Question

In the circuit shown in the figure below, I = 1 mA is a dc current and $V_{in}$(t) is a sinusoidal voltage with small amplitude :

By taking diode by its small resistance $r_d$, the cut off frequency $f_H$ is ____kHz.
(Assume $V_T=25$ mV at room temperature )

• A. 636.62

Answer 1

The correct option is A 636.62
Small signal equivalent resistance,

$r_d=\frac{\eta V_T}{I_D}=\frac{1\times 25\times 10-3}{1\times {10}^{-3}}$

$r_d=25\Omega$

$\frac{V_{0\left(s\right)}}{V_i\left(s\right)}=\frac{\frac{1}{sC}}{\frac{1}{sC}+r_d}=\frac{1}{1+r_{d}sC}$

For 3 dB cut off frequency,

the magnitude, $|\frac{V_0\left(s\right)}{V_i\left(s\right)}|=\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{1^2+r_d^2{\omega }_H^2{C}^2}}$

$r_d^2{\omega }_H^2{C}^2=1$

${\omega }_H^2=\frac{1}{r_d^2{C}^2}$

${\omega }_H=\frac{1}{r_{d}C}$

$f_H=\frac{1}{2\pi r_{d}C}=\frac{1}{2\pi \times 25\times 10\times {10}^{-9}}=636.62kHz$