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Looking for how this sound sensor works, I found this: "This sound sensor is used to detect whether there’s sound surround or not, please don’t use the module to collect sound signal. For example, you can use it to make a sound control lamp, but not as a recording device."

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Here are my opinion and doubts. Can anyone clarify this for me?

  1. C3 is used to filter VCC from the non inverting input of U1B and allow only the voltage changes of MIC1

  2. Since all this voltage difference will go to R6, why R7 is needed?

  3. Why put 2 series op amp which have the same specs?

  4. Why C2 is needed?

  5. Why it cannot capture frequency sound but it can capture intensity?

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  • \$\begingroup\$ Something is missing. After C3 you should have a dc bias by two big resistors. I would guess R7 should be connected to VCC while the node of C3 straight to U1. \$\endgroup\$ – Gregory Kornblum Apr 6 '17 at 20:57
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Here is my interpretation.

As you suspect C3 ac couples the output of the microphone, which is biased up about 1V when all is quiet, down to ground on pin 5 of U1B.

When sound happens it will force the voltage on R6 up and down around zero volts..

R7 presents a higher input impedance to the op-amp and also prevents those negative noise spikes from back biasing the op-amp.

Two amps are used, probably to provide enough gain at the frequency range required. There is sufficient gain here to cause the output to saturate with noise over a certain level.

C2. I have no idea why they decoupled the output, since that will depend on what comes next.

This circuit is not suitable for capturing the sound itself for a few reasons,

  1. The low going swing on the mic is cut off by the zero volt bias on U1B
  2. The gain is sufficient to swamp / distort the signal at high sound levels
  3. Since the signal is biased to the rail of the opamp, again distortion is generated.

It is truly set up as a sensor, through perhaps not a very elegant one, not a microphone.

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    \$\begingroup\$ Makes sense to me. Let's wait to see if someone has anything new to add it up. \$\endgroup\$ – GabrielRado Apr 6 '17 at 21:01
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If this is not an error, then

  1. C3 cuts off the bias voltage for mic. C3 forms a low pass filter with R6.
  2. Real opamp has something called bias current. If you would connect C3 directly to UB1+ pin, then C3 would be charged. Voltage at it would rise linearly.
  3. Only buffering comes to my mind. However then first op amp would be set as voltage follower. Another answer for 3 could be to have higher gain, without too much noise. Please remember that higher impedance introduces more noise.
  4. They probably assume that usage is unknown. You might want to connect the module to ADC or maybe something else.
  5. Due to bias point on UB1+, you have only positive half of the input signal. For negative half amp goes into saturation at negative supply rail (non-inverting configuration).

If this is an error, author probably would like to use differential amplifier configuration (R6 moved behind R7). However then, the circuit could be used as mic. pre-amp (with gain of 100).

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