I have designed a simple circuit that amplifies an electret microphone (this one: https://www.sparkfun.com/products/8635).

This amplified signal must be no greater than 3.3V, so I bring down the voltage to power the OP-AMP using the appropriate voltage divider.

Here is the schematic:


simulate this circuit – Schematic created using CircuitLab

Here's the problem: I hooked up an LED to Vout, and the microphone appears to literally be doing nothing. It is a very high possibility that I just simply don't understand how the microphone works. I have included the product link so that it may be useful for someone to explain the microphone to me.

Another question: the final goal of this circuit is that it feeds in human voice into the ADC of me DE0-Nano FPGA board. Would there be anything wrong with this method? What is a better way?


  • 1
    \$\begingroup\$ There's no way an LM741 will work at 3.3V. If you are using another op-amp, you should mention which one. \$\endgroup\$
    – markrages
    May 14, 2014 at 6:32
  • \$\begingroup\$ Oops! I'm using an LF412CN. \$\endgroup\$
    – thejohnny
    May 14, 2014 at 6:34
  • 2
    \$\begingroup\$ LF412 has a minimum supply voltage of 10V. \$\endgroup\$
    – markrages
    May 14, 2014 at 6:42

1 Answer 1


There are a few things wrong with your circuit.

First, you need an op-amp that will function on 5V. Half of an LM358 might work for you.

Second, you should divide the output voltage of the op-amp, not the power supply.

Third, you must bias the op-amp so that the DC value of the output is near the middle of your ADC range.

Fourth, your gain of 1,001 may be too large, and is certainly too large for a DC gain, you need to separate AC and DC feedback so that the DC gain is much less (close to one is good). Putting a capacitor in series with R2 would allow the DC gain to be 1.

Fifth, whatever op-amp you use, you must have a DC path from both inputs to a bias voltage. Connecting the non-inverting input to the capacitor means it will float to some undesirable voltage in milliseconds. A resistor to an appropriate bias voltage will do.

  • \$\begingroup\$ Thank you so much. I will try all of these suggestions once I dig up the appropriate OP-AMP. Can you clarify some things please? If my ADC takes max of 3.3V, does that mean my middle value (the value to DC bias the OP-AMP to) is 3.3/2 = 1.65V? If my signal has a minimum value of 0V, I don't understand how this biasing would help. My amplified signal would simply have a minimum value of 1.65V. What does this mean? \$\endgroup\$
    – thejohnny
    May 14, 2014 at 17:08
  • \$\begingroup\$ The signal is AC, so it has a mean value (which should be the middle of your ADC range) and it goes plus and minus from there. Suppose your ADC uses a 2.5V reference (it could be 3.3V, I have no way of knowing), so the midscale is 1.25V. If you bias so that the output of the op-amp sits at 1.25V then your AC signal can go plus and minus 1.25V peak (about 0.88V RMS if it was a sine wave). That means anything above 0.88mV from the microphone will be clipped. If you go higher or lower than mid-scale then you lose range either positive or negative. In practice the op-amp won't go quite that far. \$\endgroup\$ May 14, 2014 at 17:13

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