I was designing an audio amp for a final project for an electronics course. We were given a very vague set of requirements and told almost nothing about our input signals or desired output minus relative gains.

This was our exact prompt:

  1. Design, build, and test a practical audio amplifier that would meet the following criteria(this portion is to be built in the lab).

Input signal: Audio frequencies

Voltage gain: 100 or higher

Input Impedance: > 1 M Ohms

Output Impedance: negligible

Load impedance: 8 Ohm speaker

State any assumptions you make, use discrete components (resistors, caps, diodes, transistors, op-amps), power supply use +/- 9 V

We've covered a fair set of transistor topologies, how to calculate gains and good design practices but we did essentially nothing on push-pull amplifiers, which is what my design is based on:

chart 1 is input, 2 is output, 3 is load across 8 ohm resistor (load of output)

Reviewing some of the literature I came to realize this is more or less what a Class AB amp looks like, currently this design can do voltage gains of 1000 (something like 500-2500 if I put a variable pot 5k-15k) but one thing is bothering me. I have no idea how to calculate the current gain from the push pull, and I think that in this design there is a beta dependance on the output current. If so that's a pretty big problem since that means my current gain can wildly fluctuate and not even be the same between two transistors, and can scale unpredictably in different conditions.

My question is twofold:

  1. How do I calculate my current gain for this circuit?


  1. How can I modify this push-pull such that if there is a dependance on beta I can remove that dependence.
  • \$\begingroup\$ The push-pull stage is basically an emitter follower. The current gain is practically the same as Beta. To eliminate dependence on Beta, you'll have to ensure that the driver stage (in your circuit drawn as an opamp) can actually source the required base currents by finding the minimum value for it in the datasheet and use that in your design calculations. If you're testing this circuit in practice, you'll want to verify the opamp output voltage swing. \$\endgroup\$
    – jippie
    Jun 1, 2015 at 6:42
  • \$\begingroup\$ In our course we've taken a lot of transistor circuits and been able to make the gains dependent up on our resistors rather than the beta value. Like the equation of gain for an emitter follower be $R_E/(R_E+r_e+R_b/\beta)$. If I were to add resistors at the bases and emitters could I get a similar effect occuring \$\endgroup\$
    – Skyler
    Jun 1, 2015 at 6:56
  • \$\begingroup\$ sorry, not sure how you do latex/math markup on this SE \$\endgroup\$
    – Skyler
    Jun 1, 2015 at 6:57
  • \$\begingroup\$ latex/math markup on this SE \$\endgroup\$
    – jippie
    Jun 1, 2015 at 9:29
  • \$\begingroup\$ The only thing that matters with changing Beta is the ratio between base resistor (divided by Beta) and R(E)+r(e). If R(B)/Beta << R(E)+r(e) you accomplished your goal. In practice << means about a factor 10. In this case you can neglect the R(B)/Beta part. \$\endgroup\$
    – jippie
    Jun 1, 2015 at 9:44

3 Answers 3


Your basic concept is OK, but you need to think about some details. You are essentially using push-pull emitter followers to get a lot of current gain but otherwise not so great characteristics, then a opamp and closed loop feedback around the whole mess to fix the problems.

Again, that's not necessarily a bad concept for a assignment like this. However:

  1. You only need a gain of 100. There is no point making things more difficult by exceeding that by 10x. I'd go a bit above the minimum to make sure the specs are met, but otherwise going way beyond specs is a waste. Lots of gain has drawbacks too.

  2. Yes the feedback will fix a lot of sins of the basic emitter follower power amp. But, the opamp is a real-world device, so it won't correct for everything perfectly. Look for simple ways to make the basic power amp more linear. For example, note the instantaneous jump the opamp output voltage must make when transitioning between driving high and driving low. Think about how you might be able to lessen the two diode drop jump.

  3. Your output current capability is basically the opamp output current times the gain of the transistors. You need to look those up and then compare to the maximum your load requires.

    If what you have can't supply enough current, then you need more current gain between the opamp and the output. Think about how you can use two transistors on each side to get more current gain, but not make the voltage gap at crossover even worse than it already is.


The current gain for the transistors in the push-pull pair will essentially be their current gain as described above. In a good amplifier design it shouldn't be critical anyway because of the use of negative feedback. As described, cascading transistors (check out Sziklai pairs) could be useful to allow lower drive currents. The other important point I'd make is that the version you show is really class B because there is no DC through the output pair at 0V.


If the opamp is ideal then you don't have too manyworries BUT the expected peak current is 1 amp allowing for 1volt Vce so you need more gain in the transisters Often a darlington or similar is used to keep base currents low


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