Question

In a p-n junction diode the thickness of depletion layer is $2\times {10}^{-6}m$ and barrier potential is $0.3V$. The intensity of the electric field at the junction is:

• A. $0.6\times {10}^{-6}{Vm}^{-1}$ from n to p side
• B. $0.6\times {10}^{-6}{Vm}^{-1}$ from p to n side
• C. $1.5\times {10}^5{Vm}^{-1}$ from n to p side
• D. $1.5\times {10}^5{Vm}^{-1}$ from p to n side

Answer 1

The correct option is C $1.5\times {10}^5{Vm}^{-1}$ from n to p side

Barrier potential $V=0.3$ volts

Thickness of depletion layer $d=2\times {10}^{-6}$ m

$\therefore$ Electric field intensity at the junction $E=\frac{V}{d}$

or, $E=\frac{0.3}{2\times {10}^{-6}}=1.5\times {10}^5$ $V{m}^{-1}$ (from n to p side)