As long as you use electrolytic capacitors for that purpose, you will always have the problem of leakage current.
It is only a matter of how much.
The offset will of course be less when the output is loaded with a low-impedance load, e.g. a headphone of < 65 ohm.
You can also use so-called low leakage electrolytic capacitors, such as Nichicon UKL.
But what it will not solve is the additional distortion due to the high dielectric absorption of electrolytics.
Which is why people use MKP caps in loudspeaker crossovers.
Or they build amps without output coupling caps.
Patrick
It is only a matter of how much.
The offset will of course be less when the output is loaded with a low-impedance load, e.g. a headphone of < 65 ohm.
You can also use so-called low leakage electrolytic capacitors, such as Nichicon UKL.
But what it will not solve is the additional distortion due to the high dielectric absorption of electrolytics.
Which is why people use MKP caps in loudspeaker crossovers.
Or they build amps without output coupling caps.
Patrick
That's why I would expect the current to drop over an hour or so, that reforms the capacitor to some extent (although not quite as well as when you slowly charge the capacitor to its rated voltage and then leave its rated voltage across it for an hour). By the way, I think cbdb could be on to something.
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If you have a 1k load, try a 100R load. The offset should then be more like 4mV if it's due to leakage.
If so, burn in the capacitor by powering the circuit constantly for a day or two. The leakage should
then be much less. Off the shelf capacitors often need burn-in to meet their leakage specs, since they
can sit on the shelf for months (or years) after being built.
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Does the value of the resistor, R8 to discharge the output coupling capacitor affect the impedance seen by the amplifier ? Lets say I put a value of 10ohm, and connect a headphone of 16ohm impedance, the impedance seen by the amplifier is 6.15ohm ? Does this make it harder for the amplifier to drive the headphone ? Will it change the sound ?
Yes, it is an additional load. The effect on the circuit depends on the particular design.
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