Last edited: 2021-04-03 09:38:11

**The chemical potential has different characteristics depending on the temperature. Lets calculate and understand the approximations of the chemical potential in an electron doped semiconductor.**

Lets start with temperatures close to absolute zero. You have probably heard or read that the chemical potential must lie in the center between the donor level and the bottom of the conduction band because of charge neutrality. Lets do some math to find out why that is.

Lets use the variable $E_g$ for the bandgap energy, $E_d$ is the donor energy so the the donor level is situated at $E_g-E_d$. $\mu$ is our desired chemical potential. $n$ is the electron density in the conduction band, $n_d$ is the donor density and $n_d^+$ is the density of the ionized donors. From the Fermi-Dirac distribution we know that the probability of an electron at the energy $E_g-E_d$ and the temperature $T$ is:

$f(E_g - E_d) = \frac{1}{e^{(E_g-E_d-\mu)/(k_B T)}+1}.$So the donor electron density at the donor level is:

$\frac{n_d}{e^{(E_g-E_d-\mu)/(k_B T)}+1}.$Every donor electron not at the donor level is at the conduction band so:

$n_d^+ = n_d (1 - \frac{1}{e^{(E_g-E_d-\mu)/(k_B T)}+1}) = \frac{n_d}{e^{-(E_g-E_d-\mu)/(k_B T)}+1}.$Since we are at low temperatures $T \ll E_d/k_B$. This means $-(E_g-E_d-\mu)/(k_B T) \gg 1$ so we can make the following approximation (the 1 in the denominator is neglected):

$n_d^+ \approx n_d e^{(E_g-E_d-\mu)/(k_B T)}.$Having low temperatures also means that the electrons at the valence band don't have enough energy to make an excitation to the conduction band meaning $n \approx n_d^+$. But the electron density $n$ can also be expressed as:

$n = \frac{1}{\sqrt{2}} (\frac{m_e^* k_B T}{\pi \hbar^2})^{3/2} e^{-(E_g-\mu)/(k_B T)} = n_0 e^{-(E_g-\mu)/(k_B T)}$here is $m_e^*$ the effective mass of the electrons in the conduction band. Lets put the different expressions for $n$ at low temperature equal to each other to get the chemical potential:

$n_d e^{(E_g-E_d-\mu)/(k_B T)} = n_0 e^{-(E_g-\mu)/(k_B T)}$So now we get the chemical potential:

$\mu = E_g - \frac{1}{2} E_d - \frac{1}{2} k_B T \ln{(\frac{n_0}{n_d})}.$So when the temperature approaches zero, $T \to 0$, the chemical potential becomes:

$\mu = E_g - \frac{1}{2}E_d,$since polynomial functions dominates over logarithmic functions ($n_0$ contains $T^{3/2}$) when it comes to limits.

So that is how the chemical potential behaves at low temperatures, how about "medium" temperatures like room temperature? Well here we have more thermal energy to work with so pretty much every donor atom is ionized ($n_d^+ \approx n_d$) but the thermal energy is still not big enough to excite a significant amount of electrons in the valence band so $n \approx n_d$. This makes the math easier, we now have:

$n = n_0e^{-(E_g-\mu)/(k_B T)} = n_d,$so:

$\mu = E_g + k_B T \ln{(\frac{n_d}{n_0})},$for "medium" temperatures.

For high temperatures, like 1200 K, electron holes has manifested at the valence band and now plays the bigger role in the electron density in the conduction band. So now $n \approx p$, where $p$ is the electron hole density at the valence band. The semiconductor is said to be in an intrinsic state, meaning that it can be seen as undoped. So now we have the following relation:

$\frac{1}{\sqrt{2}} (\frac{m_e^* k_B T}{\pi \hbar^2})^{3/2} e^{-(E_g-\mu)/(k_B T)} = \frac{1}{\sqrt{2}} (\frac{m_h^* k_B T}{\pi \hbar^2})^{3/2} e^{-\mu/(k_B T)},$where $m_h^*$ is the effective mass of the electron holes. This leads to the chemical potential of:

$\mu = \frac{1}{2} E_g + \frac{3}{4} k_B T \ln{(\frac{m_h^*}{m_e^*})}.$The chemical potential is located midway between the conduction band and the valence band if $m_e^* = m_h^*$.

So there we have it! We have now calculated the chemical potential for different temperature regions for an n-type semiconductor, although we did it with some significant approximations but without those this would get a lot harder if not impossible.

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