1
$\begingroup$

I am having trouble understanding the formula for total electron concentration in the conduction band (under the Boltzmann approximation): $$n_{0}=\frac{N_{d}-N_{a}}{2}+\sqrt{\left(\frac{N_{d}-N_{a}}{2}\right)^{2}+n_{i}^{2}} $$

While I understand the steps in the derivation, I am having trouble understanding the case where $$N_d =N_a$$ in which the result is: $$n_0 =n_i $$ I will refer to the figure below to explain myself.

From my understanding, if the semiconductor was intrinsic, the electrons in the conduction band would consist only of "thermal electrons", (see figure). If I dope the semiconductor with donors, I will also have the donor electrons regardless of the presence of acceptors (as far as I understand...). The reason I say this is because the acceptors create holes, but they don't "suck up" the electrons from the conduction band (only from the valence band).

Therefore, I cannot find a reason why Na would cancel out Nd. I would expect the following result instead: $$n_0 = n_i + N_d$$

Energy-band diagram of a compensated semiconductor showing ionized and un-ionized donors and acceptors.

Improve this question
$\endgroup$
4
  • $\begingroup$ Detailed balance has to hold in all cases, including dopant compensation. $\endgroup$
    – Jon Custer
    Commented May 29, 2017 at 19:55
  • $\begingroup$ I understand that the charge has to be balanced, I don't understand why the carrier concentration remains n_i $\endgroup$
    – YoA
    Commented Jun 1, 2017 at 7:41
  • $\begingroup$ No, $np = n_{i}^{2}$, and 'detailed balance' does not mean that the charge is balanced. And your answer that you accepted below is only correct under (not always applicable) simplifying assumptions. $\endgroup$
    – Jon Custer
    Commented Jun 1, 2017 at 12:48
  • $\begingroup$ @JonCuster, The answer is correct under the simplifying assumptions that I clearly stated. Also, $n⋅p=n^2_i$ only holds under the Boltzmann approximation, which is the same approximation I made in my answer. $\endgroup$
    – YoA
    Commented Jun 1, 2017 at 13:35

1 Answer 1

Reset to default
0
$\begingroup$

As it turns out, according to this explanation (section 2.6.4.1 'Dopants and impurities'), in room temp the donor electrons more readily "migrate" (energetically) to the acceptor states than to the valence band. Therefore, contrary to the OP, donors and acceptros do 'cancel' out. This is indicated by the red arrows in the figure below.

Moreover, if: $$N_d /N_a \gg n_i $$

and the temperature is about room temp (therefore there is full ionization) then the carrier concentrations are give by:

For n-type:

$$n_0 =N_d-N_a$$

For p-type:

$$p_0 =N_a-N_d$$

which perfectly demonstrates the 'cancelling out', or compensation.

enter image description here

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.