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Differential Amplifier with Active Collector Loads

Circuit Description

Differential amplifier with active loads.

The circuit shown to the right looks considerably different from its predecessor, but is actually more similar than you might think at first glance. The emitters of Q1 and Q2 are still connected together, but we now have two mirror transistors, Q4 and Q5, in parallel feeding current to them. This means the total emitter current, IE, is now double the reference current through Q3. This is necessary because we have replaced the collector load resistors with another current mirror, this time using PNP transistors.

Here, Q6 is the PNP reference transistor, which necessarily carries the same current as Q3. Neglecting base currents, this means that the same current is applied to the collectors of Q1 and Q2. When there is no differential input, the collector currents of Q1 and Q2 match the load currents through Q7 and Q8. No extra current will flow into or out of OUT1 or OUT2.

With this circuit, the output voltages can float anywhere within the range of VIN to VCC - VBE(PNP). Generally, whatever circuit is connected to the outputs will determine the output voltages, simply because of those connections.

At the same time, the output currents behave a bit differently. If the two inputs are exactly the same, then Q1's collector current will exactly match Q7's collector current. Q2 and Q8 will also match exactly. This is the condition of perfect balance, and no current will flow into or out of OUT1 or OUT2.

Now, let's suppose the input voltage at IN1 rises slightly. This means increased input current at IN1, and a corresponding increase in Q1's collector current. However, Q7's collector current remains fixed; it is not affected by changes at the inputs. Therefore Q1 is pushing more electrons at Q7 than Q7 is ready to absorb. The result is that Q1 now pushes those extra electrons out of OUT1 — the only available place for them to go.

At the same time, the current through Q2 will decrease by the same amount. But Q8's collector current remains unchanged; it is fixed just like Q7. Therefore Q8 tries to pull the electrons it needs in through OUT2.

The advantage of using active collector loads for Q1 and Q2 rather than simple resistors is that we have, to a large extent, disconnected the output current from the output voltage. This cannot be done with just resistors, because of Ohm's Law. Instead, we can enjoy the relatively high output resistances of the transistors and still generate a large output signal current, without also requiring high voltages for +VCC and -VEE.

Of course, we don't get this advantage for free. We also get some significant disadvantages. The first of these is that it is not really possible to balance this circuit completely. The collector currents of Q1 and Q2 will be slightly lower than their emitter currents (for transistors with a β of 200, the ratio, identified as α, is 200/201 ≈ 0.995). This doesn't seem like much, but to match it, the values of β for the PNP transistors must be slightly less than 200. This level of matching isn't practical with ICs, and is beyond the bounds of reason for discrete transistors.

The other disadvantage of this circuit involves the internal output resisatances, rO, of Q1, Q2, Q7, and Q8. Q1 and Q7 will seek a balance point, adjusting the quiescent voltage of OUT1 until their internal resistances cause their collector currents to match. Q2 and Q8 will do the same with OUT2. But those two balance points will not be the same. This will result in an output voltage difference when there is no input voltage difference, and that difference will be unpredictable from one copy of this circuit to the next. Whatever circuit is connected to the outputs will affect both output voltages, but will also "see," and amplify this difference. The result is that this circuit will create its own offset that will be propogated to and through any following circuit.

These variations may seem too small to be significant, but sadly such is not the case. A differential amplifier will typically be used to amplify a very small signal from a sensing device, while filtering out surrounding noise. Any error added at this point will be amplified by every following stage, and may well overwhelm the desired signal.

The use of an active collector load is quite reasonable with a single transistor amplifier circuit, since then we don't have any requirement to closely match circuits to each other. In this application, however, we need to make the circuit balance itself, and this clearly doesn't fit the need.

Prev: Differential Amplifier with Constant Emitter Current Next: Differential Amplifier with Current Mirror Load

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