I have a class B amplifier circuit to be done in lab. Why are three pair of transistors used in this? I suppose one is a Darlington pair but what is the requirement of the third one? Also, why is there an unequal mumber of diodes even if an equal number of PNP and NPN transistors are used?
Why three pair of transistors are used in this?
I see only two pairs, the two Darlington pairs. The 2N3904 and 2N3906 are for output current limiting. These two are normally off unless so much current is flowing through the 1.2 Ohm resistors that enough voltage develops across them that the 2N3904 and 2N3906 activate. That will happen around \$V_{BE}\$ = 0.6 V. So around 0.5 A needs to flow through the 1.2 ohm resistors.
Also, why there is unequal number of diodes
Note that the bottom diode is drawn as one diode but the text says: "3x 1N4005". I guess they mean 3 diodes in series. I think that is plain wrong to draw it like that as it does not make it clear that the diodes should be in series. The reason to have 3 diodes instead of 2 diodes might be that the Darlington pair at the bottom (2N3906 + TIP32C) needs a higher biasing voltage compared to the Darlington pair at the top.
Edit the "3x 1N4005" appears to refer to all three diodes that are drawn. That would mean there is indeed only one diode at the bottom side. That means that the bottom PNP Darlington pair is biased at zero current and will only "kick in" when there is some signal. My guess is that this amplifier will have significant crossover distortion as a result of that.
In a standard class B amplifier it is typical to have the same number of diodes as there are Base-Emitter voltages stacked in the transistors. Here there are two Base-Emitter voltages stacked at both sides. Adding another diodes increases the voltage across the emitter resistor (that's the 1.2 ohm resistor) and increases the biasing (also called quiescent) current. If you do that on top and bottom sides the amplifier becomes a class AB amplifier.
The middle two transistors are just current limiters: they turn on in case of too much current through the parallel resistors and turn off their corresponding Darlington pair.
The dead-zone voltage between the push-pull Darlingtons is two diode drops each (for 2.4V in total when cold). This is brought down by three diode drops (1.8V) which is comparatively safe against being underbet by the four diode drops of the power Darlingtons even when they are running hot. So it reduces crossover distortion while still keeping class B operation.
Why the unsymmetric diode arrangement? Well, injecting in the middle of a 3 diode set is not really possible. It doesn't really matter whether you split this 1/2 or 2/1 unless you think the opAmp will get too close to the power rails to keep working: the sensing input of the opAmp is on the output rail so it will just compensate for either unsymmetry.
I am just right now fiddling with an old class B power amp in a keyboard amp that has a 3 diode drop in the push-pull transistors (one Darlington pair and one Sziklai pair) and has no diodes separating the input rails whatsoever, and it doesn't even have an opAmp but a differential transistor pair on a current source. Really contemplating adding two diodes in there, but the amp actually sounds pretty fine even now.
It probably depends on the maximum slew rate of the opAmp how much of a disruption you'll get for crossing the dead zone of the class B amp. A single diode drop is as small as you can go (unless using other diode types) and hopefully enough to stick in class B operation even when the transistors get hot and reduce their drop voltage: discrete Darlington/Sziklai pairs can be thermally separated and thus may be less susceptible to runaway temperature/dropoff voltage, assuming you don't skimp on the driver transistors' heat sink.
