Question

An abrupt $P^{+}n$ junction has depletion width of 4 $\mu m$ and built-in potential at junction is 0.8 V. Assume another abrupt $P-n^{+}$ junction has depletion width of $8\mu m$, to maintain the same built-in potential as in the case of $P^{+}-n$ junction the required doping concentration is equal to
$\text{Assume, donor concentration}{N}_D=4\times {10}^{16}c{m}^{-3}$

• A. $N_D$
• B. $2n_D$
• C. $\frac{N_D}{4}$
• D. $\frac{N_D}{2}$

Answer 1

The correct option is C $\frac{N_D}{4}$

Given for $P^{+}-n$ abrupt junction Depletion width, $W_1=4m$
Built-in potential $V_{b1}=0.8V$
For another $P-n^{+}$ abrupt junction
Depletion width $W_2=8cm$
Built in potential $V_2=V_{b1}=0.8V$
Acceptor doping concentration $N_A=?$
We know that

The depletion width,
$W=\sqrt{\frac{2{ϵ}_s{V}_b}{q}[\frac{1}{N_A}+\frac{1}{N_D}]}$
For $P^{+}-n$ abrupt junction depletion width
$W_1=\sqrt{\frac{2{ϵ}_s}{q}\frac{1}{N_D}{V}_{b1}}$
$\because for{P}^{+}-n$ junction $N_A>>N_D$
$Forp-n^{+}junction,N_D>>N_A$
$W_2=\sqrt{\frac{2{ϵ}_s}{q}\frac{1}{N_A}{V}_{b2}}$
$\therefore forp-n^{+}junction{N}_D>>N_A$
By dividing equation (i) with equation (ii),
$\frac{W_1}{W_2}=\sqrt{\frac{N_A}{N_D}}$
$\frac{1}{2}=\sqrt{\frac{N_A}{N_D}}$
$\frac{1}{4}=\frac{N_A}{N_D}$
$\Rightarrow N_A=\frac{N_D}{4}={10}^{16}c{m}^{-3}$