What is the use of resistors in series with BJTs in a push pull amplifier configuration?

Question

If you make an internet search for "Push pull BJT amplifier" you will find many different variants of the following schematics:

^{simulate this circuit – Schematic created using CircuitLab}

These three configurations are the ones I am interesting in understanding better. First, I see that this circuit is often labelled as amplifier but since the gain appears to be in the best case equal to 1 it looks to me more like a current amplifier perhaps rather than a voltage one.

Suppose that V_signal is a sinusoidal 1 Vpp (maybe even lower 100 mVpp) signal at a given frequency. My main question is as follows:

• Referring to configuration A, while I understand the role of the diodes and R1 and R2 in the biasing of the BJTs to avoid crossover distorsion, I do not understand what is the role of R3 and R4 which are in series to the BJTs. My first thought is that they should limit the current through the load but in the simulations the circuit works fine even without them so I am a bit confused on their use and how to size them.

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As a sidequestion, what are the advantages/disadvantages of configurations B and/or C over A? I can see how C is more advantageos in that it does not need a double side power supply and can work straight from 12 V but I cannot understand why I should use B which is using an additional capacitor apparently for the same result as A and C.

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Edit: I noted the circuit is not easy to visualize, so I've just added 3 screenshots (Vcc=12V, Vee=-12V)

Answer 1

You are correct that the circuit has a voltage gain of 1, and that the diodes are to minimize crossover distortion. The idea is that the diode voltage is the same as the base-emitter voltage of the transistors, so that both transistors remain biased in the "on" condition - they are both conducting. This means that there is always a theoretical current from the positive rail to the negative rail, through the pair of output transistors, which is wasted power and is dissipated as heat in the transistors and the series resistors.

With no series resistors, all of this power must be dissipated in the transistors. Consider what happens when the circuit starts to drive a load. The transistors start to heat up, and the base-emitter drop starts to decrease to a value lower than in the relatively cool diodes. This causes the current through the two transistors to increase - further increasing heat by further decreasing the base-emitter drop and increasing the through-current. So in general, make the series resistors as large as you can consistent with your output load characteristics.

As to the difference in circuit "A" and "C" from "B," think about the fact that the phase shift due to the capacitance of the diodes and transistors will not be identical. This is not going to be noticed at low frequencies, but if you crank up the frequency on your simulation, you will see if the mismatch is pronounced enough to cause any issues. If there is a problem, you can overcome some of the effect at high frequency from the diodes by using two large capacitors as shown in "B."

Answer 2

Configuration A .

This circuit is powered for a symmetrical supply, hence the output capacitor is not needed. Capacitor here only removes any residual DC-offset present at the BJT's emitters and together with the load resistance this output capacitor form a high pass filter with a corner frequency \$F_C = \frac{1}{2 \pi C_O R_L}\$.

Configuration C.

This time the circuit is powered for a single supply. So to be able to get a negative voltage at the load ("pure" AC signal) and fully "open" the PNP transistor we use a "big" output capacitor to serve as an additional DC voltage source. Without the input signal voltage at the NPN's base is equal to 6.6V and 5.4V at for PNP base. The output capacitor is charged to 6V and the input cap is also charged to 6V. Now let us apply the input signal (1Vpp = peak to peak) into the circuit. The input signal will "modulates" the DC voltage and currents in the circuit in the "rhythm" of AC input voltage. During the positive half-cycle, input DC voltage will rise from 10V to 12V and back to 10V in the "rhythm" of AC input voltage. The NPN base becomes more positive, D1 and R1 current decreases, but D2 and R2 current increasing. The NPN base current comes from 12V supply and ii is source into NPN base via R1 resistor.

For negative half-cycle, we have a very similar situation but this time PNP base becomes more negative (any voltage lower than 5.4V at PNP base), D2 and R2 currents decreases but R1 and D1 current increases.

And an important node hers is that as you can see Vin voltage source don't provide any current for NPN/PNP transistor. The base current is coming from R1 and R2. Vin only provides a current to D2 and R2 for positive half-cycle. And to D1, R1 for negative half-cycle.

Or in short:

For positive half-cycles, the upper NPN transistor delivers current to the load by charging the output capacitor. And for negative half-cycles, the lower PNP transistor discharge capacitor. And this is why we have a negative voltage at the load.

Configuration B.

The same circuit as case A. The only defenses here is that now the output transistors are now "driven directly".

And I don't see a point of having such big resistors value at the collector. They only limit output power. Instead, we move these resistors into transistor emitters. Therefore this small resistors at the emitter will help us set the DC quiescent current and provide thermal bias stability.