Physics Question on Semiconductor electronics: materials, devices and simple circuits

Question

In a semiconductor electron concentration is $7 \times 10^{13} \,cm ^{-3}$ and hole concentration is $5 \times 10^{12} \,cm ^{-3}$ . Then the semiconductor is

Answer 1

Solution

In a doped semiconductor, the number density of electrons and holes is not equal. But it can be established that $n_{e} n_{h}=n_{i}^{2}$ where $n_{e}, n_{h}$ are the number density of electrons and holes, respectively and $n_{i}$ is the number density of intrinsic carriers (ie, electrons or holes) in a pure semiconductor. In n-type semiconductor, the number density of electrons is nearly equal to the number, density of donor atoms $N_{d}$ and is very large as compared to number density of holes. Hence, $n_{e} \approx N_{a} >> n_{h}$ In p-type semiconductor, the number density of holes is nearly equal to the number density of acceptor atoms $N_{a}$ and is very large as compared to number density of electrons. Hence, $n_{h} \approx N_{a} >> n_{e}$ Since, electron concentration is $7 \times 10^{13} \,cm ^{-3}$ and hole concentration is $5 \times 10^{12} \,cm ^{-3}$ , the semiconductor is $n$ type.