View Full Version : Impedance and passive volume control
I need a bit a help understanding this.
I gather, so far, that in an ideal world a bit of kit should have a high input impedance and a low output impedance, so when connecting two bits of kit together the cable joins a low to a high impedance. The cable may influence that impedance, but that's a separate topic.
My question is, does a passive volume control affect impedance or is it 'invisible' in that consideration.
My understanding is that a passive pre provides no gain to the following amp, but that an active pre provides signal gain. I am not aware that an active pre varies the impedance. However someone might be along to say that I am talking complete bollocks:lol: (a nod to Mark Williams - The Fast Show for the last comment)
EDIT: Blimey, just read this which if it is correct I was talking doo doo :D
https://www.audiosciencereview.com/forum/index.php?threads/what-is-the-practical-difference-between-a-passive-preamp-and-an-active-one.14047/
a passive's pot will affect the impedence as its a resistor but if its not too high it shouldnt affect it too much... A pot value of 1 to 2K should be fine. the impedance ratio is normally between 5 and 10 times i think.
a passive's pot will affect the impedence as its a resistor but if its not too high it shouldnt affect it too much... A pot value of 1 to 2K should be fine. the impedance ratio is normally between 5 and 10 times i think.
So if a source having an impedance output of 1k is married to a power amp with 50K input its a good impedance match (50 times)
Now place a PVC in line with say 1 or 2 K pot. Is there now an impedance mismatch?
I would choose a potentiometer having an impedance that is the geometrical mean of the output impedance of the source and the imput impedance of the load. For the example you cite, that would be √(1000 ∙ 50000) ≈ 7000, or 7kΩ. However since this is not a standard value, either 5kΩ or 10kΩ ought to be suitable.
I would choose a potentiometer having an impedance that is the geometrical mean of the output impedance of the source and the imput impedance of the load. For the example you cite, that would be √(1000 ∙ 50000) ≈ 7000, or 7kΩ. However since this is not a standard value, either 5kΩ or 10kΩ ought to be suitable.
Getting a grip on the topic now :), but isn't it true that a volume control of say 10K changes its impedance at different volume settings?
Getting a grip on the topic now :), but isn't it true that a volume control of say 10K changes its impedance at different volume settings?
Yes but it won't necessarily be audible.
If there's a problem you'll hear it as no bass, rolled off top end, or both.
IME a passive with a buffer or an active pre with low gain are the best solutions for a digital source, sidesteps most potential issues.
Getting a grip on the topic now :), but isn't it true that a volume control of say 10K changes its impedance at different volume settings?
Yes it does, but the 10k, for example, is the impedance at the maximum volume setting. At lower a volume setting, that part of the potentiometer not tapped appears in series with the source impedance, with the combined series impedance appearing in parallel with the portion that is tapped. So the effective source impedance is reduced further.
That's great
Thank you all
Powered by vBulletin® Version 4.2.3 Copyright © 2026 vBulletin Solutions, Inc. All rights reserved.