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In a class B amplifier, both halves of the push-pull output stage are cut off at 0 input. As the input increases, the conducting half starts to conduct in a highly non-linear manner as the input increases. This is "crossover distortion". As the input increases, the conducting half starts to behave in a more linear fashion with respect to the input. To get around the crossover distortion problem, both halves of the output stage are biased so that they are conducting with zero input. This is "Class AB" operation. The higher the bias current, the lower the distortion. The price you pay for the lower distortion is higher power dissipation and lower efficiency. If you increase the bias current to the point where both output stages are conducting a current that is equivalent to the current that would flow at 1/2 output, this is "Class A" operation. In class A operation, the dissipation in the output stage is highest, and is independent of the output.
"Class AB" is so named because it is a compromise between Class A (low distortion, high dissipation) and "Class B" (high distortion, low dissipation)
One of the dangers of class AB or class A operation is the possibility of thermal runaway. This is a condition where, as the output transistors or FETs heat up, the bias current tends to increase, causing a furhter increase in temperature.......Some kind of negative feed back mechanism, such as emitter/source resistors is usually used to mitigate the thermal runaway tendency.
By varying the bias current in a class AB circuit, you can change the compromise between distortion and dissipation.
Wow guys, very detailed reply. That's more than what I can expect. Thank you all!
For the quiescent current, it may not be stable, e.g. the 0.7V voltage source V10 and V11 is variable during operation. In that case, the problem is also distortion, right?
That was a nice reply from Enrique 15! Regarding your last question: Yes, it will be thermally unstable. I think that Enrique 15 was just trying to illustrate the principle, not a practical circuit. The Vbe varies with temperature at a rate of about -2 mV/deg C. This is the reason that emitter (source) resistors are frequently used in circuits of this type. As the current increases, the voltage across the emiiter resistor increases, reducing the Vbe. In many designs, one or more diodes are used in the bias circuit, so that the voltage variation of the diodes wrt temperature matches (approximately) the Vbe variation of the transistors. Often, the diodes are placed on the same heat sink as the transistors to provide better thermal tracking. You have to be careful here, because the temperature mismatch between the diodes and the transistor junction can be large, even if the diodes are placed phisically close to the transistor package.
Thanks Kral for the explanation. I just forgot to tell this same thing in my reply.
In fact, the circuit I put (with 0.7V Power Supplies) is the basic one, that is used in books in order to explain the principles (well, at least in the book I used in college, hehehe ).
But of course, there's the problem of the temperature variation, and so, instead of power supplies, you use basically diodes, or other elements to biasing the complementary transistors. For example, another transistor, of even two more transistors.... Well, about those connections you'll understand better if you can find books about electronics. Or even in Internet. You can find a lot of stuff in the web.