I have to design a headphone amplifier circuit. We are told the impedance of the headphones and the minimum voltage they require and also the voltage source. After researching I have found that the LM386N-4 is a popular choice. (According to a datasheet of LM386N-4 - https://www.icrfq.com/part/2377557-LM386N_4.html ). I've seen many circuit diagrams but what I don't understand is how to pick the voltage gain needed for the amplifier?

I originally thought that you would need to know the input audio signal voltage and when you multiplied that by the gain that would need to give the minimum voltage required by the headphones, but it appears that what matters is the current which I don't get.

And are you able to design such a circuit without knowing about the audio input signal? If you do need to know about it what is it you need to know about it and why?

I would greatly appreciate any help as I am very lost.

Thanks.

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• You say you know all the stuff that you are asking about. I don't understand the basis of your question. – Andy aka Aug 31 '17 at 18:01
• What sort of signal would you expect to listen to using a set of headphones....? – Finbarr Aug 31 '17 at 18:08
• Consumer line level signals are 0.894V peak-to-peak. en.wikipedia.org/wiki/Line_level Input current to your amp doesn't matter, the input is supposed to be high impedance. – Dampmaskin Aug 31 '17 at 18:09
• You need to know your source because you may need to construct a specialized pre-amplifier for it that deals with its idiosyncracies (phantom voltage supply, source impedance, etc.) You then need to work out the delivered $V_{PP}$ at maximum volume for the headphones. Knowing the $V_{PP}$ output of the pre-amp, if needed, or the source and knowing the $V_{PP}$ for the headphones you can work out the required voltage gain to get from A to B. Then you may need to plan a volume control and an output stage for current compliance. – jonk Aug 31 '17 at 18:37
• you have to ensure that the impedance of amplifier's output is equal to impedance of headphones. For maximum transmission of power. – MrBit Aug 31 '17 at 19:08