I am trying to add a volume control to this audio headphone circuit I found online:

enter image description here

As you can see, there are two 470k ohm resistors on the input signal.

Here is my circuit that I created based off this one: enter image description here

I want to know if this would work or not. I have no way of testing this as its all SMD and going on a PCB. Can anyone let me know if this will work??


2 Answers 2


Change it to an active gain control. In your black (why?) circuit, remove R50/51, make the op-amp inverting, connect pin 3 to ground, and provide a linear 50k pot with one end connected to the output at pin 6, the other end connected to the input point of right of R10, and the wiper connected to pin 2. Provide 15pF or so across the pot for stability.

I agree with @AdamHaun about the excessive input impedance and filtering. Just provide C2=220nF looking into R33=47k shunt to ground, and change C4 to 2.7nF to provide a pole at 58kHz: that will be almost flat to 20kHz and still get rid of any RF nasties.

  • \$\begingroup\$ You should also add some series output resistance to protect the opamp against short circuits and low-impedance headphones. \$\endgroup\$
    – user207421
    May 3, 2015 at 1:46
  • \$\begingroup\$ So I did some research about the output impedance issue, so I added in this ferrite in parallel with the 10k: digikey.com/product-detail/en/HI0805O121R-10/240-2393-1-ND/… \$\endgroup\$ May 3, 2015 at 2:46
  • \$\begingroup\$ To limit the maximum gain, add a series resistor at the input side of the pot. \$\endgroup\$
    – user207421
    May 3, 2015 at 3:13

Your circuit will let you turn the volume down, but it won't let you amplify. If that's okay with you, your change shouldn't cause a problem.

I'm not a hi-fi audio expert, but this circuit seems overly complicated to me. The TPA6210A2 datasheet doesn't suggest any need for the elaborate pull-down network on the non-inverting input. I'm not sure that 10 MHz low-pass filter on the input is doing much good, either. (EDIT: It's actually 1.6 MHz -- my mistake.) 470k for the input pull-down seems like way too much, given that larger resistors create more noise.

  • \$\begingroup\$ Okay, so I took the 5 minutes to learn about high and low pass filters. By my calculation and checked via: learningaboutelectronics.com/Articles/… it looks like the low-pass filter is 1.59Mhz from the 1k and 100pF respectively no? If so, why would the original creator do this if 20kHz is human hearing limit??? \$\endgroup\$ May 3, 2015 at 3:41
  • \$\begingroup\$ @JohnAugust Correct. As to why, you would have to ask the original creator. It's a mystery to me. \$\endgroup\$
    – user207421
    May 3, 2015 at 3:45
  • \$\begingroup\$ EJP - Thank you, I made a little spice simulation of the circuit: s27.postimg.org/hym6ypwrn/Frequency_Sweep.png I know that still probably does not help much. I'm sure there was a reason because this persons amp (the qrv09 by Sjostrom audio). I **think the reason is because of RF filtering...? I just dont see what effect the high frequencies would have on things anyway?... \$\endgroup\$ May 3, 2015 at 4:25
  • 1
    \$\begingroup\$ Oops, I forgot the 2*pi in my calculation. 1.6 MHz is somewhat more reasonable. The actual corner frequency will also depend on the source impedance. Your potentiometer drops it to ~60 kHz at mid volume. A 10k pot would probably be better. \$\endgroup\$
    – Adam Haun
    May 3, 2015 at 4:55
  • \$\begingroup\$ @JohnAugust RF junk should be kept out of active audio devices, otherwise it can be rectified and detected by the diode junctions and appear in the audio. \$\endgroup\$
    – user207421
    May 3, 2015 at 8:01

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