# Explain how condersor microphone operates

1. Jun 9, 2012

### thisischris

The sound waves cause the felxible front plate to vibrate and change the capacitance. Moving the plates closer togther increases the capacitance. Moving the plates apart decreases the capacitance.

Explain how the sound wave produces an alternating output signal.
Heres a picture of the diagram:

I have trouble understanding/predicting the behavior of an increased / decreased capacitance would have 'across the capacitor', specially with the fact that I'd assume it wouldn't perfectly match the audio wave due to 'charging/discharging' time.

Example of 'event'
'Charges until pd across capacitor equals pd across cell' , sound wave applied forces the capacitor closer, hence pd drops across capacitor, hence increase in pd across resistor?

Could anyone mention if the above is a suitable/correct explanation?

2. Jun 9, 2012

### Andrew Mason

Think of the capacitor consisting of two membranes/plates separated by a distance s to which a constant voltage V is applied. s is much greater than the amplitude, ds, of the sound vibrations. Using the formula for charge Q on the capacitor you can show that the charge will vary in time in a way that is proportional to the amplitude of the sound vibrations:

The charge, Q, on the plates depends on the separation s according to:

Q = VC = VεA/s

dQ = VεA(1/s - 1/(s-ds)) = VεAds/(s^2-sds) ≈ Kds since s^2>>sds

So: dQ/dt ≈ Kds/dt

Since ds/dt is the speed of vibration and dQ/dt is the current, you can see that the microphone will produce a current whose amplitude is directly proportional to the speed of vibration. The speed of vibration is simply the first derivative of the amplitude so it has identical shape as the sound wave. So the current is a perfect analog of the sound.

AM