Question

When a p-n junction is reverse-biased, the current becomes almost constant at 25 µA. When it is forward-biased at 200 mV, a current of 75 µA is obtained. Find the magnitude of diffusion current when the diode is
(a) unbiased,
(b) reverse-biased at 200 mV and
(c) forward-biased at 200 mV.

Answer 1

Given:
Drift current (current under reverse bias), i₁ = 25 µA
Forward bias voltage, V = 200 mV
Net current under forward bias, i₂ = 75 µA

(a) When the p‒n junction is in unbiased condition, no net current flows across the junction.
i.e. Drift current = Diffusion current
∴ Diffusion current = 25 µA

(b) Under reverse bias, the built in the potential and applied voltage opposes the motion of the majority carriers across the junction.
Thus, the diffusion current becomes zero.

(c) Under forward bias, the voltage supports the motion of majority carriers across the junction.
Let the actual current be x.
So,
(x − Drift current) = Forward-biassed current

\[\Rightarrow x - 25  \mu \text{ A }= 75  \mu \text{A}\] 

\[ \Rightarrow x = (75 + 25)  \mu \text{ A }\] 

\[ \Rightarrow x = 100  \mu \text{ A }\]