When the pot is near the input stage and the input impedance is high ( Tube, Fet ) and Löw capacitance then the 100kOhm has the advantage that the stage before it does not get loaded much. Than can result in lower distortion when the driving stage is rather weak in clean current delivery.
That means, for high input impedance of input stage, use 100k instead of 10k. For low input impedance of input stage, use 10k instead of 100k???
Your question was with respect to distortion, if you get your signal generator and connect it to first a 10k pot then a 100k pot, you won't see much difference in THD. Distortion is more a product of non-linear gain, so if your pot was the anode load for triode/common emitter gain stage, you would get more distortion with a 10k pot as the load line is steeper and as a result gain on the positive and negative going parts of the cycle will have a bigger difference, will be less linear and as a result have more THD.
But your not likely to use a pot as a load for the above example. Took me ages to get my head round impedance, it's counter intuitive. You can't really think about your pot in isolation, think of impedance's as a potential divider, like a voltage divider and they make more sense.
On the input side you want a high Z to drive into, a low Z IP will be harder to drive as it will need more current to drive it.
On the OP side, you want your circuit to have a low OP Z, it can drive following circuits easier without distortion.
If we knew what your circuit is, we may be better able to advise you. If your pot is the first thing a voltage source sees, a 100k is better than 10k, roughly speaking.
Andy.
But your not likely to use a pot as a load for the above example. Took me ages to get my head round impedance, it's counter intuitive. You can't really think about your pot in isolation, think of impedance's as a potential divider, like a voltage divider and they make more sense.
On the input side you want a high Z to drive into, a low Z IP will be harder to drive as it will need more current to drive it.
On the OP side, you want your circuit to have a low OP Z, it can drive following circuits easier without distortion.
If we knew what your circuit is, we may be better able to advise you. If your pot is the first thing a voltage source sees, a 100k is better than 10k, roughly speaking.
Andy.
Hi Andy,
Thanks for your answer. I am going to place the pot to the input of my headphone amp. I have a 10k and 100k pot, don't know which one is better?
Class A headphone amplifier by Kevin Gilmore
Thanks for your answer. I am going to place the pot to the input of my headphone amp. I have a 10k and 100k pot, don't know which one is better?
Class A headphone amplifier by Kevin Gilmore
Meaningless question. It is very likely that they are of similar quality, so the question you should ask is: what is the appropriate pot value for my situation?edwtsui said:
You can get a first approximation by taking the output impedance of the source and multiplying by the input impedance of the load, then take the square root. Double the answer. This assumes that each are equally likely to have a non-linear impedance. If the source has a more nonlinear output impedance then increase the pot value; if the load has a more nonlinear input impedance then reduce the pot value.
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