Q: Converting 5k pot to 20k pot

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I have a very nice quality 5k stereo audio potentiometer. However the input impedance of my amp is 20k ohm. So I need a potentiometer of 20k.

With my limited knowledge of electronics I came up with the following as a solution: Simply add a 15k (metal film 1% 1/4 watt) resistor to the input of the variable voltage divider--and that's it.

Is my thinking correct? Would really appreciate some feedback thanks!
 
Very rough rule of thumb is that the pot should not be greater in resistance than the amp input. Also significantly greater than the output impedance of whatever comes before. Note I really mean 'output impedance' and not 'recommended load impedance' - people writing blurb often get these two confused.

Many devices have an output impedance in the range 100ohms to 1k, so the pot would typically be 10-50k.
 
Very rough rule of thumb is that the pot should not be greater in resistance than the amp input. Also significantly greater than the output impedance of whatever comes before. Note I really mean 'output impedance' and not 'recommended load impedance' - people writing blurb often get these two confused.

Many devices have an output impedance in the range 100ohms to 1k, so the pot would typically be 10-50k.

I agree with DF96. If your amplifier has a 20k input resistance, I would not use a pot with a higher resistance than, perhaps, 10k. Depending on the output impedance of your source, you may want to use a higher resistance pot. My experience has been that the pot should be an absolute minimum of 10X the output impedance of the source.
 
audiogeekess
I have a very nice quality 5k stereo audio potentiometer. However the input impedance of my amp is 20k ohm. So I need a potentiometer of 20k.

Not at all. Keep your high quality 5 kOhm pot and use it normally. Feeded with a low impedance source, the output at the cursor will never be no more than 1.25 kOhm (for the -6 db position) and this will drive a power amp ok. The less the impedance seen by the amp input, the better.
By the way, what's your pot ?
 
Thanks for the help :)

Turns out my pot measures 20.2k -- don't know why I thought it was 5k (that was when I measured it while it was still connected in another circuit I knew nothing about--ripped it out).

So the pot is 20.2k and the input impedance of amp is 22k. The amp used to have two cheap 20k pots in it -- one for each channel. I took them out and made a fixed voltage divider of (0k, 20.0k) -- i.e. max gain in the amp. I used a metal film resister 1/4 watt, 1% tolerance which -- both measure exactly 20.0k ohm each.

So would it sound better if I go down to 10k stereo pot or just use the one I have?
 
I have the voltage dividers fixed (0,20k) in the amp for max gain, and i need to attenuate what i feed into the back of the amp via the rca's. so use this 20.2k pot for that?

I have the fiio d3 as the source and here are the specs on that:

Specification
[FONT=宋体][/FONT]
[FONT=宋体]●[/FONT]Power supply[FONT=宋体]:[/FONT]DC 5V
[FONT=宋体]●[/FONT]Output amplitude[FONT=宋体]:[/FONT]1.6V[FONT=宋体]([/FONT]0dB Fs[FONT=宋体])[/FONT]
[FONT=宋体]●[/FONT]Frequency Response[FONT=宋体]:[/FONT]20Hz~20KHz
[FONT=宋体]●[/FONT]Signal to Noise Ratio: [FONT=宋体]≥[/FONT]90dB
[FONT=宋体]●[/FONT]Crosstalk: 75dB
[FONT=宋体]●[/FONT]Total harmonic distortion: <0.01%(10mW)
[FONT=宋体]●[/FONT]Size: 61.6mm[FONT=宋体]×[/FONT]49.1mm[FONT=宋体]×[/FONT]21mm
[FONT=宋体]●[/FONT]Weight:50g

It doesn't say anything about output impedance.. should I be able to calculate that by the output amplitude?
 
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