Doping ex. 2.3, when not to neglect n_D

Physics Q&A
1

Hi, in exercise 2 of Doping I computed the Fermi energy as
E_F = E_c - k_BT \ln \frac{N_C}{N_D-N_A}
when assuming n_D \ll N_D (subquestion 1) and
E_F = E_c - k_BT \ln \frac{N_C+N_D}{N_D-N_A}
using the simplification n_D \approx N_D e^{-(E_c-E_F)/k_BT} (subquestion 3).

How do I now determine at what dopant concentration N_D one cannot assume anymore that all donors are ionized?

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2

The solution is confusing, we should just rewrite it from scratch. This should go as follows.

  1. Take the E_F you found initially assuming n_D \ll N_D.
  2. Substitute it into the expression for n_D.
  3. Check for which T is the resulting n_D \ll N_D.

That’s all there is.

3

Oh nice, I actually did that before. But the question asks for which concentration, not which temperature, should the question be rephrased then?

I used the effective R_E for Germanium as given in the lecture notes, 0.01 eV, so E_D = E_C - 0.01 \;\mathrm{eV}. This resulted in n_D \approx N_D when k_B T \ll 0.01 \; \mathrm{eV} (so then n_D \ll N_D no longer holds). Is that what it should be?

4

It’s a bit trickier.

We have: E_F = E_C - kT \log[N_C / (N_D - N_A)] and n_D = N_D \exp[-(E_D - E_F)/kT] \ll N_D.

Substituting E_F into the second expression gives N_C / (N_D - N_A) \gg \exp[(E_C - E_D)/kT], or N_D - N_A \ll N_C \exp[-(E_C - E_D)/kT].

Then we can take kT \sim 25 meV, which roughly just gives N_D \ll N_C since E_C - E_D < kT.

5

Got it! So basically when the number of available states in the conduction band (greatly) exceeds the dopant concentration, we can take n_D \ll N_D right?

6

Exactly. This also has a very intuitive physical picture behind: it doesn’t cost much energy to ionize a dopant, and at the same time there are many more states available than there are dopants. Therefore most of the time the extra electrons would not be occupying the dopant atoms.

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