Class AB BJT biasing, diodes

Question

Every time I see such circuits, it is assumed that the voltage drop across the diodes is exactly equal to Vj * 2. But this is simply not true. The voltage between the bases of transistors is highly dependent on the value of the resistors, even if both resistors are of the same value. I can easily achieve, for example, a voltage of 1V or 2V between bases without changing the diodes, both in a real circuit and in a simulation. How it works?

Answer 1

It works 'well enough'. The diode V and transistor VBEs are similar enough so the transistors never turn fully off (which is the intent of this circuit).

Mathematically, you can show that the product of the diode currents and the product of the transistor currents are equal. What this means (if the diode currents are constant) is the the product of the transistor currents is also constant -- if one increases (say 10x), the other decreases (to 0.1x).

The effect is that it all works as a reasonable voltage follower. There is some distortion (because the VBEs change), but overall negative feedback (if used) will reduce this.

Answer 2

What you missed here is that this is a principal schematic and does not reflect the complete practical design considerations.

In practice, we don't expect the output transistors' V_BE voltages to be exactly equal to the diodes' V_F voltages, because they aren't and they will never be. And we don't expect the transistors' V_BE voltages to be exactly ±0.7V or higher/lower.

The idea is, as you already know, to bias the output transistors by making their V_BE voltages to a non-zero voltage so that a non-zero collector current flows through them without dropping them into the saturation region. Ideally, you would expect these V_BE voltages to be around ±0.6V but even at ±0.5V, there'll be a non-zero collector current. And also, the designer may need to do some adjustments during the design or the production stage depending on the practical requirements. So, the V_BE voltages can increase or decrease.

Of course, non-idealities of the base-emitter junctions may lead to crossover distortion but this is rectified by the feedback which is not shown in your schematic.

In conclusion, for a complete design, we never let the output transistors be biased just with simple silicon diodes. There are different methods for biasing (e.g. V_BE multiplier) and there'll almost always be negative feedback.

Answer 3

<<< 1 or 2 V between diodes >>>

Here what happens ... Ok, it will work ... warmly.

Answer 4

The voltage between the bases of transistors is highly dependent on the value of the resistors, even if both resistors are of the same value.

This is actually the case when the input voltage applied to the midpoint between the diodes approaches the supply rails (large input signal).

The problem of static quantities

Static resistance. The problem can be that the resistors R1 and R2 have "static" (constant) resistance. And since the voltage applied across them varies considerably, the current through them varies and this causes the voltage across the diodes to vary as well.

Static voltage. But the problem can be that the supply voltage Vcc is "static" (constant). And since the middle voltage Vmid considerably varies, the difference Vcc - Vmid = VR1 varies. This causes the current and eventually, the voltage across diodes to vary as well.

The magic of dynamic quantities

Dynamic resistance. This problem could be solved if the resistors had "dynamic" resistance that varied when the voltage across them varied. As a result, the current would stay constant and the bias voltage across diodes would not vary as well.

Let's, for example, consider the OP's circuit when the middle voltage Vmid significantly changes towards +Vcc (increases). Then the voltage drop across R1 (VR1 = Vcc - VF - Vmid), which is the difference between the two voltages Vcc and Vmid, significantly decreases... but, at the same time, R1 decreases its resistance with the same rate of change... and both the current I = VR1/R1 and the voltage VF do not change.

Such "dynamic resistors" are implemented by transistors and are called "current sources" although there is no any source (in the sense "producing power") inside them.

If you have too much imagination, you can imagine this network as consisting of elements in series with dynamic resistance - some of them (instead of R1 and R2) "current stabilizing" and others (D1 and D2) "voltage stabilizing". Each of them tries to keep stable its quantity; doing that, they interact each other.

In the example above, when Vmid increases and the difference Vcc - Vmid decreases, the current-stabilizing resistor (in the place of R1) will decrease its resistance to restore the current while the voltage-stabilizing resistor (D1) will increase its resistance to restore the voltage.

Dynamic voltage. Another clever trick can be to make the supply voltage follow the middle voltage Vmid. Then the voltage difference Vcc - Vmid will stay constant... the current will be constant... and the voltage across diodes as well. The name of this ingenious idea is "bootstrapping".

Exploring the circuit operation

I have explained in a geometrical way the operation of this circuit in my answer to the question Class AB power amplifier from "Electronic Devices" by Thomas L. Floyd.

Also, I have illustrated the circuit operation in another answer of mine to Question about Class AB amplifier.