I would like to add a A50k potentiometer to control the volume of an audio input for a small amplifier. Additionally, I used some easy passive filtering to remove the lowest frequencies and some highs from the signal.

This is the schematic:


simulate this circuit – Schematic created using CircuitLab

I understand that a potentiometer before the filter will alter the cutoff frequency of the lowpass filter. Adding the potentiometer behind the filter did not have any noticable effect.

Is it safe to assume that a potentiometer after the filter will not alter the frequency response significantly? The used Amplifier is a small PAM8403 board (see here). There is a 10k resistor at the input but I couldnt find any other information about the input impedance of the amplifier.

  • 1
    \$\begingroup\$ Depending on how you do it it may or may not have an effect. Adding a pot such that its end points are the the output of the filter and 0V will have an effect because you are loading the filter but may be small if the value of the pot is large enough. Now if I assume the input impedance of the next stage is high then the position of the wiper should adjust the volume without changing the frequency of any of the corners. If this is a 50k pot then the loading will be minimal so effect on corner frequency will be tiny. \$\endgroup\$ Mar 14, 2017 at 18:31

1 Answer 1


Is it safe to assume that a potentiometer after the filter will not alter the frequency response significantly?

If you put the pot before the circuit and it is a high impedance pot then mid-band and mid-position, you will have a fairly high impedance pot loaded by R4 and R5. This can noticeably affect things so, in this position, the pot ought to be 10k i.e. much lower than R4 or R5.

If you put the pot on the output then it needs to be high impedance. This is because the series resistor R6 will not be unduly loaded by a high impedance pot such as 100 kohm.

Two scenarios but different pot values.

A 50k pot would best suit being connected at the output.

  • 1
    \$\begingroup\$ Addition: Actually, best location is in the gain feedback loop of the amplifier itself. Adding high-Z components in the input signal line just adds more noise sensitivity. \$\endgroup\$
    – Trevor_G
    Mar 14, 2017 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.