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Sir/Madam, Kindly explain how is the following equation derived for the polarised condenser microphone? The output voltage V0 fed from the microphone to the preamplifier will be

how is it derived? Thank you,

circuit diagram equation

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On the input side of \$C_c\$, you have a voltage source (I assume of value E0) connected through \$R_c\$ to three capacitive loads \$C_s\$, \$C_t\$ and your signal \$\Delta\$C(t). In the steady state, since there's no DC path to ground and \$\Delta\$C(0)=0, the junction of \$R_c\$ and the capacitors will be E0.

However, when \$\Delta\$C(t) changes, the coulomb count in your capacitor array cannot change instantly, so as a result the voltage changes. The change in voltage is proportional to the change in capacitance, which is

\${\Delta C(t)\over C_s + C_t}\$

Multiply that by E0 and you've got the dynamic component of your waveform.

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  • \$\begingroup\$ Unlike Physics, EE uses \$ to start and end inline Mathjax. This was done to avoid breaking old posts from before MathJax was added to the site that used $ to indicate dollars. \$\endgroup\$ – The Photon Sep 25 at 16:06
  • \$\begingroup\$ So physicists are less concerned with money than engineers? :-) \$\endgroup\$ – Cristobol Polychronopolis Sep 25 at 20:12
  • \$\begingroup\$ Sure. Engineering is always about what you can do for a price acceptable to your customers. \$\endgroup\$ – The Photon Sep 29 at 14:35

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