Suppose phosphorus ($\require{mhchem}\ce{P}$) is introduced into $\require{mhchem}\ce{p-Si}$. At a given temperature (say at room temperature $300K$) is it possible to find the location of the Fermi level, if I don't know the intrinsic Fermi level position?

The information I have is:

  • The ratio of acceptors and donors i.e. $N_A/N_D$

  • The intrinsic concentration $n_i$

  • The concentration of holes in $\require{mhchem}\ce{p-Si}$ i.e. $p_o$

The general equation used for locating Fermi level is:

$$E_F-F_{F_i}=kT\ln\left(\dfrac{n_o}{n_i}\right)$$ (for $n$ type)

$$E_{F_i}-F_{F}=kT\ln\left(\dfrac{p_o}{n_i}\right)$$ (for $p$ type)

But, here I don't know the intrinsic Fermi level ($E_{F_i}$). Any way to know the Fermi level just with the given information? (Also, without looking up values from other sources)


For, n-type

$$E_F=E_C-kT\ln \frac{N_C}{N_D}$$

For, p-type

$$E_F=E_V+kT\ln \frac{N_V}{N_A}$$

So, net Fermi level would be

$$E_F=\frac{E_C+E_V}{2}-0.5kT\ln \frac{N_CN_A}{N_DN_V}$$

You can approximate $$n_i=\sqrt {N_CN_V}$$

But of course, you need to know these other parameter values to determine the Fermi level.

  • $\begingroup$ I don't know the values of $E_c$ and $E_v$. The original question was: "$P$ is introduced into a p-Si having $p_o=10^{17}$ per cc. Find the position of Fermi level at $300$ K if $N_D=N_A/5$. Given: $k_BT=26 meV$ and $n_i=1.5\times 10^{10}$ per cc". I think it should be possible to do without $E_c,E_v$, given that information. $\endgroup$ – user139621 Dec 4 '17 at 6:22
  • $\begingroup$ Well, then you can go with the original formula that you've written for Fermi shift and consider the intrinsic Fermi level to be $$\frac{E_g}{2}=0.55eV$$ and proceed ahead. $\endgroup$ – Your IDE Dec 4 '17 at 6:30
  • $\begingroup$ Intrinsic Fermi level is $\frac{1}{2}(E_c+E_v)$. Not $\frac{1}{2}(E_c-E_v)=\frac{1}{2}E_g$. $\endgroup$ – user139621 Dec 4 '17 at 6:31
  • $\begingroup$ Intrinsic Fermi level is considered at the center of the Bandgap. Perhaps, ece.umd.edu/~dilli/courses/enee313_spr09/files/… this will help you in your problem solving $\endgroup$ – Your IDE Dec 4 '17 at 6:42
  • $\begingroup$ "Intrinsic Fermi level is considered at the center of the Bandgap.". Yes. But, instrinsic Fermi level is not $E_g/2$ as you mentioned. It is $E_v+E_g/2$. And I'm not provided with $E_v$. $\endgroup$ – user139621 Dec 4 '17 at 6:44

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