Question

In an extrinsic semiconductor, the hole concentration is given to be $1.5n_i$ where $n_i$ is the intrinsic carrier concentration of $1 \times 10^{10}$ \(\text{cm}^{-3}\). The ratio of electron to hole mobility for equal hole and electron drift current is given as _____________ (rounded off to two decimal places).

Hint:

For drift currents: $J_n=q n\mu_n E$, $J_p=q p\mu_p E$. If $J_n=J_p$, then $\mu_n/\mu_p=p/n$. Use $np=n_i^2$ to relate $n$ and $p$.

Answer 1

Given: $p = 1.5\,n_i$. From mass–action law, $np = n_i^2 \Rightarrow n = \dfrac{n_i^2}{p}=\dfrac{n_i}{1.5}=\dfrac{2}{3}n_i$.
For equal drift currents, $J_n = J_p$: \[ q\,n\,\mu_n E = q\,p\,\mu_p E \Rightarrow \frac{\mu_n}{\mu_p}=\frac{p}{n} = \frac{1.5\,n_i}{(2/3)\,n_i}=1.5.\frac{3}{2}=2.25. \] \[ \boxed{\dfrac{\mu_n}{\mu_p}=2.25} \]