Advanced Solid State Physics

suppressed or at its maximum. Figure 89 shows the set up and its circuit.

island

gate 2

gate 1

source drain

V 1 Vg1 Vg2 V 2

Cg1 C 0 Cg2

Q 0

C 1 , R 1 C 2 , R 2

Figure 89: Single electron transistor with two gates and its circuit.

The total chargeQon the island is

Q=C 1 (V−V 1 ) +C 2 (V−V 3 ) +Cg 1 (V −Vg 1 ) +Cg 2 (V−Vg 2 ) +C 0 V (192)

butQneeds to be an integer numberntimes electron charge plus a background chargeQ 0 that is
caused by electric field lines that can come from any other electronic device.

So the voltages that the island is allowed to have are

V(n) = (−ne+Q 0 +C 1 V 1 +C 2 V 2 +Cg 1 Vg 1 +Cg 2 Vg 2 )/CΣ (193)

withCΣas the total capacitance of the island:

CΣ=C 1 +C 2 +Cg 1 +Cg 2 +C 0. (194)

The energy needed to add an infinitesimal chargedqto an island at voltageV(n)isV(n)dq. The
energy needed to add a whole electron at a given number of electrons on the island is therefore:

∆E=

∫−e

0

V(n)dq=−eV(n) +

e^2

2 CΣ

(195)

The extra term e

2
2 CΣis the charging energy that comes from the fact that every time you add a fraction
of an electron to the island the voltage changes.

The energy needed to remove a whole electron is analog as above:

∆E=

∫e

0

V(n)dq=eV(n) +

e^2

2 CΣ

(196)

The charging energy is still the same because the sign of the charge vanishes when it’s being squared.

You can take these formulas to calculate the energy that is needed for the four possible tunnel events:
an electron can tunnel from the first electrode on the island or from the island on the first electrode