Question

Consider a silicon sample doped with \(\rm N_D = 1×10^{15}/cm^3\) donor atoms. Assume that the intrinsic carrier concentration \(\rm n_i = 1.5×10^{10}/cm^3\). If the sample is additionally doped with \(\rm N_A = 1×10^{18}/cm^3\) acceptor atoms, the approximate number of \(\rm electrons/cm^3\) in the sample, at \(\rm T=300\ K\), will be ______.

Answer 1

Concept:

According to law of mass action \(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

Application:

The newly created semiconductor is a compensated semiconductor with doping profile, \(\rm N_A = 10^{18} cm^{-3}\) and \(\rm N_D = 10^{15} cm^{-3}\).

The hole concentration due to N_A is very large than Nd

 here, \(\rm N_A\gg N_D\)

So, we can neglect \(\rm N_D\) and the electron concentration can be found as,

using law of mass action 

\(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

\(\rm = \frac{{{{\left( {1.5 \times {{10}^{10}}} \right)}^2}}}{{{{10}^{18}}}}\)

\(\rm = 2.25 × 10^2\)