Sound detector using OpAmp
The circuit you are working with is not very good.


The microphone only gives out a few mV AC on top of the 60 mV DC you measured.
That means you have to adjust R2 very precisely. If it is a single-turn potentiometer on a 3.3 V supply then you will be at \$ \frac {60m}{3.3} = 1.8\% \$ from the end of the potentiometer. This will be very difficult to adjust as you found out.








simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. The first improvements.


It seems that 0 - 120 mV would be enough variation on the potentiometer to make it easier to adjust. This is a ratio of about \$ \frac {3.3}{0.12} = \frac {1}{27.5} \$ of the full scale. Add in R3 and make it about 27 x 10k = 270k. That might be enough to allow you to use the circuit. (You can make R4 a short-circuit, 0 Ω.)








simulate this circuit

Figure 2. Second improvement. Apply some feedback to the op-amp.

In your circuit you are using an op-amp as a comparitor. Op-amps are not well-suited for that application because on some the output "latches" or gets stuck for a short time if it is driven into "saturation" at V+ or V-. A comparitor is usually better.

We may be able to help by adding a little feedback. R5 and R6 set the gain at \$ - \frac{R6}{R5} \$ which is about -50 in this case. That means that a 10 mV swing in input voltage will result in a 0.5 V swing in output. This too should help your sensitivity.

Don't forget to add a decoupling capacitor to the op-amp V+ to ground close by.



As I said at the start, this is not a good circuit but if you try out the steps above to see if you can improve it (or make it less bad!) you will learn some useful theory.