It has been stressed out in the books that I've consulted that, for an intrinsic semiconductor, $n=p$.

However, with this in mind, they also derivate the following equation: $$E_{F_i}=\frac{E_c+E_v}{2}+\frac{3}{4}k_BT\ln\left(\frac{m^*_h}{m^*_e}\right) \quad\quad\quad\quad (1)$$ Which would be the Fermi energy level of an intrinsic semiconductor, depending on temperature. Meaning that for an intrinsic semiconductor, $E_F$ would be a little bit shifted from the center if the masses of the holes and electrons are different (in general they are).

This has implications if we want to calculate $n$ and $p$, which wouldn't be equal, because they have a dependance on this energy level. I guess that this is a contradiction, because you start with the assumption of $n=p$ but if you want to calculate them using (1), you end up with them being $n \neq p$. Why is that? Which one is correct?

Skip the following derivation if you already know the dependance of $n$ and $p$ on $E_F$.

$$n=2\int^{\infty}_{E_c} \frac{g_c(E)}{1+e^{\frac{E-E_F}{k_BT}}} \ \mathrm{d}E= 2\int^{\infty}_{E_c} \frac{g_c(E)}{1+e^{\frac{E-E_c+E_c-E_F}{k_BT}}} \ \mathrm{d}E$$ Change of variables: $x=\frac{E-E_c}{k_BT}$ and $\xi_n =\frac{E_c-E_F}{k_BT}$; and supposing that for a 2D semiconductor $g_{2D}$ is independent of E: $$n=2g_{2D}k_BT\int^{\infty}_{0} \frac{1}{1+e^{x}e^{\xi_n}} \ \mathrm{d}x$$ Same goes for p, using the same arguments, and with $\xi_p =\frac{E_F-E_v}{k_BT}$: $$p=2\int^{E_v}_{-\infty} \frac{g_c(E)}{1+e^{\frac{E_F-E}{k_BT}}} \ \mathrm{d}E =2g_{2D}k_BT\int^{\infty}_{0} \frac{1}{1+e^{x}e^{\xi_p}} \ \mathrm{d}x$$ So in the end we have $$n=F_0(\xi_n) \quad \mathrm{and} \quad p=F_0(\xi_p), \quad \xi_n \neq \xi_p$$ where $F_j(-\xi)$ is the Complete Fermi–Dirac integral

  • $\begingroup$ If the effective masses are not the same (normally true), then the Fermi energy moves with temperature. $\endgroup$
    – Jon Custer
    Commented May 16, 2020 at 2:50
  • $\begingroup$ @JonCuster I've already said that in the post, my question is if that implies p≠n. $\endgroup$ Commented May 16, 2020 at 9:36

1 Answer 1


Your equation 1 was derived with an approximation for the Fermi-Dirac integral and was derived for 3D. That is, they used $F_{\frac{1}{2}}\left(\eta_c\right) \approx \frac{\sqrt{\pi}}{2} e^{\eta_c}$, which is appropriate for many situations of interest. You're not getting the answer you expect because you are working in 2D (and not making the same approximation). For reference, see Robert Pierret's Semiconductor Device Fundamentals section 2.5.1 and 2.5.6.

FWIW, in 2D, you can do the relevant Fermi-Dirac integral exactly, so I don't think there's any need for approximation. That said, I don't know the 2D equivalent of your first equation off hand. However, it should be simple to derive by following the steps in Pierret. I'm guessing it's somewhere in John Davies' The Physics of Low-dimensional Semiconductors, but I don't have a copy of that handy.

  • $\begingroup$ I've been flicking through some pages of John Davies' book and I think that it'd certainly help me a lot since the problem I had was in 2D, thank you. The thing is that even for a 3D intinsic semiconductor, you still have to give $F_{1/2}$ an argument which is different for p and n, resulting in n≠p, and that is still bugging my mind. $\endgroup$ Commented May 16, 2020 at 23:46
  • $\begingroup$ The reason that the argument is different for electrons and holes is that a hole is an absence of an electron. So if electrons have some distribution $f\left(E\right)$, then holes have a distribution $1-f\left(E\right)$. Since the distribution is a Fermi-Dirac distribution, you can show that $1-f\left(E\right) = f\left(-E\right)$. In other words, holes have a flipped sign in the argument. That flipped sign carries over to the Fermi-Dirac integrals, and that's basically the difference in argument. For a reference on the sign flip, see J.M. Ziman's Principles of the Theory of Solids section 4.6. $\endgroup$
    – lnmaurer
    Commented May 17, 2020 at 3:28
  • $\begingroup$ I'll add that that section in Ziman is a good reference for a lot of the stuff discussed here. It covers basically the same thing as Pierret but more rigorously and compactly. (Not surprising since Ziman's book is aimed at graduate students --- basically the same level as Ashcroft and Mermin --- and Pierret is aimed at undergraduates.) $\endgroup$
    – lnmaurer
    Commented May 17, 2020 at 3:35
  • $\begingroup$ I've considered thay $1-f(E)=f(-E)$ in the derivation, notice that inside the exponential of $n$ there's $E-E_F$ and $p$ has $E_F-E$. $\endgroup$ Commented May 17, 2020 at 16:05
  • 1
    $\begingroup$ Okay, I've overcame my problem. I was supposing that $g_{2D}$ was the same in both cases, but if you have $m_e \neq m_h$ so different that they have a big enough contribution in the logarithm (the expression for $E_{F_{i}}$ is indeed different in 2D) then the difference $g_n \neq g_p$ would also be large enough to change n and p. In fact $\xi_n \neq \xi_p$ and $g_n \neq g_p$ but mixing it all up you get $n=p$, which is a beautiful answer. $\endgroup$ Commented May 17, 2020 at 21:17

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