Single Electron Box

We now focus on the derivation of the charging energy of the Single Electron Box device, which will be very useful for the analysis of Single Electron Transistors.

The Single Electron Box device is composed by an isolated metallic island (BOX) which is coupled via a tunnel junction with capacitance $C_2$ to an electrode and via a capacitance $C_1$ to a voltage source $V_G$ (Fig. 6).

The charging energy (or free energy) $F$ is defined as the energy necessary to add N electrons in the BOX, and it is equal to

$$F=E_S-L,$$ (61)

where $E_S$ is the system energy and $L$ is the work done by external forces to the system.

Figure 6: Equivalent circuit of the Single Electron Box : $C_2$ is the tunnel capacitance.

If $n$ electrons are present in the BOX, the following equations are verified

$$\left\{ \begin{array}{ll} q_2-q_1=-nq & \\ V_G=\frac{q_1}{C_1}+\frac{q_2}{C_2} & \end{array}\right .$$ (62)

where $q_1$ and $q_2$ are the charges on the $C_1$ and $C_2$ positive capacitor plates, respectively, and $q$ is the elementary unit charge. If we now define

$$C\stackrel{\mathrm{def}}{=}C_1+C_2, % &$$ (63)

we obtain

$$\left\{ \begin{array}{ll} q_2=q_1-nq & \\ C_1C_2V_G=C_2q_1+... ...1-nq & \\ C_1C_2V_G=C_2q_1+C_1q_1-C_1nq & \end{array}\right .$$ (64)

and further

$$\left\{ \begin{array}{ll} q_1=\frac{C_1C_2V_G+nqC_1}{C} & \\ q_2=\frac{C_1C_2V_G-nqC_2}{C} & \end{array}\right .$$ (65)

The system energy $E_S$ can be then computed as follows

$$E_S = \frac{q_1^2}{2C_1}+\frac{q_2^2}{2C_2}= \frac{(C_1C_2V_G)^2+C_1^2(nq)^2+2C_1C_1C_2V_Gnq}{2C_1C^2}$$

$$+\frac{(C_1C_2V_G)^2+C_2^2(nq)^2-2C_2C_1C_2V_Gnq}{2C_2C^2}$$ (66)

$$= \frac{C_2(C_1C_2V_G)^2+C_2C_1^2(nq)^2+2C_1C_2C_1C_2V_Gnq}{2C_1C_2C^2}$$

$$+\frac{C_1(C_1C_2V_G)^2+C_1C_2^2(nq)^2-2C_1C_2C_1C_2V_Gnq}{2C_1C_2C^2}$$ (67)

$$= \frac{(C_1C_2V_G)^2(C_1+C_2)+C_1C_2(nq)^2(C_1+C_2)}{2C_1C_2C^2}=$$ (68)

$$= \frac{C_1C_2V_G^2+(nq)^2}{2C},$$ (69)

while the external work $L$,

$$L=q_1V_G.$$ (70)

As a consequence the free energy reads

$$F(n,V_G) = \frac{C_1C_2V_G^2+(nq)^2}{2C}-\frac{C_1C_2V_G+nqC_1}{C}V_G$$ (71)

$$= \frac{-C_1C_2V_G^2-2C_1nqV_G+(nq)^2}{2C}.$$ (72)

The number of electrons in the BOX at the equilibrium, can be found imposing

$$\frac{\partial F}{\partial n}=0.$$ (73)

Substituting (74) in (75), we obtain

$$n=\frac{C_1V_G}{q}$$ (74)

from which we can compute the value of $F$ at the equilibrium

$$F=\frac{-C_1C_2V_G^2-C_1^2V_G^2}{2C}.$$ (75)

Since $F$ is a potential, we can refer it with respect to an arbitrary potential, so that $F$ is equal to zero at the equilibrium. In the end we obtain,

$$F(n,V_G)=\frac{(C_1V_G-nq)^2}{2C}.$$ (76)