Why are there three pair of transistors in Class B power amplifier?

I have a Class B amplifier circuit to be done in lab. Why three pair of transistors are used in this? I suppose one is a Darlington pair but what is the requirement of the third one. Also, why there is unequal mumber of diodes even if equal number of PNP and NPN transistors are used?

You are right that the main two power transistors (the TIP31C and TIP32C) are driven as Darlington pairs with their respective driver transistors (2N3904 and 3906, connected to their bases).The three diodes are there to eliminate crossover distortion, and although the diagram shows them asymmetrically, they actually apply a voltage offset symmetrically to the bases of the driver transistors.There needs to be about one diode drop of voltage (0.6V) between the base and collector of a transistor to make it conduct. Without any diodes, and with a class B output stage made with single output transistors (but see below), there would be a 1.2V "dead zone" in the input where both output transistors were turned off. This would lead to horrible sound quality - human ears are really sensitive to crossover distortion. The additional voltage keeps both transistors turned on in this crossover region.
So two of the three diodes are there to supply 0.6V of additional bias voltage to each of the driver transistor bases.
The third diode  adds an extra 0.6V so that there is a constant quiescent current through the output emitter resistors (1.2 ohm). These resistors serve three purposes. 
First, they set the quiescent current, which would be 0.6V / 2.4 ohm = 250mA with no input signal.Second, they act as stabilising negative feedback and prevent thermal runaway.Third, they act as sensing resistors for output current limiting.This is where the final two transistors come in. If the voltage across the emitter resistors gets above about 0.6V each (1.2V total), they turn on and apply strong negative feedback. This happens if the output current gets too high (about 500mA).Negative feedback is applied to the op-amp from the output terminal via the 10k resistor, which should compensate for any non-linearity in the output driver.
But! But! But! The three-diode arrangement is a classic biasing circuit for SINGLE output transistors. A Darlington pair has TWO base-to-emitter voltage drops, so I would expect there to be five diodes - one each to bias the base-to-emitter voltages of the four output transistors, and one to bias the emitter resistors. Am I missing something?

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Thank you so much for the clarification.The circuit indeed consists of three diodes in total, not five. Could there be any biasing voltage difference between TIP31C+2N3904 and TIP32CG+2N3906?

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 VBEVBE = 0.6 V. So around 0.5 A needs to flow through the 1.2-ohm resistors.Why there is an 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 diode 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.
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