simple explanation of how a transistor works in a circuit, and how to
connect transistors to create a number of different circuits. No mathematics and no complex wording.
completely different approach you can understand . . .
This eBook starts by
ON a single transistor with your finger
(between two leads) and
progresses to describing how a transistor can be connected to the supply
rails in 3 different ways.
Then it connects two transistors together DIRECTLY or via a capacitor to produce
amplifiers and oscillators.
As you work through the circuits, the arrangement of the parts are
changed slightly to produce an entirely different circuit with new
This way you gradually progress through a whole range of circuits (with
names you can remember) and they are described as if the parts are
"moving up and down" or "turning on and off."
Even some of the most complex circuits are described in a way you can
see them working and once you get an understanding, you can pick up a
text book and slog though the mathematics.
But before you reach for a text book, you should build at least 50
circuits . . . otherwise you are wasting your time.
I understand how the
circuits work, because I built them. Not by reading a text book!
From a reader, Mr Ashvini Vishvakarma,
I was never taught the influence of the coupling capacitor in
capacitor-coupled single transistor stages.
No one told me that RL of one stage delivers
the input current of the next stage.
No text book has ever mentioned these
before because the writers have never built any of the
circuits they are describing. They just copy one-another.
That's why this eBook is so informative. It will teach
you things, never covered before.
I don't talk about "formulae" or produce graphs because
transistors have a wide range of parameters - especially the
gain - and this has the greatest effect on the operation of
a circuit. It is faster to build a circuit and test a
transistor than work out the "Q-point" from a load-line.
The same with two resistors in parallel. It is faster to put
them together and measure the resistance, than look up a
You learn 10 times faster with actual circuits than
theoretical models and 10 times smarter when you know how to
course from South
Dakota School of Electronics.
These lectures cover the mathematical side of how
various circuits work.
Once you complete this eBook, the lecture notes will be
much easier to understand.
My interpretation of
the above-course is this:
It goes into far too much detail and far too much
There is very little on digital concepts and nothing on
Time would be much better spent on explaining transistor
and MOSFET behaviour in a simpler way and getting on
with digital circuitry and microcontroller projects. The
student should build at least 20 projects for the year
as this is the only REAL way to learn. I give the course
2/10. It really is a WASTED year.
You simply cannot put a transistor into a circuit and
expect it to produce the calculated results. The gain of
a transistor can be from 100 to 200 in a batch and this
changes the outcome by 50%!!
Instead of taking 30 minutes to work out the answer,
simply build the circuit and measure the REAL result.
THE NPN TRANSISTOR
There are thousands of transistors and hundreds of
different makes, styles and sizes of this amazing device. But there are only two different
types. NPN and PNP. The most common is NPN and we will cover it first.
There are many different styles but we will use the
smallest and cheapest. It is called a GENERAL PURPOSE TRANSISTOR.
The type-numbers on the transistor
will change according to the country where it was made or sold but
the actual capabilities are the
We are talking about the "common" or "ordinary" or
It is also referred to as a BJT (Bi-polar Junction
Transistor) to identify it from all the other types of
transistors (such Field Effect, Uni-junction, SCR,) but
we will just call it a TRANSISTOR.
Fig 1. NPN Transistor
Fig 1 shows an NPN
transistor with the legs covering the symbol showing the name for each
The leads are BASE, COLLECTOR and EMITTER.
The transistor shown in the photo has a metal case with a tiny tag
next to the emitter lead.
Most small transistors have a plastic case and the leads are in a
single line. The side of the transistor has a "front" or "face" with
markings such as transistor-type.
Three types of transistors are shown below:
Fig 2. NPN Transistor
Fig 2 shows two "general purpose" transistors
with different pinouts. You need to refer to data sheets or test the
transistor to find the pinout for the device you are using as there are
about 5 different pin-outs.
The symbol for an NPN transistor has the arrow on the emitter
pointing AWAY from the BASE.
Fig 3. NPN
Fig 3 shows the equivalent of an
NPN transistor as a water
valve. As more current (water) enters the base, more water flows from
the collector to the emitter. When no water enters the base, no water
flows through the collector-emitter path.
Fig 4. NPN
connected to the power rails
Fig 4 shows an
NPN transistor connected to the power rails.
The collector connects to a resistor called a LOAD RESISTOR and the
connects to the 0v rail or "earth" or "ground."
It can also be called the negative rail.
The base is the input lead and the collector is the output.
The transistor-type BC547 means a general-purpose
Sometimes a general-purpose transistor is called TUN - for
Transistor Universal NPN.
A general-purpose PNP transistor is called TUP - for
Transistor Universal PNP.
Here is a video by Ben. He
shows how to connect a solenoid to an NPN transistor:
Click at the top of the video to go to the YouTube
website to see more electronics videos.
Fig 5. NPN Transistor biased with a
"base bias" resistor and a LOAD resistor
Fig 5 shows
an NPN transistor in SELF BIAS mode. This is
called a COMMON EMITTER stage and the resistance of the BASE BIAS
RESISTOR is selected so the voltage on the collector is half-rail
voltage. In this case it is 2.5v.
To keep the theory simple, here's how you do it. Use 22k as the load
Select the base bias resistor until the measured voltage on the
collector is 2.5v. The base bias resistor will be about 2M2. This is how the transistor gets turned on by the base bias resistor:
The base bias resistor feeds a small current into the base and this
makes the transistor turn ON and creates a current-flow though the
This causes the same current to flow through the load resistor and a
voltage-drop is created across this resistor. This lowers the voltage on
The lower voltage causes a lower current to flow into the base, via the
base-bias resistor, and the transistor stops turning on a slight amount.
The transistor very quickly settles to allowing a certain current
to flow through the collector-emitter and produce a voltage at the
collector that is just sufficient to allow the right amount of current
to enter the base. That's why it is called SELF BIAS.
Fig 6. Turning
ON an NPN transistor
Fig 6 shows the transistor being turned on via a finger.
Press hard on the two wires and the LED will illuminate brighter. As you
press harder, the resistance of your finger decreases. This allows more
current to flow into the base and the transistor turns on harder.
Fig 7. Two transistors turning ON
Fig 7 shows a second transistor to "amplify the effect of
your finger" and the LED illuminates about 100 times brighter.
Fig 8. Adding a capacitor
Fig 8 shows the effect of putting a capacitor on the base
lead. The capacitor must be uncharged and when you apply pressure, the
LED will flash brightly then go off. This is because the capacitor gets
charged when you touch the wires. As soon as it is charged, NO MORE
CURRENT flows though it. The first transistor stops receiving current
and the circuit does not keep the LED illuminated. To get the circuit to
work again, the capacitor must be discharged. This is a simple concept
of how a capacitor works. A large-value capacitor will keep the LED
illuminated for a longer period of time as it will take longer to
Adding a capacitor to the
shows the effect of putting a capacitor on the
output. It must be uncharged for this effect to work. We know from
Fig 7 that the circuit will stay ON constantly when the wires are
touched but when a capacitor is placed in the OUTPUT, it gets charged
when the circuit turns ON and only allows the LED to flash.
This is a simple explanation of how a transistor works.
It amplifies the current entering the base (about
100 times) and the higher current flowing through the
collector-emitter leads will illuminate a LED or
drive other devices.
A capacitor allows current to flow through it until it
gets charged. It must be discharged to see the
Just some of the pinouts for a
transistor. You need to refer to a data sheet or test the device to
determine the pins as there are NO standard pin-outs.
Before we go any further, we need to talk about the RESISTOR.
It's a two-leaded electrical component that has resistance from a
fraction of an OHM to many millions of ohms (depending how much carbon
is in the resistor). When the resistance is very
low (small) the resistor is equal to a piece of wire and when it is very
high, the resistance is equal to . . . . . .
The value of a resistor is marked on the body with bands of
or, in the case of surface-mount resistors, a set of
numbers. These identify the value of the resistor in OHMs. When the
value of resistance is above one-thousand ohms, we use the letter "k" -
for example 1,200 ohms is 1.2k or 1k2. When the value is above
one-million ohms, we use the letter "M" - for example 2,200,000 ohms is
2.2M or 2M2. When the value is say 100 ohms we use the letter "R" -
Resistors do "all kinds of things" in a circuit. In other words, they
can join two components, separate two components, prevent a component
from getting too hot, prevent an amplifier from overloading, allow a
capacitor to charge quickly or slowly - and many more.
All these things can be achieved because a resistor has ONE SIMPLE
FEATURE . . .
A resistor limits (or reduces) the
That's all a resistor does. It
limits - or controls - or allows - a current to flow according to the
resistance of the resistor.
This simple feature of limiting the current is like a man with a
hammer - he can hammer nails, break glass, drive a pole into the ground
and lots more and a resistor can do more than 12 different "things."
When a current flows through a resistor, a voltage is
developed across it. This voltage is called the VOLTAGE
DROP. (It is also called the VOLTAGE LOST ACROSS THE
The following 3 examples will help you understand the terms VOLTAGE DROP
and VOLTAGE LOST.
In diagram A, the resistor
is only connected at one end and NO CURRENT will flow. This means the
VOLTAGE DROP across the resistor will be ZERO. 12v is present on the
lower lead of the resistor because no current is flowing.
In diagram B, the resistor is connected to a glowing lamp and current
will flow. The voltage across the resistor may be 3v. In other words,
the voltage LOST is 3v and the lamp gets only 9v. We also say the
VOLTAGE DROP is 3v across the resistor.
In diagram C, the resistor is connected across the power rails and the
voltage across it MUST be 12v. We do not talk about voltage drop or
voltage lost in this circuit because there are no other components. We
just say: the voltage across the resistor is 12v.
This will help you understand how a resistor works.
THE VOLTAGE DIVIDER Nearly ALL circuits (and
individual stages) use a VOLTAGE DIVIDER. A Voltage Divider is simply two
resistors connected in series.
However it may not be two resistors. It may be a resistor and a
transistor. A transistor is really a resistor - a variable resistor -
and they form a voltage divider with a resistor called the LOAD. Sometimes
more resistors are present
(such as resistors creating an H-bridge biasing network) and there may be more than one voltage divider in a stage.
However the same principle applies.
The principle is this:
CURRENT FLOWS THROUGH THE
COMBINATION (the current is the same for each
resistor because they are in series).
Multiply the current (in amps) by the resistance (in ohms) to get the
voltage across each resistor.
In most cases, the sum of the voltages across each resistor must
add up to the supply voltage.
Here are 2 examples of a VOLTAGE DIVIDER:
This is as far as we can go without using mathematics.
A "Stage" is a set of components with a capacitor at the input
and a capacitor on the output.
We have already seen the fact that the capacitor only has an effect on
the circuit during the time when it gets charged. It also has an effect
when it gets discharged. But when the voltage on either lead does not
rise or fall, NO CURRENT flows through the capacitor.
When a capacitor is placed between two stages, it gradually
charges. When it is charged, the voltage on one stage does not affect the voltage on
the next stage. That's why the capacitor is drawn as two lines with a
gap. A capacitor is like putting a magnet on one side of a door and a
metal sheet on the other. Moving the magnet up and down will move the
metal up and down but the two items never touch.
Only a rising and falling voltage is able to pass through the capacitor.
Fig 10. This is a
A transistor, with a capacitor
on the input and output.
10 has a capacitor on the input and output. This means the stage
is separated from anything before it and anything after it as far as the
DC voltages are concerned and the transistor will produce its own
operating point via the base resistor and LOAD resistor.
We have already explained that the value of the two resistors should be
chosen so the voltage on the collector should be half-rail voltage
and this is called the "idle" or "standing" or "quiescent"
It is the condition when no signal is being processed.
When the voltage on the collector is mid-rail, the transistor can be
turned off a small amount and turned on a small amount and the voltage
on the collector will fall and rise. (note the FALL and RISE).
Fig 11. The Input and
shows a small waveform on the input and a
large waveform on the output. The increase in size is due to the
amplification of the transistor. A stage like this will have an
amplification of about 70.
This is called "Stage Gain" or "Amplification factor" and consists of
two things. The output voltage will be higher than the input voltage and
the output current will be higher than the input current.
We will discuss the increase in current and voltage in a moment.
We need to ask: Why is the gain of the stage only 70, when a
transistor with a gain of 200 is used?
The reason is due to the base-bias resistor. It is acting as a
feedback resistor and is acting AGAINST the incoming
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.
Fig 11a. Fixed Base
Fig 11b. Fixed Base
Fig 11a and 11b shows a Common Emitter
stage with fixed base-bias. This stage produces the maximum
voltage amplification but it is very difficult to "set-up"
because the value of the base resistor will either make the
collector voltage nearly zero or full rail voltage. It is very
difficult to get the collector to sit at mid rail.
If the base resistor is a high value, the collector will sit at
rail voltage. If the base resistor is a low value, the
collector will sit a 0v.
If a transistor with a different gain is fitted, the collector
voltage will change completely.
If it sits at mid-rail, the noise produced by the transistor
will make the collector voltage rise and fall and produce a lot
It all revolves around the actual gain of the transistor and
this requires a TRANSISTOR TESTER to determine the gain.
However, this circuit can be used as an output stage and has some
It is a "Class-C" stage and means it is just at the point of
being turned on via the base-bias resistor. It
consumes the least current when "sitting around" and
is the most
Energy from a previous stage provides base current via the
coupling capacitor and the
base-bias resistor assists too.
The output waveform will be distorted at the top or bottom,
depending on the biasing and an inductor in the collector can reduce the distortion. See the article on FM Bugs (SPY BUGS) for
a Class-C output stage.
Unless you get the biasing correct, do not use this type of
stage as a general-purpose amplifier. If the transistor is
saturated (the base resistor is too low) the output will consist
of only the positive portions of the waveform and will be a lot
smaller than a self-biased stage.
shows the signal (the voltage
waveform) as it passes through 2 stages. Note the loss in
amplitude as the signal passes through capacitor C2.
CONNECTING 2 STAGES
There are 3 ways to connect two stages: 1.direct coupling - also called DC coupling
(not the coupling shown in fig 12.
Fig 12 is AC coupling). DC stands for Direct Current. I know
this sounds unusual, but it is the way to explain the circuit will
pass (amplify) DC voltages. This type of coupling will pass both AC
signals and DC voltages. When the DC voltage moves up and down
(even at a slow rate) we call it an AC voltage or AC signal or a
rising and falling voltage and when it rises and falls faster,
we call it a "signal" or waveform. 2.via a capacitor - this is also called RC coupling
(Resistor-Capacitor coupling) - only passes AC signals -
fluctuating signals - rising and falling signals. 3.via a transformer - called Transformer Coupling or
Impedance Coupling or Impedance Matching - only passes AC
Fig 12 shows two stages with a capacitor coupling the output of
the first to the input of the second. This is called
Capacitor Coupling or Resistor-Capacitor Coupling (RC
The increase in the size of the
waveform at three points in the circuit is also shown.
The waveform is inverted as it passes through each transistor and this
simply means a rising voltage will appear as a falling
voltage and after two inversions, the output is in-phase with the input.
We have already explained the fact that a capacitor only works once and
has to be discharged before it works again. When the first
transistor turns off a little, the voltage on the collector rises and
the resistor pulls the left lead of C2 UP. The right-hand lead can only
rise to 0.7v as the base-emitter voltage does not rise above 0.7v.
This means C2 charges and during its charging, it delivers current to
the second transistor.
When the first transistor turns ON, the collector voltage drops and C2
passes this voltage-drop to the base of the second transistor. But the
transistor does not provide a path to discharge the capacitor fully so
that when the capacitor gets charged again, it is already
partially charged and it cannot activate the base of the second
transistor to the same extent as the first cycle.
This means a lot of the energy available at the collector of
the first transistor is not delivered to the second stage. That's why
capacitors produce losses between stages. They are simply an
inefficient way to transfer energy. To make them efficient, they
must be discharged fully during the "discharge-part" of the
However enough is delivered to produce a gain in the second stage to get
an overall gain of about
70 x 70 for the two stages.
The value of C2 will be from 10n to 10u, and the larger capacitance will
allow low frequencies to be passed from one stage to the other.
Fig 13 provides a guide to the values of current
that will be flowing at 3 important sections of the circuit.
The input current to operate the first transistor will be about
3uA. This is worked out on the basis of the current required to
saturate the transistor with a 22k load. The collector-emitter current
5/22,000 = 200uA. If the gain of the transistor is 70, the input current
The only time when energy passes from the first stage to the second is
when transistor turns OFF. The collector voltage rises and the 22k pull
the 100n HIGH.
The maximum current that can be delivered by the 22k is 5v/22,000=
200uA. This is the absolute maximum for a very small portion of the
cycle. However it is important to realise it is not the transistor that
passes the current to the next stage but the load resistor.
The gain of the second stage is not the deciding factor for the output
current but the value of the 2k2 load resistor. This resistor will
deliver a maximum of 2,000uA (2mA) and that is how a 3uA requirement at
the input of the circuit will deliver 2mA at the output.
You can see it is not the gain of the transistors that produce
the output current but the value of the load resistors. The transistors
play a part but the limiting factor is the load resistors (and the
transfer of energy via the capacitor). This is not
always the case but applies in the above circuit.
We will now explain an emitter-follower stage and show how it works.
An EMITTER-FOLLOWER is an NPN transistor with the collector
connected to the positive rail. (You can also get PNP EMITTER-FOLLOWER stages - see below).
Both can be called a COMMON COLLECTOR stage.
Fig 14. An
The names are the SAME
Fig 14 shows an
The load is in the emitter and as the base is taken higher, the emitter
follows. But the input and output voltage signals are the SAME amplitude!
You would ask: "What is the advantage of this?"
You only need a small amount of "lifting power" to raise the base and
the emitter rises with 100 times more strength. The voltage waveform
stays the same but the CURRENT waveform increases 100 times.
The voltage on the emitter is always 0.7v lower than the base and the
base can be as low as 0.8v and as high as 0.5v less than the supply
voltage. This gives the possibilities of producing an enormous "swing."
In the common-emitter stage the transistor is only active
when the base rises from 0.55v to about 0.7v but
in the Emitter-Follower stage it rises from 0.8v to
nearly rail voltage.
This means the stage does not produce a higher output voltage but it
does produce a higher output CURRENT.
We mentioned before the current amplification of a stage was not
dependent on the transistor characteristics but the value of the load
resistor. In an Emitter-Follower stage we can quite
easily get a current gain of 100 or more.
Why do we want "Current Gain?"
We need current to drive a low resistance load such as a speaker.
Fig 15. A transistor
driving a speaker
shows an 8 ohm speaker as the load in the emitter. If the gain of the
transistor is 100, the 8R speaker becomes 8x100 = 800 ohms on the base
lead. In other words we see the circuit as "800 ohms."
See this link for
the answer to a constructor. He wanted to increase the output
from his mobile handset.
For an emitter-follower circuit, we know the base can rise and fall by an amount equal to about rail
voltage. 2. For a common-emitter stage the collector rises
and falls by an amount equal to rail voltage.
So, why not connect the two stages together without a capacitor?
We know a capacitor has considerable losses in transferring energy
from one stage to another and removing it will improve the transfer of
Fig 16. Two directly
Fig 16. We now
have two stages directly connected together.
The first transistor does not deliver energy to the second stage but the
LOAD RESISTOR does.
value of the load resistor pulls the base of the second transistor UP and
this delivers current to the second transistor and the transistor
amplifies this 100 times to drive the speaker.
Fig 17. The load resistor
and the effective load of the speaker
Using mathematics we can work out the effective load of the 8 ohm
speaker as 8 x 100 = 800 ohms. To put at least half rail voltage
into the speaker, (so the speaker can get the maximum
higher voltage and the maximum lower voltage without distorting) the LOAD
resistor has to be the same value as the "emitter follower."
This is a simple voltage-divider calculation where two equal
value resistors produce a voltage of 50% at their mid-point.
This means the LOAD resistor for the first stage has to be 800 ohms.
Fig 18. The load resistor
is 800 ohms
Fig 18 shows the circuit with 800R load resistor
in the collector of the first transistor.
The final requirement is to select a base-bias resistor for the first
stage to produce approx mid-rail voltage on the collector.
This is generally done by experimentation.
We mentioned the capacitor separating two stages cannot be
discharged fully and thus it does not provide very good transfer of
energy from one stage to the other.
An improved concept is to directly couple two stages - and remove the
This is called DIRECT COUPLING or DC coupling and the circuit will
process DC voltages (the press of your finger as shown above) and AC
voltages (as shown by the sine-wave signal shown above). When a
capacitor connects two stages they will only amplify AC signals.
There are many ways to directly connect two transistors and we will
cover the simplest arrangement. It is an extension of
Fig 18 above, because this arrangement has very
good characteristics as the two stages transfer 100% of the energy
due to the absence of a capacitor.
Fig 19 shows the previous directly-coupled circuit with a load resistor
replacing the speaker.
We have already learnt the common-emitter stage provides a voltage gain
of about 70 but the emitter-follower stage has a voltage gain of only 1.
We can improve this by putting two resistors on the second transistor
and changing the stage into a common emitter arrangement.
This time we get the advantage of the base being able to move up and
down so it matches the collector of the first transistor. It also
provides a higher voltage gain by adding a collector resistor
and taking the output from the collector. The voltage gain of
the second transistor will not be as high as the first stage but we have added the
advantage of direct coupling (called DC coupling).
The voltage gain of the second stage is the ratio of resistor A divided
by resistor B. If resistor A is 10k and resistor B is
1k, the voltage gain is 10,000/1,000 = 10.
shows biasing of the first transistor has been taken from the emitter of
the second transistor. This does not save any components but introduces
a new term: FEEDBACK
(actually NEGATIVE FEEDBACK).
Negative feedback provides stability to a circuit.
Transistors have a very wide range of values (called parameters) such as
gain and when two
transistors are placed in a circuit, the gain of each transistor can produce an
enormous final result when the two values are multiplied together.
To control this we can directly couple two transistors and take the
output of the second to the input of the first.
voltage on the base of the first transistor rises, the voltage on the
collector drops and this is transferred to the
second transistor. The voltage on the emitter of the second transistor
drops and this is fed back to the base of the first transistor to oppose
the rise. Obviously this arrangement will not work as the voltage being fed back
is HIGHER than the signal we are inputting, but if we add a 220k
we can force against the feedback signal and produce an output.
Fig 23. We have added a capacitor (electrolytic) to the emitter of the second
transistor. Let's explain how this electrolytic works.
An electrolytic is like a miniature rechargeable battery.
It charges very slowly because it is a large value.
Initially it has 0v.
The circuit starts to turn ON by current flowing through the load
resistor and this turns on the second transistor. (The first transistor is not
turned on AT ALL at the moment).
The base rises and pulls the emitter up too. And when the emitter is about 0.7v, this voltage is
passed to the first transistor via the 220k and the first
transistor starts to turn on. This causes current to flow through the
collector-emitter leads and pulls the voltage on the base of the second
transistor down to about 1.4v
This is how the two transistors settle, with the voltages shown in
The electrolytic has 0.7v on it and when a signal is delivered to the
base of the first transistor, it is amplified and passed to the emitter
of the second transistor. Normally the emitter would rise and fall as
explained in the above circuits and the result would be heard in the
speaker. But the electrolytic takes a long time to charge (and
discharge) and it resists the rise and fall of the signal.
This means the signal cannot rise and fall at the emitter.
In other words we have placed the second transistor in a stage very
similar to the first stage we described a COMMON EMITTER.
Since the emitter voltage does not rise and fall, it does not pass a
signal through the 220k to the base of the first transistor. This means
our input signal is not fighting against the feedback signal and it has
a larger effect on controlling the first transistor. This gives the
first transistor a bigger gain.
A common emitter stage has a voltage gain of about 70-100 and we now
have one of the best designs. Two common-emitter stages, directly-coupled
(DC) and with very HIGH GAIN. The feedback only controls the DC
voltages on the two transistors and does not have an effect on the AC
Fig 24a is the best circuit you can get
for amplifying a signal. The two transistors are biased via
the 470k feedback resistor so they are turned ON and ready to
amplify the signal. There is no capacitor between the two
transistors so the overall gain is very high.
Fig 24 shows typical values for biasing the two transistors.
This circuit has been tested with a speaker as the input device.
It produces 2mV with a whistle at 30cm and the output produced a
sinewave of 3,000mV (a gain of 1,500)
The component values are show in Fig 24a:
Fig 24a - the
circuit you can get.
This circuit is also
called a WIDEBAND AMPLIFIER because it
will amplify all frequencies.
From what you have learnt, you can see the
mistakes and/or the voltages in the following circuit:
The two joined transistors create a Darlington transistor
and this is just a normal transistor with a large gain.
The 330R discharges the 100u and it will only discharge it a very small
amount. This means the electro can only be charged a very small amount
during the next cycle
and the output will be very weak.
It is the 330R that determines how much (little) energy gets
delivered to the speaker. The 330R has to be 15R to nearly fully discharge
You can work out the voltage on the various points in this
circuit by referring to the examples we have already covered.
Fig 27. This is
a practical example of the circuit we have discussed. It is a
MICROPHONE AMPLIFIER (also called a pre-amplifier stage).
Here is the same circuit used as a POWER AMPLIFIER.
Both transistors are common-emitter configurations and the
circuit produces high gain due to the DC (direct) coupling.
You can create a circuit with a FIXED GAIN by selecting values
for the gain of each stage. This is calculated by dividing the
collector resistor by the the emitter resistor.
For the first stage, the gain is 22,000/220 = 100. The gain of
the second stage is 10,000/470 = 20. The gain for the two stages
is 100 x 20 = 2,000. See
for more details.
The POWER of a SIGNAL
Before we go too much further, we need
to talk about the POWER OF A SIGNAL.
What is a SIGNAL?
A Signal is an input voltage.
It may be the signal for the "input" of the amplifier in Fig 27a
above, or it may be the resistance of your finger in the
circuits above, or it may be the signal from an
electret microphone, or an unknown signal
driving a single stage shown above (as a sinewave).
A signal may be an audio waveform with a very small
amplitude or a DC voltage from a switch or a digital
signal from a chip or the output from one of the stages
In all these instances we have described the amplitude
of a signal. The amplitude is the VOLTAGE of the
But a signal consists of a VOLTAGE and comes with a
value of CURRENT. This current may be very small (such
as from an electret microphone) or it may be very high
(such as from a switch).
In most cases we do not talk about the value of current
associated with the signal. Mainly because it is a very
complex problem, matching-up the
"current-capability" of the signal with the "current
requirement" of the following stage.
At this point we will simply say that ALL signals come
with a VALUE OF CURRENT. And this is called "The Power
of a SIGNAL." In other words: The STRENGTH of a Signal"
or the "Driving capability of a signal.
We can also say a signal is "very weak" or "delicate" or
"strong" or "has good driving capability."
Some signals will drive a LED or speaker while others
need to be amplified before they can be used.
In most cases the "driving power of a signal" is unknown.
It is not provided as a specification. And yet its
value is MOST IMPORTANT. In most cases you cannot work
out the current-capability of a signal by looking at the
device generating the signal. For instance, if the
signal comes from a magnetic pick-up coil, or the output
of a pre-amplifier where the circuit is not provided.
That's why the matching of a signal to an input circuit
is so complex and is a topic for an advanced section of a
In the meantime we will assume the signal and the input
of the stage it is driving, has the appropriate input
impedance so the signal is not attenuated (reduced)
If a signal has a high current it can be connected to
a high or low impedance input and the amplitude will not
If a signal with very little current
is connected to the input of an amplifier and the input
has a low impedance, the amplitude of the signal will be
reduced. That's why the input needs to be as high
We really can't say too much more as this is a very
complex area of discussion. It is much easier to talk
about voltage levels.
USING PNP TRANSISTORS
A PNP transistor can be used
in the 2-Transistor DC amplifier studied above. It does not produce a
higher gain or change the output features of the circuit in any way but you may
see an NPN and PNP used in this configuration and need to know how they work.
Firstly we will discus how a PNP transistor works. All those things you
learnt in the first set of diagrams can be repeated with a PNP
transistor. The circuits are just a mirror-image of each other and
the transistor is simply "turned-over" and connected to the supply rail.
Study the following circuits to understand how a PNP transistor is
Fig 28. PNP Transistor
The symbol for a PNP transistor has the arrow pointing towards
Fig 29. PNP
shows the equivalent of a PNP transistor as a water
valve. As more current (water) is released from the base, more water flows from
the emitter to the collector. When no water exits the base, no water
flows through the emitter-collector.
Fig 30. PNP
connected to the power rails
shows a PNP transistor with the emitter lead connected to the power rail.
The collector connects to a resistor called a LOAD RESISTOR and the
other end connects to the 0v rail or "earth" or "ground."
The input is the base and the output is the collector.
Fig 31. PNP Transistor biased with a
"base bias" resistor and a LOAD resistor
shows a PNP transistor in SELF BIAS mode. This is called a
COMMON EMITTER stage and the resistance of the BASE BIAS
RESISTOR is selected so the voltage on the collector is
half-rail voltage. In this case it is 2.5v.
Here's how you do it. Use 22k as the load
Select the base bias resistor until the measured voltage on the
collector is 2.5v. The base bias resistor will be about 2M2.
This is how the transistor gets turned on by the base bias resistor:
The base bias resistor allows a small current to pass from the emitter
to the base and this
makes the transistor turn on and create a current-flow though the
This causes the same current to flow through the load resistor and a
voltage-drop is created across this resistor. This raises the voltage on
This causes a lower current to flow from the emitter to the base, via the
base-bias resistor, and the transistor stops turning on a slight amount.
The transistor very quickly settles down to allowing a certain current
to flow through the emitter-collector and produces a voltage at the
collector that is just sufficient to allow the right amount of current
to flow from the base. That's why it is called SELF BIAS.
Fig 32. Turning
ON an PNP transistor
Fig 32 shows the transistor being turned on via a finger.
Press hard on the two wires and the LED will illuminate brighter. As you
press harder, the resistance of your finger decreases. This allows more
current to flow from the emitter to the base and the transistor turns on harder.
Fig 33. Two transistors turning ON
shows a second transistor to "amplify the effect of your finger"
and the LED illuminates about 100 times brighter.
Fig 34. Adding a capacitor
shows the effect of putting a capacitor on the base lead. The
capacitor must be uncharged and when you apply pressure, the LED
will flash brightly then go off. This is because the capacitor
gets charged when you touch the wires. As soon as it is charged,
NO MORE CURRENT flows though it. The first transistor stops
receiving current and the circuit does not keep the LED
illuminated. To get the circuit to work again, the capacitor
must be discharged. A large-value capacitor will keep the LED illuminated for
a longer period of time as it will take longer to charge
Adding a capacitor to the
Fig 35 shows the effect of putting a capacitor on the
output. It must be uncharged for this effect to work. We know from
Fig 33 that the circuit will stay on constantly when the wires are
touched but when a capacitor is placed in the OUTPUT, it gets charged
when the circuit turns ON and only allows the LED to flash.
NPN/PNP AMPLIFIER A
2-Transistor DC amplifier can be
constructed using an NPN and PNP set of transistors.
Fig 36 shows how an NPN-PNP set of transistor is turned on.
You can think of the "turning ON" this way: The base of the NPN
get "Pulled UP" and the base of the PNP gets "Pulled DOWN."
It does not matter how you refer to the operation of the circuit, you
must be able to "SEE" how the circuit works so you can
more-complex circuit working too!
Fig 37 shows biasing on the base of the first transistor and the "in"
and "out" leads have been identified.
This circuit has a very high gain and if "general purpose" transistors
are used with a very high spread of gain for each transistor, the result
will be a very wide range of voltages on the output terminal. If
each transistor has a gain of 100, a change of 1mV on the input will
result is a voltage change of 0.001 x 100 x 100 = 10v. We don't
have a 10v supply so, this type of circuit is very UNSTABLE!
We need to design a circuit that has FEEDBACK so the output voltage will
remain within the voltage of the supply. This feedback is called
NEGATIVE FEEDBACK as it opposes an input signal to provide correction or
stability. Later we will talk about POSITIVE FEEDBACK and show what an
amazing difference it creates - the circuit behaves totally differently.
Fig 38. This circuit does
will not work because the base of the NPN transistor is not
turned on when the circuit is switched on.
This is one of the things you have to look for when designing a
Fig 39. The voltages
Fig 39 has
a voltage-divider network on the base of the NPN transistor. It
turns the first transistor ON and this turns the PNP transistor
ON until the voltage at the join of the 3k3 and 1k puts a
voltage on the emitter of the first transistor to start turning
This is a point we have to explain.
There are two ways to turn ON an NPN transistor.
1. Hold the emitter fixed and RAISE the base voltage.
2. Hold the base fixed and LOWER the emitter voltage.
In Fig 39
the base is weakly fixed by the voltage divider made up of the
1M and 220k and even though the base can move up and down a little bit, we will
assume the voltage is constant. If we raise the emitter voltage, the
transistor will be turned off. This is what the FEEDBACK voltage via the
3k3 does. It raises the emitter voltage and turns the NPN transistor OFF
slightly so an equilibrium point is reached where the two
transistors are turned on a small amount and
if one gets turned on a
little more, the other sends signal to turn it OFF. This is not a
practical circuit as an increase of 1mV on the input will produce a
large change on the output and this will be reflected back to the
emitter of the first transistor to cancel the input voltage.
Fig 40. A practical
By changing the value of the feedback resistors we get Fig 40. The values are now 10k and 100R.
This gives a ratio of 10,000:100 or 100:1 and it means the output can
rise 100mV before the emitter gets 1mv to cancel the input voltage. This
means the amplifier will have a gain less than 100 but provides a very
stable set of voltages.
Fig 40a. Another practical
Here is an amplifier with the same DC
biasing as Fig 40 but with a lower overall gain (2,200:100
or 22:1) and high-frequency feedback
(attenuation) via the 2n2 capacitor.
The CAPACITOR The capacitor is a
very complex item to discuss because it
performs so many different effects, depending on its value and
where it is placed in a circuit.
However one of the most important concepts is to see that a
signal on the left side will pass through the capacitor and
appear on the other side with about the same amplitude, if
the signal is fast-acting. In other words: high-frequency.
This is shown in the first animation where the movement of
the first person is transferred directly to the second
If the signal has a low-frequency, is will get charged and
discharged during the cycle and the amplitude on the output will be small. This is shown in the second animation where
the capacitor is charging and discharging and the second
person is seeing a small effect.
The circuit shows a Schmitt
Trigger arrangement. This is covered
The Schmitt Trigger is FAST ACTING and this means
the signal on the left-plate of the capacitor will be delivered
to the output of the circuit.
Delivering 100% of the amplitude will depend on:
1. The value of the
2. The rise-time of the signal and
3. The load on the
right-plate of the capacitor.
At the moment we just need to VISUALISE the way the capacitor
The capacitor will pass a "spike" or "signal"
from one stage to another:
The capacitor sending (transferring) a PULSE to the
This is a VERY IMPORTANT concept.
The capacitor in the diagram is
transferring the "drop in voltage" on the left-plate to the
The voltage on the input to the DIGITAL NAND GATE is initially rail
voltage and it must drop to less than 30% of rail voltage for
the gate to see a LOW.
The value of the capacitor is chosen so this will occur.
When the voltage on the left-plate drops, the capacitor will
begin to charge via the 1M resistor and the circuit-designer
must make sure the charging of the capacitor will be very small during the time
when the voltage drops so the right-plate will send a 30% rail
voltage to the gate. If the value of the capacitor
is too small, it will get charged very quickly and the right-plate will only
drop a small percentage of rail-voltage.
MEASURING THE VOLTAGE(S)
The voltage on each
line (connection) of a circuit can be measured
with a multimeter. To help you take (make) a
reading, we have written an eBook titled: Testing Electronic Components.
There is a certain amount of skill required to
take a reading and this eBook will help you
If we remove some of
the components from Fig 39 and put a LED on the emitter of
the PNP transistor we have a circuit that will illuminate the LED.
We have already talked about FEEDBACK in terns of NEGATIVE FEEDBACK to
stabilize a circuit. We will now cover a new term called
- it changes the performance of circuit completely.
It makes the circuit OSCILLATE. Negative feedback "kills" a
circuits performance - positive feedback makes it oscillate. It
increases the signal so much that the circuit becomes unstable. This is
Fig 41 shows
a circuit using an NPN and PNP connected via a 1k resistor and turned ON
via a 330k base resistor.
The LED will illuminate.
There is nothing magic about this circuit. It is simply a HIGH-GAIN,
DC-AMPLIFIER using two transistors.
The values of
current are only approximate and show how each section allows an
increasing amount of current to flow.
A current of 100mA is too high for a LED and it will be damaged.
This circuit demonstrates the possible current-flow. If this
current flows for a very short period of time, the LED will not
be damaged. Fig 42 shows how the circuit is converted to an
oscillator or "flasher."
When we connect a capacitor as shown,
an amazing thing happens. The high-gain amplifier turns into an
When the voltage on point "X" is rising, the voltage on point "Y" is
rising TOO. But point "Y" rises much higher than point "X."
This means that if we DIRECTLY join points X and Y, the voltage-rise from point Y
will push point X higher and turn the circuit ON more. This will
continue until the circuit is fully turned ON and the two transistors are
This effect is called POSITIVE FEEDBACK and the circuit will get turned
ON until it cannot turn on any more.
But we haven't joined points "X" and "Y" DIRECTLY
(we have used a capacitor) so we have to start again and
explain how the circuit works.
When the power is applied, the 10u gradually charges and allows a voltage
to develop on the base of the NPN transistor. When the voltage reaches 0.6v, the
transistor turns ON and this turns on the PNP transistor.
The voltage on the collector of the PNP transistor increases and this
raises the right side of the 10u electrolytic and it firstly pushes its
charge into the base of the NPN transistor. Then the 330k takes
over then it continues to charge in the opposite direction via the base-emitter junction of the
NPN transistor. This causes the two transistors to turn ON more. This
keeps happening until both transistors cannot turn ON any more and the
10u keeps charging. But as it continues to charge, the charging current
eventually drops slightly and this turns off the first transistor
slightly. This gets passed to the PNP transistor and it also turns off
slightly. This instantly lowers both leads of the 10u and both
transistors turn OFF.
The 10u is partially charged and it gets discharged over a long period
of time by the 330k resistor and when it starts to charge in the
opposite direction, the base of the first transistor sees 0.6v and the
cycle starts again.
The end result is a very brief flash and a very long pause (while the
capacitor starts to charge again).
As you can see, there is very little difference between the high-gain DC
amplifier we discussed above and the oscillator circuit just described.
That's why you have to be very careful when looking at a circuit, to
make sure you are identifying it correctly.
is the same circuit with the components re-arranged. It is a high-frequency oscillator with an inductor as the load and when the circuit
turns off, the inductor produces a high voltage in the opposite
direction to the supply voltage and this is high enough to illuminate a LED. The
LED will not illuminate on the 1.5v supply so when the LED illuminates,
you know the circuit is
is the same arrangement of the two transistors we have just studied,
but with a third transistor above the two.
We have already seen the importance of charging a capacitor (and then it
must be discharged so that the re-charge will produce a "current-flow.")
That's what the two transistors in the output are doing.
The top transistor charges the electrolytic and the bottom transistor
In the process, the charging and discharging current flows through the
speaker to produce audio.
We have already studied the two lower transistors. The BC327 turns ON
and allows current to pass through the emitter-collector leads and this
discharges the electrolytic.
The top transistor is an emitter-follower and it turns
ON when the
bottom two transistors are effectively "out of circuit."
The base is pulled to the supply rail by the 1k and the emitter follows.
In other words the collector-emitter leads allow current to flow and
this charges the electrolytic. The charging current flows through the
GAIN OF AN EMITTER FOLLOWER STAGE
We have seen the need to provide
current into and out of a speaker to move the cone. This is because
current produces magnetic flux and many items work on magnetic flux,
such as: motors, relays and speakers. And some items need a lot of
current to be activated - especially globes.
Most transistors will provide a CURRENT GAIN of 100 when up to
25% of their rated current flows, but only a gain of 50 for the next 25%
increase in current
and a gain of 30 for the next 25% increase in current and a gain of only about 10
when the maximum allowable current flows.
That's why you have to understand transistor
data-sheets. The gain of a transistor is very low when
maximum current flows.
There is a hidden factor with motors and globes. They take 6 TIMES more current for a globe to start glowing or to start a
motor revolving. This is because the resistance of a cold globe is only one
sixth of its glowing resistance and a motor has a very low resistance
until the back emf (electro-motive force - another name for voltage) produced by the armature, reduces the current-flow.
This means you have to design a circuit that will deliver up to 6 times
the operating current, so these items will turn on.
We explained the 800R LOAD resistor provides the turn-on current for the
speaker in the following circuit. When the BC547 turns off, the current
through the 800R is amplified by the emitter-follower
transistor to drive the speaker. This is a very wasteful
way of operating a circuit as current is always flowing
through the 800R and during part of the cycle, this
current is not achieving any result.
We can design a circuit where this current is provided by a
This is important when we are providing high currents as a
transistor can be turned on to deliver the current and turned off when the current is not required,. This saves energy and
We will look at the following 2-Transistor DC amplifier driving a speaker
(taken from Fig 18) and modify the circuit.
Fig 45. An
emitter-follower driving a speaker
Fig 45. The
drives a speaker.
Fig 46. We replace the speaker with a motor.
We replace the LOAD resistor with a transistor and add a
resistor called a: Current Limiting Resistor. It is designed to limit the current between the first and
second transistors as these will turn ON and allow a very high
current to flow if the resistor is not included.
Fig 48. The current
required by the motor is 300mA. The emitter-follower will have a
gain of 10 and the gain of the other two transistors produces
the set of conditions shown on the diagram.
You can see that very little input current is required to activate the
motor when 3 transistors are used.
Fig 49. The input
current can be supplied from a voltage-divider using a pot (to
adjust the setting) and a Light Dependent Resistor.
We cannot use only 2 transistors as the LDR cannot supply 1mA under low-level light conditions and
that's why 3 transistors are needed.
Fig 50. Dancing Flower
Fig 50. Here is a
commercial version of a 3-transistor circuit.
This circuit was taken from a dancing
flower. A motor at the base of the flower has a bent shaft up
the stem and when the microphone detects music, the shaft makes
the flower wiggle and move.
The circuit will respond to a whistle, music or noise.
The circuit uses a different arrangement to our 3-transistor
design and we will discuss the differences.
It is very easy to get a change in voltage
from an input device such as an LDR or electret microphone.
Simply add a LOAD resistor and "tap off" the change in voltage
at the join of the two components.
There is also a very small change in CURRENT at the join of the
two components (but we normally refer to the change in voltage).
We can amplify this voltage via two transistors to get a voltage
equal to rail voltage. This is not a problem for 2 transistors.
But we also need to amplify the CURRENT to operate a motor. We
cannot get enough CURRENT GAIN with 2 transistors and that's why
we need 3 transistors.
The change in voltage must be passed through 3 transistors to
get the CURRENT GAIN
required by a motor.
Both circuits (Figs 49 and 50) appear to perform the same but you need to look at
the voltage drop across the leads of the output transistors to see how the
two circuits compare.
There are two important values for a FULLY-TURNED-ON transistor:
Fig 51. The characteristic voltage drops across a
Fig 52. The voltage losses
across the output transistor
The emitter-follower design
(the first circuit) has a total
voltage drop of 0.8v and the motor will see a maximum of
2.2v. The motor in the common-emitter
design will see a maximum of 2.8v.
You can see the advantages and disadvantage of each design.
Because the emitter-follower has a 0.6v drop between base and
emitter, it is generally used in a PUSH-PULL arrangement as
we will see in Fig 53, to charge and discharge the electrolytic or in
an H-Bridge to drive a motor forward and reverse as shown in Fig
54. But when a common-emitter stage is used, the output voltage
as a LINEAR AMPLIFIER
The EMITTER FOLLOWER stage can also be called a LINEAR AMPLIFIER
as the output follows the input voltage EXACTLY except it is
about 0.6v lower than the input. The output has about 100 times
more current capability than the input and this gives it the
name AMPLIFIER. See
Emitter-Follower for circuits.
A Linear Amplifier can amplify the current from a pot to
create a very simple Motor Speed Controller or LED Illuminator:
The actual result in increasing the speed of the motor or the
brightness of the LED will not seem to be linear because they do
not respond in a linear way to an increase in voltage. The pot
also has to be linear to produce a linear output.
Motor Speed Controller
PUSH-PULL STAGE also called PUSH-PULL AMPLIFIER
We have studied the emitter-follower in Figs 45
We have also shown how to connect a PNP transistor to
the power rails. (It is basically a mirror-image of the
NPN transistor.) Combining these facts we can produce a
circuit consisting of two emitter-followers as shown in
The top emitter follower is an NPN transistor and the
lower emitter-follower is a PNP transistor. The is
called a PUSH PULL output stage or PUSH PULL
or Complementary-Symmetry output stage.
voltage will be 3.2v
The Push-Pull stage can be connected
to the output of a micro to get a higher DRIVE CURRENT,
as shown in Fig 52aa.
The Push-Pull stage does not increase the voltage - the
voltage is slightly lower than the microcontroller
supply voltage but the current will be increased by a
factor of about 100 to 200.
The output voltage is reduced by 0.6v due to the
base-emitter voltage-drop plus the output of the micro
is about 0.3v less than the supply rail.
This occurs for both HIGH and LOW, making the output 0.3
+ 0.6 + 0.3 + 0.6 = 1.8v less than the supply = 3.2v
voltage will be 4.4v
The output voltage of the micro can be increased to 4.4v
by placing the PNP transistor above the NPN transistor
as shown in Fig 52ab.
Both transistors will turn ON during the time when the
micro is changing from HIGH to LOW but since this is
very brief, (less than 1/10th microsecond) they will not
Fig 52ac. Quiescent current zero
when micro output HIGH
In the circuit above, the output of the
micro cannot be turned off as one transistor will be
The solution is to "AC couple" one of the transistors as
shown in Fig 52ac and this will allow the micro
to go HIGH and the output section will turn off.
The 1u will have an impedance of about 200 ohms when the
circuit is operating at 1kHz.
Fig 52b. Push-Pull Current
Fig 52b shows a very clever variation on the Push-Pull
circuit described above.
It uses a low-value resistor between the collector of
the driver transistor and output. This resistor
transfers the low-level signals directly to the speaker.
As the signal-level increases, the output transistors
come into operation.
This arrangement removes cross-over distortion and uses
It is called CURRENT DUMPING.
Lifting the Input line will raise the
output line and it will have "100 times more strength."
Lowering the input line will make the output line go
down with "100 times more strength."
In other words this circuit turns a "weak line into a
This feature is also called IMPEDANCE MATCHING.
The circuit is also called a PUSH PULL OUTPUT as one
transistor "pushes energy" into a device
to the output) during one half of a cycle while the
other transistor will "pull energy" out of a
device. This is one of the ways to charge and discharge
a capacitor on the output and any device connected to
the other side of the capacitor will see the AC waveform
and become active. This is shown in Fig 53:
Fig 53 PUSH-PULL to charge/discharge the 100u electrolytic
Fig 54 PUSH-PULL
driving the motor forward/reverse
Fig 54aa PUSH-PULL Amplifier
Fig 54aa is a
3-Transistor Push-Pull amplifier.
When the supply is turned on, current flows though the 8R
speaker and through R4 to the base of T2. This pulls the base of
T2 towards the 9v rail and the transistor rises to nearly the 9v
rail. The voltage on the emitter of T2 is 0.6v lower than the
base and this pulls the emitter of T3 towards the 9v rail. The
base of T3 is 0.6v lower than the emitter.
This is as far as we can go with the current-path at the moment
and we now have to go to T1.
The join of the two emitters has a voltage near the 9v rail and
this voltage is passed to the base of T1 via the 82k resistor.
The 82k resistor forms a voltage divider with 12k and the
resulting voltage at their join is sufficient to put 0.6v on the
base of T1. This turns ON T1 and the voltage between collector
and emitter drops to a low value. The exact value will be shown
in a moment.
We can now go back to the base of T3 and continue the
current-path (also called the voltage path) from the 9v rail to
the 0v rail.
T1 pulls the base of T3 towards the 0v rail.
We now have three transistor that all turn on. They are not
fully turned on but partially turn on.
The exact amount of “turn-on” for each of the transistors is due
to the 83k and 12k biasing components and diodes D1 and D2.
Here’s how the DC coupled amplifier self-adjusts to a state
called the QUIESCENT STATE. This is the state where some of the
components adjust the “turn-on” of other components and the
circuit reaches a point where the voltages settle down and reach
a stable value and the current is a constant minimum value.
The voltage at the midpoint of the two output transistors is
fairly high and this creates a slightly higher voltage on the
base of T1. This turns on T1 slightly more and the voltage on
the collector drops. This lowers the voltage on the base of T3
and the emitter voltage drops. This lower voltage is passed to
the base of T1 and the transistor turns OFF slightly.
This is how the three transistors adjust themselves to a final
The exact final voltage is called a DESIGN VOLTAGE and designer
of the circuit want the voltage on the join of the two emitters
to be half-rail-voltage.
This allows the circuit to rise and fall and reproduce a
waveform without clipping or cutting off the top or bottom of
To get the circuit to sit with the output (the join of the two
emitters) at 4.5v, the values of R2 and R3 have been selected.
We now have the circuit sitting, ready to amplify a signal.
The output stage is called PUSH PULL because one transistor
pushes current through the winding of the speaker via the 100u
electrolytic and the other transistor pulls current through the
speaker via the electrolytic.
You could connect the speaker directly to the output of the
stage and remove the electrolytic. The circuit would work just
However if the speaker is connected directly, a voltage of 4.5v
will be paced across the speaker and this voltage will cause a
current to flow in the winding of the peaker (the voice coil)
and the cone will be pulled in. If we try to reproduce a
waveform, the cone is already partially pulled-in and it will
not reproduce half of the waveform.
In addition, this constant current will heat up the voice coil.
By adding the 100u, we remove the Dc component of the output and
only the AC (waveform) will be passed to the speaker.
Now we have to understand how an electrolytic passes energy
(current) to the speaker.
If you connect an electrolytic and speaker directly to a supply,
you will hear a “plop” This is the electrolytic charging and the
charging current flows through the speaker and produces the
But after a very short time the electrolytic is charged and no
ore current flows.
Even if you remove the supply and connect it again, no sound
will be reproduced because the electrolytic is already charged.
The only way to hear another plop, is to remove the components
and short between the power leads.
When the supply is re-applied, you will hear another plop.
To get sound from the circuit, this is what it has to do.
Firstly it has to charge the electrolytic. Then it has to
discharge the electrolytic.
As you can see from the circuit, the lower transistor charges
the electrolytic and the top output transistor discharges the
Now we have to drive the two transistors so that they charge and
discharge the electrolytic.
To charge the electrolytic, T1 turns ON and pulls T3 towards the
This is the easy part.
How do you pull T2 UP so that it discharges the electrolytic?
This is how it is done. It is very clever.
Connected between T2 and T3 are two diodes. Each if these diodes
has a voltage drop of 0.6v.
This voltage drop is exactly the same voltage as between the
base and emitter of the two transistors in the output.
This means we can directly pull on the base of the top
transistor, just like we are directly pulling on the base of the
lower output transistor.
Now we have a situation where we can pull down on both
transistors and this will turn ON the lower transistor and turn
OFF the upper transistor.
This is done when T1 turns ON.
When T1 turns OFF, the top transistor is pulled HIGH via the
That’s how it works.
Fig 54a Two
Push-Pull circuits driving the primary of a transformer
Fig 54ab. A
High-current Driver stage - faulty design
Fig 54ab shows an actual
high-current driver stage of a 500 watt inverter, taken from the
The designer of the circuit has tried to provide a high-current
capability for the 2N6277 by driving its base via a 2N3055 and
TIP122. Theoretically the base current for the TIP122 will be
only a few milliamps as the gain of the Darlington transistor
and 2N3055 will deliver a high base-current to the output
However this circuit is a faulty design.
For the 2N3055 to deliver current into the base of the 2N6277,
it must have a collector voltage that is higher than the
And for the TIP122 Darlington transistor, it must have a
collector voltage that is higher than its emitter.
The minimum collector-emitter voltage for a Darlington
transistor is 2v.
The base-emitter voltage for a 2N6277 is about 1.8v to 3.5v (use 2.1v) and for a
2N3055 it is about 0.7v.
54ab-1. A High-current Driver stage - improved design
This means the TIP122 can only turn on when the collector
voltage is 0.7v + 2.1v + 2v = 4.8v.
This means the
collector of the 2N6277 cannot be less than 4.8v.
This faulty design can be fixed by taking each of the
transistors to the supply-rail via a suitable resistor.
The collector-emitter saturation voltage for the 2N6277 is
between 1v - 3v.
This means the transformer sees a higher voltage.
This improvement will make an enormous difference in the
output capability of the circuit and reduce the heat
generated in the output transistor(s).
When using a 2N6277 transistor on
each leg of the output, the base must receive about 5 amps to
fully saturate the transistor for 40 amp collector current.
The circuit on the left is an ideal way to drive an inductor.
The transistor will handle 40 amps to produce a 500 watt
The voltage on the collector will be about 1.6v so that for a
12v supply, the inductor will see 12v - 1.6v = 10.4v.
Fig 54ab-3. This
arrangement is a very
However when you
drive an output transistor as shown in Fig 54ab-3, two
To deliver 5 amps to the base of the 2N6277, the TIP122
transistor has a saturation-voltage across its collector-emitter
leads of about 4v.
We will explain this in a moment. Firstly e have to go to the
2N6277 and cover the fact that the base-emitter voltage will be
about 3v for a collector current of 40 amps.
The TIP122 is now sitting 3v above the 0v rail and the collector
must see a voltage of 7v so that it can deliver 5 amps to the
base of the 2N6277.
This means the collector of the 2N6277 cannot go below 7v.
In other words we are losing 7v from the 12v supply and only 5v
will be available for the inductor.
This method of driving an output transistor is a very bad
Fig 54b shows a free-running
multivibrator configured so the transistors drive a
transformer in Push-Pull
THE TOTEM POLE OUTPUT STAGE A slightly different push-pull output stage
can be created with two NPN transistors. It is called a
Totem Pole Output stage.
Fig 55a. When
the input is less than 1v, the output is pulled high via the 1k
resistor and the "strength" of the "pull-up" will be 1,000/100 =
approx 10 ohms.
When the input reaches 1.4v, the output is pulled low via the
lower transistor and will about 0.2v from the 0v rail. The
"strength" of the "pull-down will be about equivalent to a 10
This is about the same as the output driving capability of a
normal Push-Pull arrangement, however there is a mid-point where
both transistors are turned on at the same time and this
produces a large current that can overheat the transistors or
Fig 55b "Open Collector" Output
Fig 55b. The
circuit Fig 55a above is used in many applications because it
will drive the output HIGH and LOW. In other words
a transistor will pull the output HIGH and the other transistor
will pull it LOW. The output of many (most) integrated circuits
will SINK (pull the output LOW) and SOURCE
(pull the output HIGH). This is called PUSH-PULL or
more-accurately: TOTEM-POLE output.
When it is SINKING, current flows through the load (from the
supply) and the output acts like a "switch" to connect the load
to the 0v rail.
When it is SOURCING, the output delivers the current to the LOAD
and the load is connected to the 0v rail.
But if the output can only SINK as shown in Fig 55b,
it is called "OPEN COLLECTOR."
This means a LOAD must be connected to the supply rail and the
output will "switch" (connect) it to the 0v rail.
(actually Bridge Biasing)
Another way to
connect a transistor to produce a "stage" is called a BRIDGE. It
consists of 4 resistors:
Fig 56. A BRIDGE arrangement
consisting of 4 resistors
Fig 56. We have already studied the purpose of Ra and Rb
to produce a voltage on the base of the transistor. If they are
the same value, the base voltage will be half the supply. We
also know the emitter voltage will be 0.7v lower than the base.
This will produce a current through Re and the same current will
flow in Rc. We can now work out the voltages on the three leads
of the transistor.
But that's not the point of our discussion at the moment.
We want to know how to work out the values of Ra, Rb, Rc and Re.
There are two types of "bridges."
1. A small-signal bridge and
2. A medium or high-power signal bridge.
A small-signal bridge deals with signals that do not have much input-current. We have already learnt the ability of a stage to pass a
CURRENT from one stage to the next stage depends on the value of the
LOAD resistor (for the common-emitter stages we have covered).
If this current is very small, we do not want to attenuates it (reduce
it) by making the input of our bridge stage LOW IMPEDANCE (low
resistance). If the values of Ra and Rb are low, any signal being
applied to this stage will be partially lost (reduced - attenuated) by
the value of the voltage-divider. That's why the resistors have to be
as high as possible.
They are generally about 470k to 2M2.
Suppose we make Ra = 1M and Rb = 470k.
Biasing the BASE
Fig 57. The base is
biased at about 1/3 rail voltage.
The emitter will be about 0.7v below the base voltage so the
collector can produce a swing of about 50% of rail voltage.
This is the normal way to bias this type of stage.
Fig 57a. The emitter resistor provides NEGATIVE FEEDBACK
Fig 57a. In the
Bridge Circuit, 4 resistors bias the transistor and Re is
the EMITTER RESISTOR.
It is also a NEGATIVE FEEDBACK resistor and works like this:
When the voltage on the base rises by 10mV, the transistor turns
on more and the current through the collector LOAD resistor Rc
increases and the same current flows through the emitter
This causes a slightly higher voltage to appear across this
resistor and the voltage on the emitter rises.
We have already discussed how to turn ON a transistor or turn
OFF a transistor and when the voltage on the emitter increases,
the transistor is turned OFF slightly. This means the 10mV rise
on the base may be offset by a 2mV rise on the emitter and the
transistor will not be turned on as much. This is the effect of
The gain of the
stage is the ratio of Rc/Re If Rc=22k and Re=470R the gain is 46.
It does not matter if the transistor has a gain of 200 - the stage is
limited to a gain of 46. The actual DC voltage on the leads of the
transistor depends on the quality of the transistor (its gain) and we
will not be concerned with these values as the stage will have a
capacitor on the input and output and it will be biased by the 4 resistors.
A stage-gain of 46
Fig 58. shows a stage
and Re=470R, producing a stage-gain of 46. The actual voltage on
the collector will depend on the gain of the transistor.
Fig 59. A
stage-gain of 100
If we use the
values: Rc=22k and Re=220R
the gain will be 100.
A stage-gain of
200 or more
Fig 60. If
we add an electrolytic across the emitter resistor, the emitter
will not move up and down when a signal is processed and this
makes the transistor similar to a common-emitter stage. The
transistor will now have a stage-gain similar to its
specification. It may be 200.
The gain of the stage will also depend on the frequency. It will
have a higher gain with high frequencies as the
(resistance) of the 10u will be lower at high frequency.
However the capacitor on the input will produce losses from one
stage to the other and the capacitor on the output will reduce
the gain of this stage.
That's why it is very difficult to specify the gain of this and
any other stage.
In most cases you can count on a gain of 50 to 70 when a stage
is incorporated in a multi-stage design.
Fig 61. A medium-power bridge
When we add the electrolytic, the gain of the stage is not dependent on the values of Rc and Re,
and we can reduce the value Rc (the resistor on the collector) so the stage will pass a higher current
to the following stage.
This stage is called a medium-signal stage. The stage will also have a higher gain at high frequencies.
The electrolytic is called a BY-PASS capacitor because any
signal that appears on the emitter is passed (sent) to the 0v
This capacitor can also be called a SHUNT capacitor as it
"shunts" (sends) the signal to the 0v rail. In other
words, the electro connects the emitter to the 0v rail just like
a very low value resistor (about 10R).
ADJUSTING (SETTING) THE STAGE GAIN
EMITTER DEGENERATION - or EMITTER FEEDBACK
Fig 61a. "emitter resistor" adjusts the gain of
Fig 61a. The
gain of a stage can be adjusted (or SET) to a particular value
by adding an emitter resistor. We have seen in Fig 58, the gain
of a stage is determined by the ratio of:
the resistor in the collector/ the resistor in the emitter. Increasing the
value of the resistor in the emitter, decreases the gain of the
In Fig 57a, we saw this as NEGATIVE FEEDBACK.
This effect is also called EMITTER DEGENERATION as it
reduces the gain of the stage.
On Page 2 of this eBook you will find a program where you can design your own
Transistor Amplifier: Design Your Own Transistor Amplifier
It uses the circuit in Fig 61a to adjust the gain of the
The components in the
rectangle are not really needed when the resistor called:
emitter resistor is used. They only adjust the "setting of
the transistor" slightly up or down between the supply rails.
Fig 61aa. The electrolytic increases the gain at high
shows two circuits with an electrolytic and resistor in
Why have these components been added?
Firstly they will reduce the gain of the stage in circuit "A"
but the high frequencies will be amplified more than the low
frequencies. This is because the capacitive-reactance
(resistance) of the electrolytic will be low at high frequency
and prevent the emitter rising and falling and gives the stage a
higher gain at high frequencies.
In circuit "B" the electrolytic also allows the circuit to
produce a higher gain at high frequencies without changing the
DC biasing arrangements of the 4 resistors.
MORE DETAILS ON THE GAIN OF
Fig 61b. Three circuits
with the same gain.
The three circuits above have (approximately)
the same gain
(amplification). The gain will be about 70-100. Even though the
transistor may have a DC gain of 200-400, the base-bias resistor
is acting AGAINST the incoming signal and this creates a
reduction in gain.
However we are looking at the idle-current (quiescent-current)
and aiming to reduce this to a minimum for long battery life.
The first circuit takes a high
quiescent current because the load resistor is 1k. The second
circuit takes about one-tenth the current and the third stage
Minimum quiescent current is necessary when designing a battery
operated project. But you must also take other things into
account - such as the ability of the stage to deliver the
maximum signal to the next stage.
In this article we will explain the fact that a stage passes
energy (signal) to the next stage via the output capacitor and
the value of the load resistor.
The first circuit above is capable of delivering a high signal to a
next stage in your project whereas the second circuit has only
one-tenth the capability of delivering a signal. And the third
stage delivers even less.
If you don't match-up the driving capability of one stage with
the next, the gain of the stage will be very low and you will
wonder why the project is not working.
Start with a "high-current-stage" (circuit1) and gradually
increase the value of the load resistor and change the value of
the base-bias resistor to get mid-rail voltage on the collector.
A point will come where the transfer of signal is a maximum and
you will have achieved the minimum current for the stage with
In general, a higher rail-voltage will produce higher gain and
this is most-noticeable when increasing rail voltage from about
3v, to 6v to 9v.
Connecting a small-signal stage to a medium-signal stage:
Fig 62. Connecting a small-signal stage to a medium-signal stage
Fig 62. When
describing small-signal and medium-signal stages we are
referring to the size of the waveform (voltage waveform) and
are capable of transferring. The two values normally go together.
In most cases the voltage AND current increase as it
progresses though each stage.
Both stages in Fig 62 produce a high gain but the final gain will
depend on the amount of energy each capacitor will
For instance, the 22k will pull the 10u high but the 47k
discharges the 10u and so it will be partially charged for the
next cycle. This means the energy transfer will only be
equivalent to a load resistor of 47k.
COMMON BASE AMPLIFIER
We have discussed the
importance of matching the output impedance of one stage to the input
impedance of the next stage. When the two are equal, the maximum energy
Suppose you want to match a very low resistance device (such as speaker
or coil) to the input of an amplifier. The speaker may be 8 ohms and the
input impedance of the common-emitter amplifiers we have described are about
500R to 2k.
The two can be connected via a capacitor but we have already mentioned
how a capacitor transfers only a small amount of energy when the two
impedances are not equal. And when the two impedances are so mismatched
as 8:2,000, the transfer may be very poor.
The answer is to use a stage that has a very low input impedance.
That's a COMMON BASE amplifier.
Fig 63. The
common-base amplifier (Common-Base stage) accepts a low value of
resistance on the input and produces a high gain. Since the
input is directly coupled to the transistor, there are no
We have already mentioned two ways to turn ON an NPN transistor.
1. Hold the emitter fixed and RAISE the base voltage.
2. Hold the base fixed and LOWER the emitter voltage.
We are using the second option. The base is held rigid (as far
as signals are concerned) and any rise or fall in voltage on the
emitter appears on the collector with a voltage increase.
Fig 64. Dynamic
Fig 64. This circuit
converts an ordinary speaker into a very sensitive microphone.
The fact that the load resistor is (a low) 2k2, means the stage has a
good capability of delivering energy to the next stage.
We have already discussed the fact that the "load" resistor
determines the capability of the stage to pass energy to the
Here are the details of the gain to expect from the stage:
The impedance of the speaker is 8 ohms. Suppose we generate a
voltage of 1mV from the speaker. This voltage will produce 1/8mA
in the emitter-line.
The collector current is almost the same as the emitter-current
and thus the voltage produced across the 2k2 will be 2,200/8 =
275mV. Thus the gain of the circuit is 275.
We are already assuming the voltage on the collector is 3v and
the 47k has been selected to create this 3v. The collector can
increase by 2.75v and decrease by 2.75 before clipping occurs
and thus the speaker can produce a 20mV p-p before clipping.
Fig 64a. Common Base
and Common Emitter stages directly coupled together
Fig 64a. This circuit
adds a Common Emitter stage to the Common
Base shown in Fig 64 to produce a DC coupled (Directly Coupled)
amplifier with very high gain.
The common-emitter transistor can be called a BUFFER stage as it
provides a lower impedance output than the first stage.
In Fig 71ac, (below) the output of the second transistor has
been taken back to the input to produce an improvement called a
BOOTSTRAP Circuit to create a higher gain.
Fig 65. Hum Detector
Fig 65. This circuit
picks up mains hum via a coil. The common-base first stage has
very high gain.
And we can see a common-emitter stage plus a 3 transistor DC
amplifier driving a speaker.
All the things we have learnt, put into a single circuit.
Fig 65aa. The common-base amplifier can be found in many FM transmitter
circuits. The electret microphone and 22n capacitor do not form
part of this discussion, but the tuned circuit made up of the 8
turn coil and 10-40p capacitor form a TANK CIRCUIT and this
will also be covered.
We will start the operation of the circuit with the 4k7
base-bias resistor turning ON the transistor.
The 1n capacitor is designed to hold the base rigid and at the
moment it charges as the base voltage rises to turn on the
As the transistor turns ON, two things happen.
Current flows through the 330R emitter resistor (and a voltage
develops across it). And current flows through the 8 turn coil.
The coil produces magnetic flux. We call this expanding flux and
we draw arrows coming out of the coil. There is a very small
voltage produced across the coil during this time and the
voltage gradually increases. This means the voltage on the
collector is becoming less than the 3v rail voltage. (The 22n is
designed to hold the rail voltage rigid.)
This voltage is being passed though the 4p7 and is lowering the
voltage on the emitter.
There are two ways to turn on a transistor.
1. Raise the base voltage with respect to the emitter or
2. Lower the emitter voltage with respect to the base.
This is what the circuit does.
The base is held rigid via the 1n and the emitter voltage is
being lowered. This action turns on the transistor more and more
until it is fully turned ON. At this point the flux being
produced by the coil is a maximum but it is not increasing. This
means the voltage on the collector is not reducing. In other
words it is remaining stationary at some voltage that is lower
than rail voltage. This means the pulse of energy through the
4p7 does not push the emitter voltage lower and the transistor
is turned off a small amount by the increasing voltage-drop
across the 330R.
The current through the coil reduces and the magnetic flux
surrounding the coil starts to collapse.
This produces a voltage across the coil THAT IS IN THE OPPOSITE
The voltage on the collector starts to rise and this action is
passed through the 4p7 to the emitter.
The voltage on the emitter rises and the transistor starts to
This action continues until the transistor is fully turned off.
This all happens very quickly and the magnetic flux collapses
very quickly and cuts the turns of the coil to produce a voltage
that is much higher than the original voltage across the coil.
The ratio of the original voltage to the final voltage is called
the "Q" of the coil and it can be 10 or even 100 times
higher than the original and provides
the signal that is passed to the antenna.
The capacitor across the coil simply charges and discharges
during the cycle and the delay it creates produces the frequency
of operation of the circuit.
There are a number of ways to bias the base of a
transistor so it is turned on a small amount or just at the point of turning on.
There are reasons why a transistor is biased in different ways.
If is it biased so it is just at the point of turning ON, it does
not consume any current when in quiescent mode (idle mode) and is ideal
for battery operation.
However the transistor will not amplify the first part of a waveform as
it will be less than the 0.6v needed to start to turn the transistor ON.
If it is turned ON so the collector is half-rail voltage, it will
amplify both the positive and negative parts of the waveform.
If it has a resistor in the emitter, the current into the base will
never damage the transistor. This is not strictly "base-biasing" but
Fig 65a. Four
ways to bias a transistor
The voltage on the collector of a transistor using Fixed Base Bias
will alter according to the actual gain of the transistor. This is not a
reliable way to bias a transistor. Feedback Bias. The collector voltage is set by selecting the
value of the two resistors in this diagram and if different transistors
are used, the collector voltage will not alter as much as the Fixed Base
Bias arrangement. Feedback base Bias is also called SELF BIAS.
It gets negative feedback via the feedback resistor.
Voltage-Divider Bias is also called BRIDGE BIAS and produces a very
stable collector voltage over a range of transistor parameters and
temperature ranges. Emitter-feedback Bias uses a resistor in the emitter to allow the
base to rise above 0.7v without damaging the transistor. The emitter
resistor is also called EMITTER DEGENERATION or EMITTER
FEEDBACK. It produces negative feedback.
Negative feedback is STABILISATION FEEDBACK.
shows a transistor with a gain of 175 and 350 in a
circuit. The collector voltage is 3.5v ofr the first
transistor. If a transistor with
a gain of 350 is placed in the same circuit, the collector
voltage will fall to 1.75v. If the stage has a capacitor on
the output and the waveform is less
than rail-to-rail, the different gains will not affect the
overall amplification of the stage. Although
the gain of the two transistors is different, the approx
gain that will be produced by this circuit is about 70 and
this applies to a transistor with a gain of 175 or 350.
The BASE BIAS resistor produces negative feedback and it
reduces the effective gain of the circuit to about 70.
In conclusion: The gain of the
circuit is about 70.
The collector voltage will fall when a high-gain transistor
If the circuit is producing a waveform of about 1500mV, the
output will be the same for either transistor.
If the waveform of these circuits is viewed on a CRO, the first transistor will be
producing the 1500mV waveform towards the top of the screen
on the CRO and the high-gain transistor will produce the
waveform towards the lower part of the screen.
When this waveform is passed though an output capacitor it
does not matter where the waveform is generated on the CRO.
Fig65c shows a
common-emitter transistor and two BASE BIAS
This arrangement is the simplest and best way to create a
mid-voltage on the collector and maintain the same
mid-voltage for a range of transistors with different gains.
However the circuit requires 5 times more energy from a previous
stage to achieve the same amplitude as the arrangement in Fig
This is the same as saying the stage will produce a gain of:
70/5 = 14.
In conclusion: To achieve
a constant mid-rail voltage you have lost 80% of the gain and
created a stage that takes more current.
That's why this form of biasing has not been covered in this
Now go to: Configurations - summary of features of Common Emitter, C-Collector, and Common Base
Here are a number of
circuits using the stages we have covered:
Fig 66. 4-Transistor Amplifier
Fig 66. This
4-transistor amplifier uses the minimum of components and has
negative feedback via the 3M3 to set the voltages on all the transistors.
It is actually 3 stages and that is why the feedback can be
taken from output to input.
Transistors 3&4 are equivalent to a single transistor called a
Darlington transistor and this is covered in Fig 71.
Fig 67. This Hearing
Aid uses the 3-transistor DC amplifier covered above, (with some
Fig 68. A
3-transistor amplifier operating on 1.5v
Fig 69. This Hearing
Aid circuit uses push-pull to reduce the quiescent current and
also charge/discharge the electrolytic feeding the 8R earpiece.
Fig 70. This Hearing Aid circuit has the first transistor
turned on via a 100k and 1M resistors. Connected to this supply is a
transistor that discharges the biasing voltage when it sees a
signal higher than 0.7v This reduces the amplitude of the
signal being processed by the first transistor and produces a constant volume amplifier.
How does reducing the voltage on the base of the first
transistor reduce the gain of the first stage?
When the voltage delivered by the 100k and 1M resistors on the
base of the first transistor is REDUCED, the current (energy)
being delivered to the base is reduced and thus more energy has
to be delivered by the 100n capacitor. This causes a larger
signal-drop across the 100n coupling capacitor (discussed in Fig
71c below) and thus the amplifier produces a reduced
This is along the same lines as changing from a "Class-A"
amplifier to a "Class-C" amplifier (as shown in Fig 107a) where
a "Class-C" amplifier gets ALL its turn-on energy from the
There are two types of Darlington transistors. One type is
made from two NPN or PNP transistors placed "on-top" of each other as shown in
Fig 71 and Fig 71aa:
Fig 71. Two
NPN transistors connected as shown in the first diagram are equal to a single
transistor with very high gain, called a DARLINGTON.
The second diagram shows the symbol for an NPN Darlington Transistor
and the third diagram shows the Darlington as a single transistor
(always show a Darlington as TWO transistors.) One difference
between a Darlington and a normal transistor is the input voltage must rise to 0.65v +
0.6v5 = 1.3v before the NPN Darlington will turn ON fully.
Fig 71aa. shows two
transistors connected to produce a single
transistor with very high gain, called a PNP DARLINGTON.
The second diagram shows the symbol for a PNP Darlington Transistor
and the third diagram shows the Darlington as a single transistor.
The input voltage must fall 0.65v +
0.6v5 = 1.3v before the PNP Darlington will turn ON fully.
The other type of Darlington transistor is called the Sziklai Pair.
It has an advantage:
Fig 71ab. shows a
NPN and PNP
transistor connected to produce a single
transistor with very high gain, called a
The second diagram
shows a PNP
and NPN transistor connected to produce a single
transistor with very high gain, also called a Sziklai Pair.
The advantage of this arrangement is the input voltage only
needs to be
0.6v5 for the
Sziklai Pair to turn ON fully.
Note: Some Darlington transistors have
inbuilt resistors and this reduces the input impedance
enormously. Two separate transistors in Darlington
configuration will have an input impedance of about 300k. The
Darlington in the diagram has in input impedance of about 8k.
In Fig 71abab we have a
transistor and two transistors in Darlington configuration.
But they do not perform the same in reality.
The difference is most noticeable when the load current is high.
shows the difference. The voltage on the collector of a Darlington
transistor will be much higher than a normal transistor
(carrying the same current). This is one of the characteristics
of a Darlington transistor that has to be recognised.
shows voltages when the transistors are carrying full current
and the different output voltage is considerable.
A Darlington transistor has a much-higher collector-emitted
voltage than a normal transistor.
shows voltages when two transistors are connected to produce a
These voltages are only approximate but show how the output
voltage is created.
This means the load gets pulled down to 0.8v to 1.5v above the 0v rail,
for a Darlington configuration,
whereas a normal transistor will pull the load down to about
0.2v to 0.5v.
When the current is 1amp to 10amp, this arrangement will have a
voltage-drop of about 2.2v to 2.5v and at 10amp, this represents
a loss of 25 watts!! Not a good design.
All transistors produce different results however the
voltage across the Darlington is always higher than the second
circuit in Fig 71abab-1.
The losses in a Darlington can be as high as 400% more - it all
depends on how the two transistors are connected.
shows voltages when two transistors are connected in
Darlington configuration as emitter-followers.
The load loses at least 1.35v and this is considerable when the
supply voltage is small.
It also means a lot of heat will be lost in the driver
This is a very inefficient way to drive a high-power load and
should be avoided. (use an NPN common-emitter driving a PNP
output transistor as shown in:
The Driver Stage
shows two transistors connected as a Sziklai Pair.
This produces 0.9v across the output transistor.
THE "SUPER-ALPHA" CIRCUIT
also known as the:
THE "HIGH INPUT IMPEDANCE" CIRCUIT
shows two transistors "on top of each other"
called a DARLINGTON
This arrangement produces a very high input impedance of
about 200k and only a very small current is required to produce
a "swing" on the output.
The circuit is commonly called a SUPER ALPHA PAIR and the input voltage must rise to 0.65v +
0.6v5 = 1.3v before the circuit will start to turn on.
The actual high impedance only occurs when the Darlington
pair is just starting to turn on (when the voltage is
1.3v). Below this voltage the impedance is infinite (but of no
use). Above 1.3v, the Darlington needs slightly more current and
the input impedance is slightly less.
"CURRENT BUFFER" CIRCUIT
shows a CURRENT BUFFER stage. Both the EMITTER FOLLOWER
and COMMON EMITTER stages can be used as a
CURRENT BUFFER and both have the same current amplifying
A current buffer simply assumes you have a waveform with
sufficient voltage but not enough current to drive a LOAD.
stage can be
connected directly to a previous stage, this makes it the better
"VOLTAGE BUFFER" CIRCUIT
shows a VOLTAGE BUFFER stage. You can also say it is a
VOLTAGE FOLLOWER as the output voltage follows the input
You need to define why you need a Voltage Buffer.
In most cases a device (or circuit or stage) will produce a voltage but very little
current and if this is connected to another circuit, the output
will be reduced (attenuated). To prevent this, an EMITTER FOLLOWER can be used as a
VOLTAGE BUFFER as the output follows the input EXACTLY but
0.6v lower than the input.
stage provides added
current so the voltage from the source is not attenuated.
A Voltage Buffer and Current Buffer circuit can be
identical. It's all in the way you describe your requirements.
"VOLTAGE AMPLIFIER" CIRCUIT
shows a VOLTAGE AMPLIFIER stage. It is really a
common-emitter stage with another name. The circuit can have a
base-bias resistor or it can be removed.
The actual voltage gain of the circuit is unknown and will
depend on the transistor and surrounding components.
However this is a Voltage Amplifier stage and
Fig 71abb above can also be classified as a Voltage
You can call a circuit by a name that describes what it is doing
in a project.
THE BOOTSTRAP CIRCUIT
interesting circuit is the Bootstrap Circuit. It uses
positive feedback to achieve very high gain.
The two transistor circuit shown in Fig 71ac has a gain of
approx 1,000 and converts the very low output of the speaker
into a waveform that can be fed into an amplifier.
The circuit is simply a common-base stage and an
But the output of the emitter-follower is taken back to the
input of the same stage and this is the Bootstrap feature. It is
like pulling yourself UP by pulling your shoe laces.
When the voltage from the speaker reduces by 1mV, the transistor
turns ON a little more and pulls the collector voltage
This action takes a lot of effort and to pull it lower, requires
more energy from the speaker.
In the Bootstrap circuit, the first transistor pulls the 10k
down and this pulls the emitter-follower transistor down. At the
same time the 22u is pulled down and it pulls the 10k down to
assist the first transistor. In other words the first transistor
finds it much easier to pull the 10k resistor down.
When the first transistor turns off, the 2k2 pulls the 10k
resistor UP and it is aided by the 22u.
The end result is a very high output voltage swing.
Fig71acc shows a Sound
Activated Switch using a BOOTSTRAP arrangement for the first
The first transistor is biased ON via the 3M3 and 47k. This
collector voltage will be very low and the second transistor
will be biased OFF and the third transistor will also be OFF.
The relay will not be activated.
When the electret microphone receives audio in the form of a
CLAP, the peak will not have any effect on the first transistor
as it is already saturated, but the falling part of the waveform
will reduce the voltage on the base and allow the transistor to
turn off a small amount.
This will turn ON
the second transistor and the voltage on the collector will
The 4u7 is connected to this point and it will fall too and
reduce the voltage on the base of the first transistor
considerably. This will turn the first transistor off more and
the process will continue and turn on the relay.
But during this time the electrolytic is discharging, then
charging via the 3M3 and eventually it charges to a point where
the base of the first transistor sees a voltage above 0.7v and
it is turned on again.
The collector voltage of the second transistor rises and this
turns on the first transistor fully and the two transistors swap
states. The relay turns off.
If the microphone continues to produce negative (or falling
waveforms), the relay will continue to remain energised.
The 4u7 has the effect of a "snap action" where the circuit very
quickly changes from one state to the other. It is very similar
to the action you get with a SCHMITT TRIGGER.
This circuit is not a LATCH. The relay does not stay energised.
It is only energised for a short period of time. Fig 71acc is an example of
There are two types of FEEDBACK.
Positive and Negative.
Positive feedback delivers a signal that makes the circuit
produce a larger signal. This feedback may be in the form of a
signal that moves in the negative direction. We are not talking
about the signal being in the positive or negative direction, we
are talking about the effect it has on the circuit.
Negative Feedback is where the signal passes back to a previous
stage to reduce the amplitude of the signal. In most cases the
feedback signal has a greater effect on the peaks (and troughs)
and these normally represent noise or distortions in the signal.
In this way the quality of the signal is improved.
Clap Switch with 15-second Delay Designed 12-11-2011
- C Mitchell
Fig71acd shows a 3 transistor circuit using a piezo
to detect the noise of a clap. The first two transistors form a
high-gain amplifier, studied in Figs 40 & 40a.
The voltage across the 33k resistor is kept below 0.7v by adding
the 1M5 and 1M voltage-dividing resistors to the base of the
first transistor and this sets the voltages for the first two
The sound of a clap produces a waveform across the 33k to turn on
the third transistor and this pulls the 100u down via the 100k,
to turn ON the BC557.
This keeps the 2nd and third transistors turned ON and
illuminates the LED for about 15 seconds.
100u charges via the 100k and the emitter-base junction of the
BC557 and initially this current is high. But gradually the 100u
becomes charged and the current-flow reduces and eventually the BC557
cannot be kept ON.
It turns OFF and the third transistor turns OFF too.
The negative end of the 100u rises and takes the positive end
slightly higher too. The 100u discharges through the 27k, 100k
and 10k resistors.
The circuit takes about 20 seconds to reset after the LED goes out.
During this time the circuit will not respond to another clap.
The quiescent current is about 20uA, allowing 4 AA cells to last a
This circuit is very clever in that it uses the middle
transistor TWICE. It is equivalent to having 4 transistors.
The first two transistors form a high-gain amplifier and the
middle and third transistors form a delay-circuit using a
BOOTSTRAP arrangement discussed above.
As we mentioned at the beginning of this eBook, three
directly-coupled transistors can produce an enormous gain and
you have to be very careful that unwanted feedback (sometimes
called motor-boating) does not occur. We have avoided this by
keeping the voltage across the 33k below 0.6v so the third
transistor is only turned ON when noise is detected. The
second and third transistors then turn into a switch to keep the
LED illuminated and the 100u creates a time-delay.
THE "LOW IMPEDANCE" CIRCUIT (stage)
A circuit or "stage" can be classified as LOW IMPEDANCE.
This can refer to its INPUT IMPEDANCE or its OUTPUT
IMPEDANCE or BOTH.
We have already covered this type of circuit but have not
specifically referred to it as LOW IMPEDANCE.
Low Impedance generally refers to a component on the input
or output that is less than 500 ohms. The circuit can also
be called "Impedance Matching" or a "Driver Stage" and the
following two circuits can be classified as "Low
The input impedance of the common-base
stage is very low.
impedance of the emitter-follower stage is very low.
The input impedance is 100 times greater than the output.
100 x 8R = 800R. The input impedance can also be classified
as LOW IMPEDANCE.
A low-impedance circuit (such as Fig 64) can employ
non-screened, long leads between the speaker and input of the circuit
without the problem of noise, hum or spikes being picked up.
This is one of the reasons for using a low-impedance circuit. It does
not pick up noise. The reason it does not pick up noise is this:
Noise consists of high-voltage spikes that have low current. This type
of waveform does not have the "strength" to raise and lower the signal
on a low-impedance input and thus it does not appear on the input.
THE "HIGH IMPEDANCE" CIRCUIT (stage)
A circuit or "stage" can be classified as HIGH IMPEDANCE.
This can refer to its INPUT IMPEDANCE or its OUTPUT
IMPEDANCE or BOTH. High Impedance generally refers to a component on the input
or output that is higher than 1M or a set of components that
cause the transistor to take very little current. This
type of circuit is very unstable and prone to
interference and noise and spikes from external sources.
In addition, the voltages on the transistor will change
with temperature and the gain of the transistor.
The following circuit has very high value resistors on
the first transistor and this allows it to detect very
small changes in voltage due to very small changes in
current-flow in the components in the circuit.
The first two transistors form a very
high-gain amplifier. If the 100p is removed, the circuit
will not work. If a capacitor is placed on the base of
the first transistor, the circuit will not work. The
circuit must be kept as shown.
The first two transistors form a very unusual
The circuit is not really an "oscillator" but a circuit
with high instability. It's the same instability as
"motor-boating" or "squeal." The feedback is the 3M3
on the base of the first transistor. It delivers the
signal from the output to the input. The circuit needs
"noise" to start its operation and it can sit for 5
seconds before self-starting.
Let's look at how the two transistors are connected.
They are directly connected (called DC connection) and this forms a circuit with very high gain (about 250
x 250 = about 60,000). Transistors can achieve very high
gain when lightly loaded. Both transistors are arranged as
Here is the amazing part of the circuit. The 100p is
acting as a miniature rechargeable battery. It takes
time to charge and discharge and produces the timing
(the frequency) for the oscillator.
To start the discussion we consider the 100p is holding
the emitter of the first transistor "rigid." This makes it a common-emitter
stage for a PNP transistor.
The transistor will produce a very small amount of junction-noise
and because the 2M2 collector-load is such a high value,
the noise will be passed to the base of the second
transistor. We will assume the first transistor turns ON
a small amount due to this junction-noise. This will
make the collector voltage rise and this will be passed
to the base of the middle transistor.
This will turn on the middle transistor and the voltage
on the collector will fall.
The base of the first transistor is connected to this
via a 3M3, and the base voltage will fall.
The emitter is being held "up" by the 100p and because
the base-voltage drops, the transistor turns on more. It
gets current to turn on from the energy in the 100p and
this allows the middle transistor to turn ON more. This
action continues with both transistors turning ON more
The energy to keep the transistors turning ON comes from the 100p and
the voltage on this capacitor drops. Eventually the
voltage falls to a point where the first transistor
cannot supply energy to the base of the second
transistor and the collector voltage rises. This makes
the base of the first transistor rise and it gets turned
off a small amount. This action turns off the middle
transistor slightly more and eventually they are both
turned off. The 100p is charging during this time via
the 3M3 and eventually the emitter rises to a point
where the first transistor gets turned ON a small amount
to start the next cycle.
There are a couple of features you have to understand
with this circuit, (the first transistor) because it
uses very high value resistors.
The feedback signal will pass through the 3M3 resistor
to the base of the first transistor with very little
attenuation (reduction) because the base presents a very
high impedance due to the fact that the transistor is
very lightly loaded and the base requires very little
Normally a 100p could not be used to create an audio
frequency as it provides very little energy and be able
to only produce a very high frequency. But when
the timing resistor is a very high value (in this case
the 3M3 on the emitter) it will take a long period to
charge and discharge and an audio frequency can be
The 100p sees a waveform of nearly 7v during its charge
and discharge cycles.
covered many circuits and one question you will be asking is:
"How do you select the resistor for the collector load?" or "Why
is the voltage on the collector equal to half-rail?"
The answer comes under the heading: VOLTAGE DIVIDER.
When two equal resistors are placed in series, the voltage at
their join is equal to half the rail voltage. But if the values
are different, the voltage is either higher or lower. The
circuit is still called a VOLTAGE DIVIDER.
The load resistor is one resistor and the transistor is the
other resistor. But the other resistor does not have to be a
transistor, it can be a relay, globe, motor or Light Dependent
We will take the simple case of a Light Dependent Resistor
The LDR we are using is 300k in the dark and 470R when
An LDR must be connected in series with a resistor to create a circuit
called a VOLTAGE DIVIDER so the voltage at the join of the two
components can be connected to a detecting-circuit. This voltage
is called the "pick-off" voltage. The detecting-circuit in this
case needs to see about 8v when the LDR is not illuminated and
about 1v when illuminated.
Let us select 900k for the LOAD
RESISTOR. The voltage at the join will be 3v when the LDR is not
Here's how to work it out:
(we have selected values for easy calculating.) The total
resistance is 1,200k This means each 100k will have 1 volt
This means 300k will have 3v across it and
900k will have 9v across it.
The resistance of the load resistor is too
See next diagram . . .
100k load resistor. The voltage at the join will be 9v.
Here's how to work it out:
The total resistance is 400k. Each 100k will have 3v across it.
The "pick-off" voltage will be 9v
when the LDR is not illuminated. .
But if the LDR sees a low level of illumination in a
normally-lit room, its resistance
will be about 50k, so we need to change the values to reflect this.
See next diagram . .
Choose 22k for the LOAD RESISTOR.
voltage at the join will be about 8v when the LDR is faintly
This is called a "realistic assessment." You have to
consider a small amount of light will be present in a normal
environment and this will reduce the resistance of the LDR from
300k to about 50k.
Now we will determine the "pick-off voltage" when the LDR is
brightly illuminated from say a torch.
voltage at the join will be about 0.5v when the LDR is fully
The 22k load resistor produces the required 8v and 1v values for
the "pick-off' voltage.
The same VOLTAGE DIVIDER
principle applies to this transistor stage.
The transistor and 2M2 are effectively equal to 22k since the
"pick-off voltage" is half rail voltage.
When the collector voltage sits at half-rail, the signal
can extend in the positive direction and negative direction by
the maximum amount.
To achieve mid-rail voltage, the base-bias resistor is selected
to get 6v on the collector.
There is no formula to achieve mid-rail voltage as it will
depend on the gain of the transistor and most of them come
from a batch with a wide range of values. Simply select a
base-bias resistor that provides half-rail voltage.
The stage is classified as a self-biased stage as a capacitor is
added to the input and output, when the stage is added to a
The voltage at the join of two equal
resistors is half-voltage, but if the resistors are
different-value, the voltage needs to be calculated. Here is a
Voltage Divider Resistance Calculator
Voltage Divider Resistance Calculator
Enter any of the three values, and then click the Calculate button.
Vin = Input voltage
Vout = Output voltage
On the next page we
continue our coverage of the transistor (called a Bipolar Junction
- BJT or "normal" or "standard or "common
transistor") in amplifying circuits, including
oscillators . . .