Question

An abrupt pn junction with a built-in voltage of 0.7 V has a junction capacitance of 2pF at zero bias. What bias must be applied to obtain a 1pF capacitance (include both magnitude and the sign of the bias)?

Answer 1

To reduce the capacitance of a pn junction from 2pF to 1pF, a reverse bias of 2.1 V must be applied. This bias will ensure the capacitance is halved under reverse bias conditions.

Step-by-step explanation:

To find out what bias must be applied to a pn junction to change its capacitance from 2pF to 1pF, we can use the relationship between capacitance and bias voltage for a pn junction. The capacitance C of a junction varies with the applied voltage V as:

C = C0 / (1 + V/Vbi)^(1/2)

Where C0 is the zero-bias capacitance, Vbi is the built-in voltage, and V is the applied bias voltage (positive for reverse bias and negative for forward bias).

We are given that C0 = 2pF and Vbi = 0.7 V, and we want to find V when C = 1pF. Substituting the given values and solving for V:

1pF = 2pF / (1 + V/0.7V)^(1/2)

This implies:

(1 + V/0.7V) = (2pF/1pF)^2

(1 + V/0.7V) = 4

V/0.7V = 3

V = 3 × 0.7V

V = 2.1V

A reverse bias of 2.1 V must be applied to obtain a 1pF capacitance. The sign of the bias is positive, indicating a reverse bias condition.