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The background charge of the dielectric surrounding the QD is indicated by <math>q_0</math>. <math>n_{\rm S}</math> and <math>n_{\rm D}</math> denote the number of electrons tunnelling through the two tunnel junctions and the total number of electrons is <math>n</math>. The corresponding charges at the tunnel junctions can be written as:
:<math>q_{\rm S} = C_{\rm S} V_{\rm S}</math>
:<math>q_{\rm D} = C_{\rm D} V_{\rm D}</math>
:<math>q = q_{\rm D} - q_{\rm S} + q_0 = -ne + q_0,</math>
where <math>C_{\rm S}</math> and <math>C_{\rm D}</math> are the parasitic leakage capacities of the tunnel junctions. Given the bias voltage, <math>V_{\rm bias} = V_{\rm S} + V_{\rm D},</math> you can solve the voltages at the tunnel junctions:
:<math>V_{\rm S} = \frac{C_{\rm D} V_{\rm bias} + ne - q_0}{C_{\rm S} + C_{\rm D}},</math>
:<math>V_{\rm D} = \frac{C_{\rm S} V_{\rm bias} - ne + q_0}{C_{\rm S} + C_{\rm D}}.</math>
The electrostatic energy of a double-connected tunnel junction (like the one in the schematical picture) will be
:<math>E_C = \frac{q_{\rm S}^2}{2 C_{\rm S}} + \frac{q_{\rm D}^2}{2 C_{\rm D}} = \frac{C_{\rm S} C_{\rm D} V_{\rm bias}^2 + (ne - q_0)^2}{2(C_{\rm S} + C_{\rm D})}.</math>
The work performed during electron tunnelling through the first and second transitions will be:
:<math>W_{\rm S} = \frac{n_{\rm S} e V_{\rm bias} C_{\rm D}}{C_{\rm S} + C_{\rm D}},</math>
:<math>W_{\rm D} = \frac{n_{\rm D} e V_{\rm bias} C_{\rm S}}{C_{\rm S} + C_{\rm D}}.</math>
Given the standard definition of free energy in the form:
:<math>F = E_{\rm tot} - W,</math>
where <math>E_{\rm tot} = E_C = \Delta E_F + E_N,</math> we find the free energy of a SET as:
:<math>F(n, n_{\rm S}, n_{\rm D}) = E_C - W = \frac{1}{C_{\rm S} + C_{\rm D}} \left( \frac{1}{2} C_{\rm S} C_{\rm D} V_{\rm bias}^2 + (ne - q_0)^2 + e V_{\rm bias} C_{\rm S} n_{\rm D} + C_{\rm D} n_{\rm S} \right).</math>
For further consideration, it is necessary to know the change in free energy at zero temperatures at both tunnel junctions:
:<math>\Delta F_{\rm S}^{\pm} = F(n \pm 1, n_{\rm S} \pm 1, n_{\rm D}) - F(n, n_{\rm S}, n_{\rm D}) = \frac{e}{C_{\rm S} + C_{\rm D}} \left( \frac{e}{2} \pm (V_{\rm bias} C_{\rm D} + ne - q_0) \right),</math>
:<math>\Delta F_{\rm D}^{\pm} = F(n \pm 1, n_{\rm S}, n_{\rm D} \pm 1) - F(n, n_{\rm S}, n_{\rm D}) = \frac{e}{C_{\rm S} + C_{\rm D}} \left( \frac{e}{2} \pm (V_{\rm bias} C_{\rm S} + ne - q_0) \right),</math>
The probability of a tunnel transition will be high when the change in free energy is negative. The main term in the expressions above determines a positive value of <math>\Delta F</math> as long as the applied voltage <math>V_{\rm bias}</math> will not exceed the threshold value, which depends on the smallest capacity in the system. In general, for an uncharged QD (<math>n = 0</math> and <math>q_0 = 0</math>) for symmetric transitions (<math>C_{\rm S} = C_{\rm D} = C</math>) we have the condition
:<math>V_{\rm th} = \left|V_{\rm bias}\right| \ge \frac{e}{2 C},</math>
(that is, the threshold voltage is reduced by half compared with a single transition).
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In the case where the permeability of the tunnel barriers is very different <math>(R_{T1} \gg R_{T2} = R_T),</math> a stepwise I-V characteristic of the SET arises. An electron tunnels to the island through the first transition and is retained on it, due to the high tunnel resistance of the second transition. After a certain period of time, the electron tunnels through the second transition, however, this process causes a second electron to tunnel to the island through the first transition. Therefore, most of the time the island is charged in excess of one charge. For the case with the inverse dependence of permeability <math>(R_{T1} \ll R_{T2} = R_T),</math> the island will be unpopulated and its charge will decrease stepwise.{{citation needed|date=January 2020}} Only now can we understand the principle of operation of a SET. Its equivalent circuit can be represented as two tunnel junctions connected in series via the QD, perpendicular to the tunnel junctions is another control electrode (gate) connected. The gate electrode is connected to the island through a control tank <math>C_{\rm G}.</math> The gate electrode can change the background charge in the dielectric, since the gate additionally polarizes the island so that the island charge becomes equal to
:<math>q = -ne + q_0 + C_{\rm G}(V_{\rm G} - V_{2}).</math>
Substituting this value into the formulas found above, we find new values for the voltages at the transitions:
:<math>V_{\rm S} = \frac{(C_{\rm D} + C_{\rm G}) V_{\rm bias} - C_{\rm G} V_{\rm G} + ne - q_0}{C_{\rm S} + C_{\rm D}},</math>
:<math>V_{\rm D} = \frac{C_{\rm S} V_{\rm bias} + C_{\rm G} V_{\rm G} - ne + q_0}{C_{\rm S} + C_{\rm D}},</math>
The electrostatic energy should include the energy stored on the gate capacitor, and the work performed by the voltage on the gate should be taken into account in the free energy:
:<math>\Delta F_{\rm S}^{\pm} = \frac{e}{C_{\rm S} + C_{\rm D}} \left( \frac{e}{2} \pm V_{\rm bias}(C_{\rm D} + C_{\rm G}) - V_{\rm G} C_{\rm G} + ne + q_0 \right),</math>
:<math>\Delta F_{\rm D}^{\pm} = \frac{e}{C_{\rm S} + C_{\rm D}} \left( \frac{e}{2} \pm V_{\rm bias} C_{\rm S} + V_{\rm G} C_{\rm G} - ne + q_0 \right).</math>
At zero temperatures, only transitions with negative free energy are allowed: <math>\Delta F_{\rm S} < 0</math> or <math>\Delta F_{\rm D} < 0</math>. These conditions can be used to find areas of stability in the plane <math>V_{\rm bias} - V_{\rm G}.</math>
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