Modeling of the contact between the metal tip and n-type semiconductor as a schottky barrier and...
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![Page 1: Modeling of the contact between the metal tip and n-type semiconductor as a schottky barrier and tunneling current calculation](https://reader033.fdocuments.net/reader033/viewer/2022052912/55a155991a28abc42c8b484d/html5/thumbnails/1.jpg)
- Schottky Barrier and the Tunneling Current Calculation
Dongwook Go
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SEM (Scanning Electron Microscope)
Source : http://lamp.tu-graz.ac.at/~hadley/sem/index/index.php
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4-point Measurement EBIC (Electron Bean Induced Current)
Source : http://lamp.tu-graz.ac.at/~hadley/sem/index/index.php
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Source :
Alexander Schnabel, Stephan Stonica,
Experimental laboratory exercise report, TU Graz
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Theory and Modeling
Numerical Calculation Results and Discussion
- Square Barrier
- Spherical Schottky Barrier
Conclusion
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Source : http://web.tiscali.it/decartes/phd html/node3.html
Schottky Contact
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Poisson Equation
• Potential Energy Barrier
Schrodinger Equation
• Transmission Coefficient
Fermi-Dirac Function
• Current Density
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Depletion approximation𝜌 = 𝑒𝑁𝑑
Multiply 𝑟2 and integrate
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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
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𝑟𝑑 can be calculated from the condition
𝑉𝑠 : intrinsic potential drop at the contact 𝑉𝑏𝑖 : bias voltage
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Calculation of 𝑟𝑑 for various bias voltages. 𝑉𝑠 is set to be -1 V.
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Potential barrier shapes for various bias voltages. 𝑉𝑠 is set to be -1 V.
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To be solved :
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,where
The solution for the Schrodinger equation at the constant potential = plane wave
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Plug in the initial condition at x=a (𝐴𝑇 is set to be 1)
and integrate the Schrodinger equation numerically.
So we get , and
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From the form of the solution,
Get 𝐴𝐼, and 𝐴𝑇
And the transmission probability is
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Normalized solution on the left/right side
Periodic boundary condition
with
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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
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If we change the expression into the integral form
Change the integration variable into the energy,
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Similarly, the current density for the left moving electron is
The total current density is the difference of the two
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𝑈0= 2 eV
𝜉𝑟= 2 eV
𝜉𝑙= 2 eV
𝑇=300 K
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𝑈0= 2 eV
𝜉𝑟= 2 eV
𝜉𝑙= 2 eV
𝑎= 1 nm
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𝜉𝑟= 2 eV
𝜉𝑙= 2 eV
𝑎= 1 nm
𝑇= 300K
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𝑈0= 2 eV
𝜉𝑙= 2 eV
𝑎= 1 nm
𝑇=300 K
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𝑈0= 6 eV
𝜉𝑙= 2 eV
𝜉𝑟= 2 eV
𝑇=300 K
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𝑈0= 6 eV
𝜉𝑙= 2 eV
𝜉𝑟= 2 eV
𝑎=1 nm
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Source : http://web.tiscali.it/decartes/phd html/node3.html
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𝜉𝑙= 2 eV
𝜉𝑟= 2 eV
𝑎=1 nm
𝑇=300 K
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𝜉𝑙= 2 eV
𝜉𝑟= 2 eV
𝑎=1 nm
𝑇=300 K
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𝑈0= 6 eV
𝜉𝑙= 2 eV
𝑎= 1 nm
𝑇=300 K
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*Thermoionic Emission Dominates
Conventional Schottky diode behavior
Small 𝑈0Large 𝜉Large 𝑎High 𝑇
*Tunneling Dominates
Reverse rectifying behavior
Large 𝑈0Small 𝜉Small 𝑎Low 𝑇
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