emitter-circuit and
common-collector-circuit. After that you get a dialog, where you
have to put in all parameters of your amplifier.
The following 3 images show the layout of the circuit you will
For the selection of the transistor-type you can click on the
button: "Select transistor type from ", and you will see a
list of all supported transistor types. TransistorAmp supports
some thousand transistor types - even some Germanium
transistors. Select your desired transistor type and click OK. The selected transistor type will be displayed in
the dialog. Both NPN and PNP transistors are supported.
When you have completed your input in the dialog, click
OK and see the result. You see a window with your input data,
the circuit, the component values and the most important
parameters of the operation point. If you want to change your
design, you only need to click again on "New Amplifier" and the
circuit in the pull-down-menu. Your previous input data will be restored
in the input dialog and you can change one or
Note: for the Common-Collector amplifier: "Collector current in A"
means: "Collector current in Amps." For 2mA, insert 0.002 etc.
Decibels are defined as ten times the log
of a power ratio. This calculator converts between decibels,
voltage gain (or current), and power gain. Just fill in one field
and the calculator will convert the other two fields.
dB= 20log(V1/V2)= 10log(P1/P2)
When you are satisfied with the result, click on:
"Result - Save". TransistorAmp saves all data
in the result window to an HTML-file. You can open this file with
a browser (e.g. Firefox or Internet Explorer), inspect it and
After reading all these discussions you will realise
how little has been provided in the average text book on
designing transistor stages. That's why these articles
were produced. They explain this extremely important
topic in a completely new way - one that even I can
understand . . .
50 years ago I heard an engineer say: It is only a 3
leaded device, like a triode valve: "It will never
On the next page we cover connecting a "normal" or
"standard or "common
- called Bipolar Junction Transistor (BJT)
to a Field Effect Transistor (FET).