Electron and hole concentration

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 replies · 2K views
NerdyGuy
Messages
1
Reaction score
1
Homework Statement
Physics, Semiconductor, carrier concentration
1. part unsolved
2. part solved, but not sure
Relevant Equations
##n = N_C \exp \left( - \frac{E_C - E_F}{k_B T} \right)##
##n_i (T) = \sqrt{N_C N_V} \exp \left( - \frac{E_g}{2 k_B T} \right)##
I can't solve the following exercise:

Assume for a certain non-degenerate semiconductor sampe at T = 300 K an intrinsic carrier concentration ##n_i = 2 \cdot 10^{13} \frac{1}{cm^3}## and the band effective densities of states ##N_C = N_V = 10^{19} \frac{1}{cm^3}##.
1. Determine the electron and hole concentrations n and p.
2. Find the position of the Fermi level in respect to the conduction band.

For part 1 I tried:
$$n_i (T) = \sqrt{N_C N_V} \exp \left( - \frac{E_g}{2 k_B T} \right) \\
= ... \approx 0.68 eV$$
But here I'm not sure if this is necessary and how to continue. Can anybody please help me?

My calculation for 2 is:
$$n = N_C \exp \left( - \frac{E_C - E_F}{k_B T} \right) $$
$$\Leftrightarrow E_C - E_F = k_B T \ln \left( \frac{N_C}{n} \right) $$
$$\Leftrightarrow E_C - E_F = 1,38 \cdot 10^{-23} \frac{J}{K} 300 K \ln \left( \frac{10^{19} \frac{1}{cm^3}}{2 \cdot 10^{13} \frac{1}{cm^3}} \right) $$
$$\approx 0.34 eV$$

Can anyone confirm this? Is the last step correct, where I set ##n = n_i = 2 \cdot 10^{13} \frac{1}{cm^3}##?

Best regards

NerdyGuy
 
Last edited by a moderator:
Reply
  • Like
Likes   Reactions: Delta2
on Phys.org
First, welcome to PF!

For part #1, you did not state what values ##n## and ##p## are. Is this an intrinsic sample? If it is you can answer part #1 trivially. For part #2, you are correct only if this is an intrinsic semiconductor.
 
Last edited: