# Variable capacitors (capacitive microphone)

Why in this circuit $$V_{out}=\frac{C_1}{C_1+C_2} V_{cc}$$ and not $$V_{out}=\frac{C_2}{C_1+C_2} V_{cc}$$?

The instructor drew also this picture: but nothing is reported on x-axis and y-axis. Can you help me to understand its meaning? Many thanks simulate this circuit – Schematic created using CircuitLab

Your instructor is right :) Impedance of capacitor is inversely proportional to it's capacitance. $$V_{out}=V_{cc}\frac{\frac{1}{j\omega C_{2}}}{\frac{1}{j\omega C_{2}}+\frac{1}{j\omega C_{1}}}=V_{cc}\frac{C_{1}}{C_{2}+C_{1}}$$

The picture illustrates transient response of the divider. Verical is Vout and horisontal is time. At the first moment output is determined by C1/C2 ratio, but after long time (after relaxation) it determined only by R1/R2 ratio.

$$V_o = \frac{\dfrac{1}{C_2}}{\dfrac{1}{C_2} + \dfrac{1}{C_1}} V_{cc}$$

• A few more words of explanation around this would have made it a good answer to part of the question.
– RoyC
Dec 18, 2017 at 10:04