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Short answer

64.253.163.66

Posted by lipmanl on December 5, 2006 at 10:30:08

In Reply to: Function of coupling cap value in SE amplifier? posted by mbhcid on December 5, 2006 at 08:49:52:

Xc=1/(2*Pi*f*C).
Longer answer:
That means that the reactance of the capacitor is inversely proportional to frequency and to the value of the capacitor. So the value of the capacitor you choose affects the low end of the frequencies that get passed to the output transformer. As you descend into bass, the capacitor offers increasing reactance, which as a first approximation you can think of as resistance) to the bass part of the signal (the mids and highs are unaffected). The lower in frequency you go, the more the bass gets attenuated, and the more phase distortion occurs. So you want to choose a capacitor the is big enough to pass the lowest frequency of interest with minimal attenuation and phase shift.

How do you do that? The simplest, nonmathematical answer is that if your amp is unusually stable, you can use a humongous capacitor and pass very close to DC. If the amp motorboats or exhibits some other oscillation, back off the value. You can parallel capacitors.

For a more complete answer, see any book on electronics. I highly recommend "The Art of Electronics," by I forget who. It's popular, common, readily available, and is extraordinarily well-written and approachable.

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