There are many situations where we want to amplify a small difference between two signal levels and ignore any ‘common’ level both inputs may share. This can be a achieved by some form of Differential Amplifier. Figure 1·7 shows an example of a differential amplifier stage that uses a pair of bipolar transistors.
This particular example uses a Long-Tailed Pair connected to a Current Source. We can understand how the circuit works simply from an argument based upon symmetry. For simplicity in this diagram we represent the Current Source by the standard symbol of a pair of linked circles. We will look at how a current source actually works later on. For now we can just assume it is an arrangement that insists upon always drawing a fixed current which in this case has the value .
Assume that and that the two transistors are identical. Assume that we start with the two input voltages also being identical. The circuit is arranged so that . By symmetry, when it follows that , hence .
Now assume we increase, say, by a small amount. This will mean that the transistor, Tr1, will try to increase its current level and hence lift the voltage present at its emitter. However as it does this the base-emitter voltage of Tr2 will fall. Since the current in a bipolar transistor depends upon its base-emitter voltage the result is that the current, , rises, and falls. The sum of these currents, , does not alter very much, but the balance between the two transistor currents/voltages changes. The result is that the rise in , keeping fixed, causes more current to flow through and less through . The reduction in means the voltage drop across will reduce. hence the voltage at its lower end moves up towards .
To evaluate this more precisely, we can assume that for each transistor in the pair the collector-emitter current is related to the base-emitter voltage via an expression
i.e. we can say that
where is the emitter voltage which the two transistors have in common and and are the base voltages. The value represents the voltage-current gain of each transistor. The sum of these two currents always has a fixed value imposed by the current source, but any difference between them will be such that
where vanishes as it is common to both of the initial expressions.
Since we can say that the above is equivalent to saying that
When only concerned about a.c. signals we can ignore the constant and say that this becomes
where is one of the small-signal h-parameter values for a bipolar transistor, represents the change in the current in produced when we change the input voltage so that their imbalance from being equal is . Since this change in current appears across , and the potential at the top of this is held fixed at it follows that the lower end of which we are using as the output will change in voltage by
i.e. the stage has an effective voltage gain of .
Differential amplifiers are particularly useful in three applications:
- When we have an input which has come from some distance and may have had some added interference. Using a pair of wires to send the signal we can then take the difference in potential between them as the signal and reject any ‘common mode’ voltages on both wires as being induced by interference.
- In feedback arrangements we can use the second input to control the behaviour of the amplifier.
- When we wish to combine two signals we can feed one into one transistor, and the second signal into the other.
Most Operational Amplifier integrated circuits have differential amplifier input stages and hence amplify the difference between two given input levels. Many use bipolar pairs of the kind shown in figure 1·7, but similar arrangements using Field Effect Transistors are also often used.
The Current Source
Having made use of one, we should finish with an explanation of the Current Source. Figure 1·8 shows a simple example.
A positive voltage is applied to the base of a bipolar transistor, but this base is also connected to a lower potential via a series of three diodes. The potential across such a diode when conducting current will tend to be around 0·6 V. Since there are three of these, the base of the transistor will sit at a potential about 1·8 V above the lower end of . The base-emitter junction of a bipolar transistor is also a diode, hence it will also drop about 0·6 V when conducting. As a result, there will be 1·8 - 0·6 = 1·2 V across . This means the current through this resistor will be . since the transistor will probably have a current gain of over × 100 this means that more than 99% of this current will be drawn down from the transistor’s collector.
Hence approximately, we can say that
in practice the current will vary slightly with the chosen values of and . However provided the current through the diodes is a few mA the above result is likely to be reasonably accurate. Provided that is much larger than 1·8 V the exact value does not matter a great deal. The circuit therefore tends to draw down much the same current whatever voltage appears at the transistor’s collector – assuming that it, too, is more than a couple of volts. So the circuits acts to give a ‘fixed’ current and behaves as a steady current source. Although this example uses a bipolar transistor, as with many other forms of circuit we can make similar arrangements using other types of device.
You should now know the difference between some of the basic types of amplifier and the kinds of electronic ‘building blocks’ that can be used as amplifier stages. It should also be clear that the range of tasks and signal details/levels varies enormously over a wide range of applications. You should now also know how the basic amplifier arrangements such as Long-Tailed Pairs, etc, work.
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