Question

If the reverse bias voltage of a silicon varactor is increased by a factor of 2 , the
corresponding transition capacitance

Answer 1

Capacitance:

$C=\frac{q}{U}$

$C_1=\frac{q}{U_1}$

$C_2=\frac{q}{U_2}$

where $U_2=2U_1$ by the condition.

Divideng the equations, we get:

$\frac{C_2}{C_1}=\frac{U_1}{U_2}=\frac{U_1}{2U_1}=\frac{1}{2}$

Capacitance will decrease by 2 times.