When a p – n junction is reverse–biased, the current becomes almost constant at 25 $\mathrm{\mu }$A. When it is forward–biased at 200 mV, 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 

 

In unbiased condition drift current=diffusion current and net current is zero.

So diffusion current is also  $25 \mathrm{\mu A}$

In the reverse biased condition diffusion current becomes zero.

In forward bias= current (i)= diffusion current(i₁) -drift current (i₂)

$\mathrm{i}=75 \mathrm{\mu A}, {\mathrm{i}}_1=? , {\mathrm{i}}_2=25 \mathrm{\mu A}$

${\mathrm{i}}_1=\mathrm{i}+{\mathrm{i}}_2=75+25=100 \mathrm{\mu A}$