Question

The delpetion layer in a p-n junction diode is ${10}^{-6}m$ wide and its knee potential is $0.5V$, then the inner electric field in the depletion region is:

• A. $5\times {10}^{6}V/m$
• B. $5\times {10}^{-7}V/m$
• C. $5\times {10}^{5}V/m$
• D. $5\times {10}^{-1}V/m$

Answer 1

Answer: (C)

The correct option is C $5\times {10}^{5}V/m$
In both forward biasing and reverse biasing, applied potential establishes an internal electric field which acts against or towards the potential barrier. This internal electric field is weakened or stronger at the junction. In forward biasing knee voltage is the forwards voltage at which the current through the junction starts to increase rapidly. Once the applied forward voltage exceeds the knee voltage, the current starts increasing rapidly. In forward biasing condition, the inner electric field is given by
$E=-\frac{△V}{△r}$
or $|E|=\frac{△V}{△r}=\frac{5\times {10}^{-1}}{{10}^{-6}}$
$=5\times {10}^{5}V/m$