I am interested in the following compressor, which I reverse-engineered from a pcb (there might be some mistakes, but I don't think so)


(open the schematic in a separate window to see the small text). The lightning sign is +9V DC and Vin is only AC, with frequencies 20 Hz < f < 800 Hz. The chip is a TL074CN quad op-amp, and there's a jfet which says K192A.

I guessed this is a compressor because it makes sense in this application to have one (e.g. make the bass more rich), but I am otherwise clueless as to what this part of the circuit actually does.

I suspect it does amplify the signal a bit, could you tell me how much? I am in principle only interested in the linear range of the compressor, as my point is to bypass it, but I need to match the input and the output's magnitude. Of course, if you want to tell me exactly what the circuit does, please by all means, go ahead :)

EDIT: To clarify: In my previous question (see Short circuit or disconnect to bypass a compressor), I wanted to bypass this compressor. In this question, I am interested in its response, particularly in the linear range.

  • \$\begingroup\$ Perhaps you could redraw the schematic for better readability by using the integrated tool, it would help getting better response. \$\endgroup\$
    – clabacchio
    Commented Feb 24, 2014 at 11:00

1 Answer 1


Connect the JFET gate down to 0V to disable the compression that the JFET brings about. As for gain, I can't tell you because the component references as too small to list and ditto the values. But I can say the the lower right-hand op-amp has the answer to the gain it brings about before the compression threshold starts to have its effect.

enter image description here

One error on the circuit - where I've put "shorting this kills compression" - the short should go to 0V not 4.5V.

  • \$\begingroup\$ I replaced the picture with a better one. How did you draw your conclusions? Out of habbit of these modules? Otherwise could you expand a little? \$\endgroup\$
    – yannick
    Commented Feb 23, 2014 at 19:30
  • \$\begingroup\$ @yannick I've added a modified circuit with my notes on - this should help. \$\endgroup\$
    – Andy aka
    Commented Feb 25, 2014 at 23:27

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.