Understanding Current Gain in Push-Pull Amplifiers

AI Thread Summary
Current gain in Class AB push-pull amplifiers is primarily determined by the relationship between emitter current and base current, expressed as the ratio i_e/i_b. This ratio can be approximated as h_fe + 1, where h_fe represents the transistor's beta. In the case of Darlington pairs, the "+1" can often be disregarded, simplifying calculations. However, for single transistors operating at high currents, this additional factor may become significant, potentially lowering the gain to around 10. Understanding these nuances is crucial for accurate amplifier design and performance assessment.
saad87
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I'm trying to figure out the currubt gain for a class ab push pull amp. I know that it's basically two emitter followers, so am I correct in assuming that the current gain is just Beta?
 
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the current gain is basically the emitter current over the base curent

\frac{{{i_e}}}{{{i_b}}}\quad = \quad {h_{fe}} + 1

With darlingtons you can forget the 1, but it can be significant with single transistors at high currents where the gain is maybe not more than 10.
 
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