Question

A n-type silicon with doping carrier concentration $N_D={10}^{16}c{m}^{-3}$ is given. If this sample is uniformly illuminated, the steady state minority concentration in the sample will be $4\times {10}^{3}c{m}^{-3}.$ Then the rate at which electron-hole pairs (EHPs) are generated is
(Assume intrinsic carrier concentration $n_i=1.5\times {10}^{10}c{m}^{-3},$ minority carrier life time ${\tau }_p={10}^{-5sec}$

• A. $1.00\times {10}^{10}c{m}^{-3}/s$
• B. $2.00\times {10}^{10}c{m}^{-3}/s$
• C. $1.75\times {10}^{10}c{m}^{-3}/s$
• D. $1.25\times {10}^{10}c{m}^{-3}/s$

Answer 1

The correct option is C $1.75\times {10}^{10}c{m}^{-3}/s$

$\text{We know that},$
$P=p_0+G_L{\tau }_p$
where,
$P=\text{Steady state minority concentration}$
$P_0=\text{Thermal equilibrium concentration of holes}$
${\tau }_p=\text{Carrier life time}$
$G_L=\text{Generation rate of EHPs}$
but $p_0=\frac{n_i^2}{N_D}=\frac{(1.5\times {10}^{10}{)}^2}{{10}^{16}}=2.25\times {10}^{4}c{m}^{-3}$
$but{G}_L=\frac{P-P_0}{{\tau }_p}=\frac{(4\times {10}^4)-(2.25\times {10}^4)}{{10}^{-6}}$
$\therefore G_L=1.75\times {10}^{10}c{m}^{-3}/s$.