capasitor microphone?

Question

one of the earliest high quality mic. was d capasitor mic. consisting of a flat circular disk of metal foil separated by a thin insulating washer around its periphery from a flat conducting sheet. the foil and plate form a capasitor. the plate is grounded and the foil is raised to a potential V through a resistor. if the resistor is 10x10^6 ohm and if the foil diameter is 75mm, its separation from the plate 1mm, potential 100V, what rms voltage change when 500Hz vibrates the foil through a total amplitude of 40micrometer at its midpoint. assume that the midpoint amplitude of the foil is twice the average displacement across the entire foil surface.

Answer 1

If the voltage is maintained on a capacitor, its charge is V*C. The capacitance varies with displacement. The capacitance between two parallel plate is e*A/d, where A = area and d=distance between the plates. In the case of the microphone, we can approximate the capacitance by using the average distance as if it were between parallel plates, The microphone capacitance is then varying in a sinusoidal manner with peak-to-peak amplitude of e*A/(d-.5*s) - e*A/(d+.5*s) where s = the average displacement from vibration . (Average is given as half the mid-point deflection of 40um.) Therefore the capacitance is

C(t) = e*A*.5[1/(d-.5*s) - 1/(d+.5*s)] * sin(2*pi*f*t) where f=freq,

The charge variation is Q(t) = V*C(t) = V*e*A*.5[1/(d-.5*s) - 1/(d+.5*s)] * sin(2*pi*f*t). The current is d/dt [Q(t)] = 2*pi*f*V*e*.5[1/(d-.5*s) - 1/(d+.5*s)] * cos(2*pi*f*t). The signal voltage is the load resistance times this, or Vs(t) = R*2*pi*f*V*e*.5[1/(d-.5*s) - 1/(d+.5*s)] + cos(2*pi*f*t)
The RMS voltage of a sine or cosine wave is .707 times the peak amplitude, so

Vs(RMS) = .707* R*2*pi*f*V*A*e*.5[1/(d-.5*s) - 1/(d+.5*s)]

The construction description implies that there is air between the capacitor discs, so e = e0 = 8.85*10^-12 farad/meter. d = 2mm; s = 20um (average full displacement); A = pi*(D^2)/4, D=75mm; f = 500Hz; V=100v; and R = 10*10^6 ohm.

Using Mathcad to keep the units straight, I get Vs(RMS)=0.217volts

EDIT

The equation can be simplified: 1/(d-s) - 1/(d+s) = [(d+s) - (d-s)]/(d^2-s^2) = 2*s/(d^2-s^2). Since s<
Vs(RMS) = .707* R*2*pi*f*V*A*e*(s/d^2)

The actual result should take into account the impedance of the capacitor compared to the resistor. In this case, the nominal capacitance is 19.5pF and its reactance at 500Hz is 16*10^6 ohm. This is close enough to the load resistance (10*10^6 ohm) not to be ignored; that is, not all the current generated by the capacitor flows through the resistor. The circuit impedance magnitude (R in parallel with C) is 8.5*10^-6ohm, and the output voltage is then 0.185 volt RMS