Semiconductors — Question 3

Question 3

Compensating dopants reduce majority density while preserving bulk neutrality

A semiconductor has fully ionized donors $N_D=8.0\times10^{21}$ and acceptors $N_A=3.0\times10^{21}$ m$^{-3}$. Its intrinsic density is $n_i=1.0\times10^{16}$ m$^{-3}$. Use $n-p=N_D-N_A$ and $np=n_i^2$.

1. Find leading majority and minority densities and identify the material type.

2. Write the full charge-neutrality relation and estimate the fractional error in the majority-density approximation.

Question 3 - Solution

Compensating dopants reduce majority density while preserving bulk neutrality

(a) Method and calculation

$$D=N_D-N_A=5.0\times10^{21}\,\mathrm{m^{-3}}\gg n_i.$$$$n\simeq D,\quad p\simeq n_i^2/D=2.0\times10^{10}\,\mathrm{m^{-3}}\quad\text{(n-type)}.$$

(b) Interpretation and further calculation

$$\rho_q=e(p+N_D-n-N_A)=0.$$$$n=D+n_i^2/D+\cdots,\quad(n-D)/D\simeq(n_i/D)^2=4.0\times10^{-12}.$$

Physical check

Compensation changes the net dopant imbalance; it does not require a macroscopically charged crystal.