When a p – n junction is reverse–biased, the current becomes almost constant at 25 $\mathrm{\mu }$A. When it is forward– biased at 200mV, a current of 75$\mathrm{\mu }$A is obtained. Find the magnitude of diffusion current when the diode is
a) unbiased 
b) reverse–biased at 200mV and
c) forward–biased at 200mV

p-n junction is reversed biased.

$\begin{array}{r}{\mathrm{i}}_1=25 \mathrm{\mu A}\\ \mathrm{V}=200 \mathrm{mA}\\ {\mathrm{i}}_2=75 \mathrm{\mu A}\end{array}$

(a) unbiased condition.
Diffusion current = Drift current = i₁

∴ Diffusion Current = 25μA
(b) Reversed biased condition at 200 mV
On the reverse biasing, the diffusion current becomes zero.
(c) Forward biased condition at 200mV
At forward biased, the actual current be i
So, 

i - Drift Current= Forward biasing current.

$\begin{array}{rcl}\mathrm{i} - {\mathrm{i}}_1 & =& {\mathrm{i}}_2\\ \mathrm{i}-25 \mathrm{\mu A}& =& 75 \mathrm{\mu A}\end{array}$

$\mathrm{i} = 75 + 25 = 100 \mathrm{\mu A}$