tunnel through the barrier because below all the states on the right side are filled and above there are
no electron states occupied on the left side. So we should integrate over pairs of states where the one
on the left is occupied and the one on the right is empty. The tunnelling rate finally looks like
ΓL→R=∫ 2 π
~
|〈l|HT|r〉|^2 DL(EL)f(E−EL)DR(ER)(1−f(E−ER))dE (185)whereDL,R are the density of states of the metal on the left and right side andf(E−EL)is the
Fermi function that describes the occupation of the states on the left and(1−f(E−ER))is the Fermi
function of the empty states on the right. The density of states for a metal is approximately constant,
especially for small voltage ranges this approximation is okay. So we can pull the density of states out
of the integral and get
ΓL→R=
2 πt^2 DL(EL)DR(ER)
~∫
f(E−EL)(1−f(E−ER))dE (186)withtas the transfer integral (overlap of the states on the left and right side). (We assume the same
for the transfer integral as for the density of states.) The integral over the Fermi function of the
occupied states on the left times the Fermi function of the empty states on the right can be calculated
with e.g. Matlab which leads us to
ΓL→R=
2 πt^2 DL(EL)DR(ER)
~ER−EL
exp(
−ERkB−TEL)
− 1. (187)
The difference in energy equals the electron charge times the applied voltage on the tunnel junction:
ER−EL=eV.The final solution for the tunnel rate looks like this
ΓL→R=V
eR1
exp(
−keVBT)
− 1. (188)
At low temperatures the exponential term can be neglected and the tunnel rate only depends onVR
but at high temperatures electrons are thermally activated across the barrier and the temperature
dependent term becomes important again. A tunnel rate from right to left can be calculated equiv-
alently. If you bias the tunnel junction so that the electrons will move from left to right, at finite
temperature there are still some electrons that can move in the opposite direction because the Fermi
function allows some electrons on the right to occupy states that are empty on the left. The resulting
tunnelling current is the difference of the tunnelling rates times the electron charge
IL→R=−|e|(ΓL→R−ΓR→L). (189)The temperature dependent parts cancel out and the current only depends onV andR:
IL→R=V
R
(190)
with R as
R=~
2 πe^2 Dl(El)Dr(Er)t^2(191)
So the tunnelling current is proportional to the applied voltage.
If you combine two tunnel junctions you get a single electron transistor.