I'm a bit lost trying to solve an issue I'm having with the attached microphone preamp circuit. If anyone recognizes what this problem is, I'd be very grateful for any help - thanks so much in any case!

The issue: as I turn the 50k pot the gain increases and everything seems to be working fine, but the sound disappears from when the knob is around 85% - 100% turned. When the input source is a square wave I can hear the sound of the pulse-width becoming narrower at the moment when the sound disappears. Unfortunately I don't have access to an oscilloscope, so I cannot see precisely what the output signal looks like.

Some guesses I've had were: some inaudible high frequency oscillation begins, or an issue with the power supply since +/-5v is the minimum suggested operating voltage for the NE5532.

The supply voltage is +/-5v, which is being generated from a 7805 (from a 9v DC input), then -5v is generated by an LM2662. On the multimeter these read as +4.99 and -4.98 on the NE55332's supply pins.

Here are the datasheets for the NE5532, and the LM2662:



PCB Layout

preamp circuit


  • \$\begingroup\$ quite possibly the pot itself is faulty, worn track and it goes open circuit near one end. Then the gain becomes very high, the the amp output goes to one rail or other and no AC comes out. clunK! then no sound. Do you have a multimeter? \$\endgroup\$
    – danmcb
    Commented Sep 27, 2021 at 13:55
  • 2
    \$\begingroup\$ You don't have the same resistance to ground from each input of the NE5532, the - input: 510 ohm when at highest gain but + input: 1 Mohm! See this video to learn why you might want to fix that: youtube.com/watch?v=TxBJb-Z0XFI The result could be that the opamp is not outputting 0 V DC when the gain is high and there is no signal. You can check that: don't apply any signal, set the gain to maximum and measure the DC voltage at the NE5532's output. Is is colde to 0 V? \$\endgroup\$ Commented Sep 27, 2021 at 14:02
  • \$\begingroup\$ Hey, @Bimpelrekkie thanks so much for the info! For your check, did you mean to leave the input floating, or to ground it? when it's floating, at power up i see -5.00V volts that slowly 0v (a capacitor somewhere is draining), after this it seems to stay at around 0v (between -.1 and .1 let's say) - when the input is ground the result seems more or less the same - is that normal? \$\endgroup\$
    – otivlasc
    Commented Sep 27, 2021 at 14:22
  • \$\begingroup\$ @Bimpelrekkie I have noticed something else while measuring the output with no signal at the input - when I turn the pot down the output voltage swings to around +1.4v which slowly drains to 0, when I turn the pot up the voltage swings to around -2.5v that then returns to 0 - is that behavior explainable by what you were mentioning? (I'm watching the video as well, thanks so much for that!) \$\endgroup\$
    – otivlasc
    Commented Sep 27, 2021 at 14:30
  • 1
    \$\begingroup\$ What is R6 * Ibias(IN-) * DC gain? The symptoms you describe suggest it may equal or exceed the supply voltage. \$\endgroup\$
    – user16324
    Commented Sep 27, 2021 at 16:04

1 Answer 1


I think your main issue is that your amplifier has a DC gain that is much higher than 1 (one). For audio amplifiers you're generally concerned with frequencies between 20 Hz and 20 kHz so amplifying 0 Hz (DC) is not needed and also it creates problems!

I propose this change to make the DC gain = 1:


simulate this circuit – Schematic created using CircuitLab

The component I added is C2.

I mentioned in the comments that you don't have the same resistance to ground from the + and - inputs of the opamp. This will cause DC offset.

The nice thing about making the DC gain = 1 like this is that this DC offset will not matter much anymore! DC offset is a DC voltage, which, in your original circuit, is amplified.

When you add C2 like I suggest, this amplification becomes 1 (one) and therefore is doesn't matter so much anymore.

  • \$\begingroup\$ hello! I can't thank you enough for putting time into this, and for the excellent suggestions - It's a little bit difficult for me to implement this exact change to the circuit quickly because it's on a circuit board- if the issue is DC offset, is AC coupling the output with a series capacitor + resistor to ground sufficient to determine if DC offset is causing issue? \$\endgroup\$
    – otivlasc
    Commented Sep 27, 2021 at 15:23
  • \$\begingroup\$ No, a series C you already have (C2), and it prevents the DC from reaching the output connector (U1). But at C2, the "damage is already done". What I would do to confirm that DC offset is the issue: short the input to GND and set the gain to max. Then measure the DC voltage at the output of the opamp and wait for the voltage to settle. Then repeat but with gain at minimum. With gain at mimimum the voltage should be much smaller. \$\endgroup\$ Commented Sep 27, 2021 at 15:38
  • \$\begingroup\$ Another way to confirm is to temporarily "hack" the circuit and change it to my proposed circuit. Instead of using one capacitor for R2 and R3, you can also give each resistor its own cap. Desolder R2 and solder it back on the PCB but with one side "sticking up" then solder a 1 uF cap between the R2 and the GND pad. Do the same for R3. Then try the circuit. \$\endgroup\$ Commented Sep 27, 2021 at 15:42
  • \$\begingroup\$ I understand now, before I measured the output after the AC coupling cap, not the op-amps output itself. Now that I measured the op-amps output correctly, can see that the voltage goes from -.18V to -3.60V (at the highest gain)! I will try to make the change on the circuit board next- thanks so much again! \$\endgroup\$
    – otivlasc
    Commented Sep 27, 2021 at 15:55
  • 1
    \$\begingroup\$ I put a 510 ohm resistor in parallel with R6 Yes, that would make the resistance to GND from + and - inputs more equal and reduce the offset. Then even when at full gain, the output of the opamp will not saturate at -3.6 V. \$\endgroup\$ Commented Sep 27, 2021 at 19:35

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