Question

In a $p-n$ junction diode, the thickness of deplection layer is $2 \times 10^{-6}\, m$ and barrier potential is $0.3\, V$. 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

Answer: (C)

Here, the barrier voltage, $V=0.3$ volt
and the width of depletion layer,
$d=2 \times 10^{-6} m$
$\therefore $ Electric field at the junction,
$E =\frac{V}{d}=\frac{0.3\, V }{2 \times 10^{-6} m } $
$=1.5 \times 10^{5} V / m$
The direction of electric field is from $n$ -type to $p$ -type.