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2.7 The Double Tunnel Junction

   Consider two tunnel junctions in series biased with an ideal voltage source as shown in Fig. 2.7. The charges on junction one, junction two, and on the whole island can be written as
q_1=C_1V_1, \quad q_2=C_2V_2 \quad\text{and}\quad q=q_2-q_1+q_0=-ne+q_0,
respectively, with n1 the number of electrons that tunneled through the first junction entering the island, n2 the number of electrons that tunneled through the second junction exiting the island, and n=n1-n2 the net number of electrons on the island.
Figure 2.7: Two tunnel junctions in series biased with an ideal voltage source. The background charge q0 is non-integer, and n1 and n2 denote the number of tunneled electrons through junction one and junction two, respectively.

A   background charge q0 produces generally a non-integer charge offset. The background charge is induced by   stray capacitances that are not shown in the circuit diagram Fig. 2.7 and  impurities located near the island, which are practically always present. Using (2.32) and Vb=V1+V2gives
V_1=\frac{C_2V_b+ne-q_0}{C_{\Sigma}}, \quad
V_2=\frac{C_1V_b-ne+q_0}{C_{\Sigma}} \quad\text{with}\quad
With (2.33) the  electrostatic energy stored in the double junction is
In addition, to get the free energy one must consider, as in (2.11), the work done by the voltage source. If one electron tunnels through the first junction the voltage source has to replace this electron -e, plus the change in   polarization charge caused by the tunneling electron. V1 changes according to (2.33) by $-e/C_{\Sigma}$ and hence the polarization charge is $-eC_1/C_{\Sigma}$. The charge q1 gets smaller, which means that the voltage source `receives' polarization charge. The total charge that has to be replaced by the voltage source is therefore $-eC_2/C_{\Sigma}$ and the work done by the voltage source in case electrons tunnel through junction one and junction two is accordingly
\begin{gather}W_1 = -\frac{n_1eV_bC_2}{C_{\Sigma}}\quad\text{and}\quad
The    free energy of the complete circuit is
At zero temperature, the system has to evolve from a state of higher energy to one of lower energy. At non-zero temperatures transitions to higher energy states are possible, but have exponentially reduced probability (see 2.23). The change in free energy for an electron tunneling through junction one and two is given by
\Delta F_1^{\pm}=F(n_1\pm 1,n_2) - F(n_1,n_2) = \frac{e}{C_{\Sigm...
... \frac{e}{C_{\Sigma}}
\left(\frac{e}{2}\pm (V_bC_1-ne+q_0)\right).
The probability of a tunnel event will only be high, if the change in free energy is negative - a transition to a lower energy state. This is a direct consequence of (2.23). The leading term in (2.37) and (2.38) causes $\Delta F$ to be positive until the magnitude of the bias voltage Vb exceeds a threshold which depends on the smaller of the two capacitances. This is the case for all possible transitions starting from an uncharged island, n=0 and q0=0. For symmetric junctions (C1 = C2) the condition becomes $\vert V_b\vert>e/C_{\Sigma}$. This suppression of tunneling for low bias is the  Coulomb blockade. The Coulomb blockade can be visualized with an energy diagram Fig. 2.8.
Figure 2.8: Energy diagram of a double tunnel junction without and with applied bias. The Coulomb blockade causes an energy gap where no electrons can tunnel through either junction. A bias larger than e/C overcomes the energy gap.
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Due to the charging energy of the island, a  Coulomb gap has opened, half of which appears above and half below the original Fermi energy. No electrons can tunnel into the island from the left or right electrode, or out of the island. Only if the bias voltage is raised above a threshold can electrons tunnel in and out, and current will flow.

The same fact, the existence of a Coulomb blockade, is clearly visible in the     IV-characteristic, Fig. 2.9.

Figure 2.9: IV-characteristic of a double tunnel junction. The solid line gives the characteristic for q0=0 and the dashed line for q0=0.5e. The Coulomb blockade is a direct result of the additional Coulomb energy, e2/2C, which must be expended by an electron in order to tunnel into or out of the island.

For low bias no current flows. As soon as Vb exceeds the threshold the junction behaves like a resistor.

However, the   background charge q0 can reduce, or for $q_0 = \pm (0.5+m)e$ even eliminate the Coulomb blockade. This suppression of the Coulomb blockade due to virtually uncontrollable background charges is one of the major problems of single-electron devices. It will be addressed in more detail in Section 5.2.3.

If an electron enters the island via junction one, it is energetically highly favorable for another electron to tunnel through junction two out of the island. Thus an electron will almost immediately exit the island after the first electron entered the island. This is a    space-correlated tunneling of electrons (see Fig. 2.10). After a varying duration another electron might first exit the island via junction two and again, immediately a new electron will tunnel through junction one, entering the island and establishing charge neutrality on the island. If the transparency of the tunnel junctions is strongly different, for example $R_{T1} \ll R_{T2} = R_T$, a  staircase like IV-characteristic appears, as shown in Fig. 2.11.

Figure 2.10: Space-correlation of tunneling in a double tunnel junction. A tunnel event in junction one is immediately followed by an event in junction two and vice versa. The duration between those tunnel pairs is varying.

Figure 2.11: For strongly differing tunnel junctions a staircase like IV-characteristic appears. Depending on which tunnel junction is more transparent, and the direction in which the charge carriers flow, the island will be populated or depleted by an integer number of carriers with increasing bias voltage. If carriers enter the island through the more transparent junction and leave through the opaque one the island will be populated with excess carriers. If they have to enter through the opaque junction and leave through the transparent one, the island is depleted of carriers.
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Carriers enter the island through the first tunnel junction and are kept from the high resistance of the second junction from immediately leaving it. Finally the carrier will, due to the high bias, tunnel out of the island, which quickly triggers another electron to enter through junction one. For most of the time the island is charged with one excess elementary charge. If the bias is increased more electrons will most of the time populate the island. The charge characteristic is shown on the right side of Fig. 2.11. If the  asymmetry is turned around and the second junction is more transparent then the first one, $R_{T2} \ll R_{T1} = R_T$, the island will be de-populated and the charge on the island shows a descending staircase characteristic. Carriers are sucked away from the island through the transparent junction and can not be replenished quickly enough through the opaque one. However, the IV-characteristic does not change.

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Christoph Wasshuber