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BASIC
ELECTRONICS COURSE
Page 43
INDEX
A CLEVER
CIRCUIT
The very clever circuit below converts an analogue signal into a digital signal. It looks very
simple, and it is. But it's very difficult to work-on and diagnose,
(if you get a fault) as you will see in a minute.
HOW IT
WORKS
The way the circuit works is simple. The first transistor is biased ON and
this means the second transistor is not conducting. The voltage across the
LOAD resistor is zero, so that any device connected to the output (in
other words the device will be the resistance of the LOAD) will not
be operating. An output device can be a LED, globe, motor, relay, an input line of a gate or
microcontroller, or any other device.
The advantage of this circuit is the rapid response of the output. The output goes from
one state to the other during the time when the signal is rising only a
few millivolts and this will take a very short period of time.
The circuit is not a Schmitt Trigger, however it has the advantage of the
speed of a Schmitt Trigger with a lot fewer components.
We said the circuit is difficult to diagnose because you would expect the
collector of the first transistor to change appreciably when it is
detecting an input signal. But this is not the case.
Since it is connected to the base of the output transistor, the collector
of the first transistor cannot rise above 0.7v, even when it is not
conducting, because the base-to-emitter voltage of the second transistor
prevents any rise above 0.7v.
When the first transistor is turned on, the collector will fall to about
0.2 to 0.4v and because this voltage is lower than that required by the
second transistor to turn on, it is controlled.
But this means the collector of the first transistor will only change between
0.2v and 0.7v during its two states of operation and this is a very small
change.
If you connect a voltmeter to the collector, you may be tricked into thinking
the circuit is not working. It is not until you realise the voltage change
is extremely small, that diagnosis can begin.
You will need to set a voltmeter to a very low range to detect the
difference between the two states. The ON state will be less than
0.6v and the OFF state will be above 0.6v That's how close the two will be!
The animation below shows the to transistors in the ON and OFF states.
When a transistor is ON, it is equivalent to a low-value resistor and when
it is OFF, it is equivalent to a high-value resistor.
The voltage across resistor R and the LOAD also
change during the two states. The animation below shows the voltage
across these two components. Don't go by the animation for swing
on resistor "R" because it sees less than 0.3v change from one
state to the other! On the other hand, the voltage across the LOAD will
change by almost full rail voltage!
DESIGNING
THE CIRCUIT
Designing the circuit consists of working out the value of the three
resistors:
1: The base-bias resistor for the first transistor,
2: The load/turn-on resistor
3: The LOAD resistor.
The values can be worked out by mathematics or simply determined by
experimentation. You can spend lots of time working out the values by
computation, but what values of current-flow are you going put into the
equations? It's much better to generate the values by experimentation
because there are lots of hidden tricks in driving different devices and
the only way you will find out is to actually drive them in
reality.
For instance, relays, motors and solenoids have back EMF (a high voltage
produced by the device) that can destroy the output transistor if it is
not protected.
The easiest way of protecting the transistor is to provide a
reverse-biased diode across the device. Other devices take a high start-up current and unless this
is factored into the design, the circuit will not work properly.
We start by deciding on the load we wish to drive. If it is a LED, we only
need about 20mA. If it is a relay we need about 50 - 150mA. If the load is
a globe or motor, we need to be aware of a very important fact. A globe
requires about 6 times more current to get it to turn on. This is because
the cold resistance of the filament is much lower than its operating
resistance. Once it is
turned on, the current requirement is as stated on the globe. A motor
requires at least 4 times the running current to get it to start. In both
cases this is called start-up current and is especially noticeable
when you are trying to drive larger globes and motors.
Due to the design of the circuit, you can see current flows through
resistor "R" whenever power is applied to the circuit and
this makes it a very wasteful design if you are using batteries. The
circuit is only really suitable for low-current loads.
But if you are using a plug-pack, consumption will not be a factor.
Firstly we need to decided on the current-value of the "LOAD."
For a LED it will be 20mA, for a globe it can be from 100mA to
1,000mA (1A) and for other devices it can range from a few mA to
1amp or more.
Once we know the current through the LOAD, we can work out the value
of "R" in the diagram above.
The way the circuit works is this: Current flows through R and through
the base-emitter circuit of the output transistor, to turn the transistor
ON. The current through R is multiplied by the "gain
of the transistor" and this current will pass through the LOAD
(and collector-emitter circuit). If
this current is exactly what is needed by the device, the circuit will
work perfectly. This is called the DEVICE CURRENT. If the current
is not enough, the globe will not turn on fully or the motor will
not reach full-speed.
This is a very dangerous situation as the transistor will get much
hotter than expected and may overheat. This is a topic for further
discussion, so we will concentrate on driving the transistor FULLY
in this discussion.
Now, here's the unusual part. Suppose the current through the
collector-emitter of the transistor is mathematically MORE than required
by the device! The result is the device will only allow the DEVICE
CURRENT to flow. The additional current capability of the transistor
is "up-your-sleeve" (available - but not used at the
moment).
Why do we need additional current-capability?
If we are driving a globe
or motor, the starting current is 4 - 6 times the operating current and
the value of R must be worked out using these values.
During start-up, high current will
flow through the LOAD and as soon as the globe or motor starts-up, the current
will drop. But the high base-current will remain ALL THE TIME.
We will give an example to show what we mean:
If a globe requires 100mA for its operation, the start-up current
will be as high as 500mA. The gain of a transistor under
high-current conditions will be about 50, so the base current will
be 500/50 = 5mA.
Suppose the rail voltage is 10v. The value of can be determined from
Ohm's Law. The value of resistance = 2k. You can use 1k8 or
2k2.
The current through "R" flows all the time. When the first
transistor is not turned on, the current will flow into the base of
the second transistor and activate the LOAD.
When the first transistor is turned ON, the current will flow
through the collector-base leads of the first transistor and since
the voltage between base-emitter of the second transistor will be
less than 0.6v, the output stage will NOT be activated.
We can now work out the value of the base-bias resistor. The
collector current for the first transistor is 5mA. Suppose the gain
of the first transistor is 100, the base current will be 5/100mA or
50 micro-amps.
This current can be obtained from the base-bias resistor or from an
external source that delivers the current to the base. The value of
base-bias resistor = 200k. You can use 180k or 220k.
The current to activate the circuit (the input current) must be a
minimum of 50uA (50 microamps) and must be higher than 0.7v, to turn
on the first transistor.
All values of current depend on the voltage of the rail as well as
the gain of the transistor(s).
QUESTIONS
Question 151: What is the base-emitter turn-on voltage for a transistor?
Question 152: What is the voltage across the collector-emitter junction when
a transistor is turned ON?
Question 153: In the circuit above, titled: "CONVERTS ANALOGUE
SIGNALS TO DIGITAL," the two transistors are directly coupled.
Explain how the first transistor turns the second transistor ON and OFF.
Question 154: If the collector-emitter voltage of a turned-on transistor
was 0.8v, how would the circuit above work?
Question 155: In the circuit above, if the first transistor turned
ON? What is the state of the second transistor?
Question 156: Describe how to change the state of the circuit:
Question 157: Why is the circuit a "wasteful" design?
Question 158: Why is the circuit only suitable for low-current loads?
AN
IMPROVED DESIGN
In the next circuit we have an improved design. The circuit turns ON
a lamp (or other device) when a very small current is delivered to
the input. In this case the current is provided by a finger,
touching two TOUCH-PADS. The circuit also has another feature. It
will "latch-on" and the globe will remain illuminated. A
second touch pad is provided to turn the circuit off.
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The TOUCH-PADS deliver current from the power rail to the
input of the circuit, via a moist finger. The finger acts as a very high
resistance but it will allow enough current to flow to turn the circuit
ON.
As the circuit turns ON, voltage is developed across the globe and this
is passed to the "front-end" via a 4M7 FEEDBACK resistor. This
resistor effectively takes the place of your finger so that when the
finger is removed, the circuit will stay active.
To turn the circuit OFF, Q1 is made active by the OFF touch-pad and this
transistor removes the "turn-on" voltage to transistors Q2 and
Q3. This turns off the output transistor Q4 and the voltage across the
globe reduces. When the finger is removed from the OFF pads, the circuit
has completely turned off and absolutely no current will be drawn.
The only current taken when the circuit is off is called LEAKAGE CURRENT
and this will be LESS than 1uA and will flow through the junctions of the
BC 557 when the transistor is OFF.
This circuit is a vast improvement over the previous design and it can be used
for all types of applications where an electronic "latch" is required.
There are a number of technical features in this simple circuit and these
will be discussed on the next page.
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