Quantum Capacitance of 2D Material - I

Now, we will derive an expression for the quantum capacitance of a 2D material. Consider a neutral isolated sheet of degenerately doped 2D semiconductor of area $A$. Let the geometric capacitance of the sheet per unit area be $C_0$. The equilibrium condition of the sheet is shown on the left side of the figure below. The density of states $D(E)$ is plotted as a function of energy $E$. The area under the blue curve in the figure represents the total number of free electrons per unit area, $N_0$.

As we have seen in the chapter on carrier statistics, for degenerately doped 2D semiconductors, the number of free electrons can be written as

$$N_0=D_0(E_F-E_0)$$

where $D_0$ is the density of states, $E_F$ is the Fermi energy and $E_0$ is the first subband energy. Now, suppose we add $\delta N$ electrons per unit area to the sheet. This means we add an additional charge of $\delta \sigma =e\delta N$ to the sheet. This will lead to a change in electrostatic potential $\delta {\varphi }_e=e{C}_0\delta N$. A change in electrostatic potential is depicted as an upward shift in the bands by $e\delta {\varphi }_e$ on the right side of the figure above. Therefore the first subband $E_0$ is at $E_0+e\delta {\varphi }_e$ now.

What about the change in the Fermi level ? Let the Fermi level shift by $\delta E_F$. The new Fermi level is at $E_{F1}=E_F+\delta E_F$. The new carrier concentration $N_1=N_0+\delta N$ is shown as the blue shaded area in the figure above. We can write it as

$$N_1=N_0+\delta N=D_0(E_F+\delta E_F-(E_0+e{\varphi }_e))$$

From this equation, it is very easy to conclude that -

$$\delta E_F=e{\varphi }_e+\frac{\delta N}{D_0}$$

This equation shows that the shift in Fermi energy ($\delta E_F$) is not equal to the shift electrostatic potential energy ($e{\varphi }_e$). There is an additional shift of $\delta N/D_0$. However, in the limit that the density of states is very high, $D_0\to \mathrm{\infty }$, the shift in Fermi level is nearly equal to the shift in electrostatic potential energy.

$$\delta E_F\approx e{\varphi }_e(D_0\to \mathrm{\infty })$$

We have already seen the physics behind this previously. A high density of states at a given energy implies that we can pump more electrons to a given energy. In contrast, a low density of states implies we need to go to higher energy levels to add more electrons.

Classical physics assumes a very high density of states. In such a situation, we can see that the system’s capacitance is nearly the same as that of its geometrical/electrostatic capacitance.

$$C=e\frac{\delta q}{\delta E_F}\approx \frac{\delta q}{{\varphi }_e}=C_0$$