Which Choice Effectively Increases Tunneling Current?

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Homework Help Overview

The discussion revolves around the factors affecting tunneling current in electronic devices, specifically through a potential-energy barrier. The original poster seeks to understand how to effectively increase the tunneling current by manipulating the width and height of the barrier.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of reducing the width and height of the barrier, referencing equations related to the tunneling coefficient. Questions arise about the comparative effects of these changes on the tunneling current.

Discussion Status

Some participants have provided equations and suggested analyzing the effects of halving the height and width of the barrier. There is ongoing inquiry into how to calculate and compare the resulting tunneling coefficients, with no clear consensus reached yet.

Contextual Notes

Participants express confusion regarding the calculations and the implications of their findings, indicating a need for further clarification on the mathematical relationships involved.

tkm2002
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Homework Statement


Suppose a tunneling current in an electronic device goes through a potential-energy barrier. The tunneling current is small because the width of the barrier is large and the barrier is high. To increase the current most effectively, what should you do?
a. Reduce the width of the barrier.
b. Reduce the height of the barrier.
c. Either choice (1) or choice (2) is equally effective.
d. Neither choice (1) nor choice (2) increases the current

Homework Equations



The Attempt at a Solution


should I use this equation?
T ≈ e^(-2CL) , L is he width of the barrier.
c^2 = 2m(U-E)/(h/2pi) , U is the height of the barrier ; E is the energy of electron ;m is the mass of electro; h is plank constant

I am not understand why the answer is a.
 
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tkm2002 said:
should I use this equation?
T ≈ e^(-2CL) , L is he width of the barrier.
c^2 = 2m(U-E)/(h/2pi) , U is the height of the barrier ; E is the energy of electron ;m is the mass of electro; h is plank constant

I am not understand why the answer is a.

Yes, use those equations. What happens if you cut 'U' in half? What happens if you cut 'L' in half?...So which has the greater effect?
 
gabbagabbahey said:
Yes, use those equations. What happens if you cut 'U' in half? What happens if you cut 'L' in half?...So which has the greater effect?

I do not know how to calculate and compare it
 
Call the initial U, U_0 and the initial L, L_0...what does that make the initial transmission coefficient? What doe the transmission coefficient become when you plug in U=U_0/2 and L=L_0? How about when you plug in U=U_0 and L=L_0/2
 
gabbagabbahey said:
Call the initial U, U_0 and the initial L, L_0...what does that make the initial transmission coefficient? What doe the transmission coefficient become when you plug in U=U_0/2 and L=L_0? How about when you plug in U=U_0 and L=L_0/2

But I do not how to compare them
http://img98.imageshack.us/img98/6779/89062472.jpg

T_0 = e^2CL
T of L_0/2 = T_0^1/2
I do not how to write T of U_0/2 in terms of T_0
 
Last edited by a moderator:
Well, the largest effect that the first one will have is to reduce the exponent by a factor of \sqrt{2} (when E=0)...While the second one reduces the exponent by a factor of 2...

P.S. You are missing the negative signs in your exponents!:wink:
 
gabbagabbahey said:
Well, the largest effect that the first one will have is to reduce the exponent by a factor of \sqrt{2} (when E=0)...While the second one reduces the exponent by a factor of 2...

P.S. You are missing the negative signs in your exponents!:wink:

I am not very understand.
Can you express the equation?
 

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