Doped Semiconductors — Question 5

Question 5

Partial donor ionization requires a self-consistent population

Neglect holes and acceptors. In a supplied donor-occupation model, $N_D^+=N_D/[1+2e^{(E_F-E_D)/(k_BT)}]$, and $n=N_ce^{-(E_c-E_F)/(k_BT)}=N_D^+$. Define $A=2e^{(E_c-E_D)/(k_BT)}/N_c$.

1. Derive a quadratic equation for n and its physical solution.

2. If $AN_D=100$, find the ionized fraction and compare with the deep-freeze approximation $n\simeq\sqrt{N_D/A}$.

Question 5 - Solution

Partial donor ionization requires a self-consistent population

(a) Method and calculation

$$e^{(E_F-E_D)/k_BT}=(n/N_c)e^{(E_c-E_D)/k_BT}\Rightarrow An^2+n-N_D=0.$$$$\boxed{n=\frac{\sqrt{1+4AN_D}-1}{2A}}.$$

(b) Interpretation and further calculation

$$\boxed{n/N_D=(\sqrt{401}-1)/200=.09512}.$$$$An\gg1:\quad\boxed{n/N_D\simeq(AN_D)^{-1/2}=.100}\quad(5.12\%\text{ high}).$$

Physical check

The degeneracy factor is part of the supplied occupation model. Assuming complete ionization here would overestimate the mobile electron population by more than a factor of ten.