Question

A silicon diode in an adapter rectifying circuit has a carrier density of (10⁻²¹) in p material and (10²² ,m⁻³) in n material. The temperature of the charger changes from (25^∘textC) to (40^∘textC). Find the change in barrier potential of the diode.

A) (0.56 ,eV)
B) (0.34 ,eV)
C) (0.22 ,eV)
D) (0.45 ,eV)

Answer 1

The change in barrier potential of the diode is approximately 0.34 eV.

Step-by-step explanation:

The change in barrier potential of a diode can be determined using the equation:

ΔVb = (ΔT / T) * (Vt)

Where:

ΔVb is the change in barrier potential

ΔT is the change in temperature

T is the initial temperature

Vt is the thermal voltage given by Vt = k * T / q

Using the given values, we can calculate:

ΔVb = (40 - 25) / (25 + 273) * (k * (25 + 273) / q) = 0.34 eV

Therefore, the change in barrier potential of the diode is approximately 0.34 eV