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I have tried simulating a simple class B amplifier, and I can't seem to get any gain (in fact, due to the crossover distortion, the gain is less than 1)

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For example, in this diagram the resistors are biasing the input signal, but I see no control over the gain. Is a class B amplifier only for current amplification?

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This is not a conventional (or at least linear) class-B amplifier, as there is no global negative feedback loop. That's where the gain would be controlled, if it was there. You would also need diodes or a bias spreader transistor instead of Rb1/2 in a practical linear design. –  EJP Jul 5 at 0:37
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The basic problem is that you misunderstand "gain". You are looking at voltage amplification only, not power amplification. This circuit has a voltage gain just below 1, but does have significant power gain.

Each top and bottom half is just a emitter follower. At first approximation, the B-E drop is fixed, but the emitter current is β+1 times the base current. If these transistors have a gain of 50, for example, then for every mA you put into the base, you get 51 mA into the load. Overall, the voltage gain is about 1 and the current gain is about the gain of the transistors, so the power gain is also about the gain of the transistors.

Another way to look at this power gain is as a impedance converter. The input and output signals are about the same in voltage, but the impedance of the output signal is reduced by the gain of the transistors. For example, if the load is a 8 Ω speaker and the transistors have a gain of 50, then the input of this amplifier has a impedance of about 400 Ω. That is a lot easier for the previous stage to drive than a 8 Ω load. No matter how you look at it, in this example the gain is about 10*Log10(50) = 17 dB.

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So the current gain is only a factor of \$\beta\$? I have heard that the \$\beta\$; value varies significantly in the manufacturing process. Won't this cause a mismatch in current between the +ve and -ve cycles?? –  tgun926 Jul 4 at 11:53
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@tgun: Yes, transistor gain can vary widely between parts. You design the circuit to work with whatever the minimum guaranteed gain is, preferably up to infinite gain. In the example I used, if the transistors have a minimum guaranteed gain of 50, then the minimum imput impedance is 400 Ohms. You design the previous stage to be able to drive that, but with no harm if the impedance happens to be higher. A little negative feedback around the whole amp will also help deal with variations and assymetries due to part variances. –  Olin Lathrop Jul 4 at 12:04
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It's not that it's class B, it's that it's a common-collector (emitter follower) push-pull configuration.

That gain of the common-collector configuration, whether single-ended or push-pull, is less than 1.

The voltage amplification must come from previous stages.

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So if I'm amplifying a 20V P-P signal as the input to the class B stage, how can I know the transistors will operate in the linear region? IIRC with the common emitter amplifier the input voltage range was kept small so that the transistor operates in the linear range. –  tgun926 Jul 4 at 10:59
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@tgun926, the common-collector is a large signal amplifier. –  Alfred Centauri Jul 4 at 12:21
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