- Schottky Barrier and the Tunneling Current Calculation

Dongwook Go

SEM (Scanning Electron Microscope)

Source : http://lamp.tu-graz.ac.at/~hadley/sem/index/index.php

4-point Measurement EBIC (Electron Bean Induced Current)

Source : http://lamp.tu-graz.ac.at/~hadley/sem/index/index.php

Source :

Alexander Schnabel, Stephan Stonica,

Experimental laboratory exercise report, TU Graz

Theory and Modeling

Numerical Calculation Results and Discussion

- Square Barrier

- Spherical Schottky Barrier

Conclusion

Source : http://web.tiscali.it/decartes/phd html/node3.html

Schottky Contact

Poisson Equation

• Potential Energy Barrier

Schrodinger Equation

• Transmission Coefficient

Fermi-Dirac Function

• Current Density

Depletion approximation𝜌 = 𝑒𝑁𝑑

Multiply 𝑟2 and integrate

Divide by 𝑟2 and integrate

At the end of the depletion layer (𝑟𝑑), E=0

The ground of the potential is set to be zero at the end of the depletion layer, 𝑟𝑑

𝑟𝑑 can be calculated from the condition

𝑉𝑠 : intrinsic potential drop at the contact 𝑉𝑏𝑖 : bias voltage

Calculation of 𝑟𝑑 for various bias voltages. 𝑉𝑠 is set to be -1 V.

Potential barrier shapes for various bias voltages. 𝑉𝑠 is set to be -1 V.

To be solved :

,where

The solution for the Schrodinger equation at the constant potential = plane wave

Plug in the initial condition at x=a (𝐴𝑇 is set to be 1)

and integrate the Schrodinger equation numerically.

So we get , and

From the form of the solution,

Get 𝐴𝐼, and 𝐴𝑇

And the transmission probability is

Normalized solution on the left/right side

Periodic boundary condition

with

The current density for this quantum state is

The total current density that flows from left to right is

: the probability that the quantum state on the left side is occupied

: the probability that the quantum state on the right side is empty

If we change the expression into the integral form

Change the integration variable into the energy,

Similarly, the current density for the left moving electron is

The total current density is the difference of the two

𝑈0= 2 eV

𝜉𝑟= 2 eV

𝜉𝑙= 2 eV

𝑇=300 K

𝑈0= 2 eV

𝜉𝑟= 2 eV

𝜉𝑙= 2 eV

𝑎= 1 nm

𝜉𝑟= 2 eV

𝜉𝑙= 2 eV

𝑎= 1 nm

𝑇= 300K

𝑈0= 2 eV

𝜉𝑙= 2 eV

𝑎= 1 nm

𝑇=300 K

𝑈0= 6 eV

𝜉𝑙= 2 eV

𝜉𝑟= 2 eV

𝑇=300 K

𝑈0= 6 eV

𝜉𝑙= 2 eV

𝜉𝑟= 2 eV

𝑎=1 nm

Source : http://web.tiscali.it/decartes/phd html/node3.html

𝜉𝑙= 2 eV

𝜉𝑟= 2 eV

𝑎=1 nm

𝑇=300 K

𝜉𝑙= 2 eV

𝜉𝑟= 2 eV

𝑎=1 nm

𝑇=300 K

𝑈0= 6 eV

𝜉𝑙= 2 eV

𝑎= 1 nm

𝑇=300 K

*Thermoionic Emission Dominates

Conventional Schottky diode behavior

Small 𝑈0Large 𝜉Large 𝑎High 𝑇

*Tunneling Dominates

Reverse rectifying behavior

Large 𝑈0Small 𝜉Small 𝑎Low 𝑇