Trouble replicating a calculation in Silicon Nanoelectronics

In summary: The problem must be with the value of the effective mass. It looks like the author used 0.2 times the mass of an electron as his value for effective mass. If you use this value, then the transmission probabilities are 10 nm and 5 nm, respectively.In summary, the transmission probability for a rectangular-shaped potential barrier with width d and height ϕ is 1 × 10^-3 and 1 × 10^-6 at a barrier height of 100 mV.
  • #1
Achmed
10
0
Trouble replicating a calculation in "Silicon Nanoelectronics"

I'm reading the textbook "Silicon Nanoelectronics" and I've encountered an equation for the transmission probability, which you can see among the attachments.

In this equation, T is the transmission probability for a rectangular-shaped potential barrier with width d and height ϕ, where m* is the effective mass of silicon and q the elementary charge.

They go on to say that "From Equation (3.1), the barrier width
giving transmission probabilities of 1 × 10^-3 and 1 × 10^-6 at a barrier height of 100
mV can be estimated to be 10 and 5 nm, respectively."

I really, really want to replicate this estimation/calculation but I can't seem to do in. When I plug in the given numbers (ϕ = 100 mV, d = 5 or 10 nm, q = elementary charge, ħ = reduced Planck constant, and m* = effective mass silicon), I can't seem to get even remotely close to the listed probabilities. Perhaps the problem is the effective mass of silicon? I am not certain what value I should take, but I went with 0,2 times the mass of an electron (see: http://ecee.colorado.edu/~bart/book/effmass.htm)

Can anyone please show me how the writer approximately got 10nm and 5nm using the equation and the given probabilities? I know it's not a precise calculation, nor a precise equation, but I'd still like to see how he got this estimation.
 

Attachments

  • Formula Transmission Barrier.jpg
    Formula Transmission Barrier.jpg
    8.3 KB · Views: 391
Physics news on Phys.org
  • #2
I think you have the transmission probabilities swapped. The wider barrier will give a lower probability. Using your numbers I got 9.6 nm and 4.8nm for transmission probabilities of 1E-6 and 1E-3, respectively. Why don't you post your calculations and we'll see if we can find what is wrong.
 
  • #3
phyzguy said:
I think you have the transmission probabilities swapped. The wider barrier will give a lower probability. Using your numbers I got 9.6 nm and 4.8nm for transmission probabilities of 1E-6 and 1E-3, respectively. Why don't you post your calculations and we'll see if we can find what is wrong.
That's what I tought, too: It must be a sloppy mistake by the author (it is written this way by the author).

I don't know what I'm doing wrong, but I can't really show you anything insightful. I just plugged in the numbers into the formula, and my calculator keeps giving me 0 as the answer.
 
  • #4
I can't give you any more help unless you show an attempt. Show what numbers you are plugging in and some intermediate steps in the calculation.
 
  • Like
Likes 1 person
  • #5
phyzguy said:
I can't give you any more help unless you show an attempt. Show what numbers you are plugging in and some intermediate steps in the calculation.
q = 1.60217657 × 10^-19
ϕ = 100 x 10^6

(qϕ)^1/2 ≈ 4 x 10^-6

2m* = 0.4 * 9.10938291×10^ −31 ≈ 3.64 x 10^-31
ħ^2 = (1.05457173 × 10^-34)^2 ≈ 1,1 x 10^-68

(2m*/ħ^2 )^1/2 ≈ 5.75 x 10^18

-2 * 5.75 x 10^18 * 4 x 10^-6 ≈ -4.6 x 10^(13)

10^-3 = exp[-4.6 x 10^(13)d]

ln(10^-3) = -4.6 x 10^(13)d

d ≈ 1.5 x 10^-13 , which is obviously extremely incorrect.
 
Last edited:
  • #6
Achmed said:
ϕ = 100 x 10^6

That is 100 megavolts. That would be a huge barrier. You were talking about 100 millivolts before.
 
  • Like
Likes 1 person
  • #7
Cthugha said:
That is 100 megavolts. That would be a huge barrier. You were talking about 100 millivolts before.

Wow... that solves it.
 

1. Why is it important to replicate calculations in Silicon Nanoelectronics?

Replicating calculations in Silicon Nanoelectronics is important because it allows for validation and verification of the results. This helps to ensure accuracy and reliability of the data, which is crucial in scientific research.

2. What are some common challenges encountered when replicating calculations in Silicon Nanoelectronics?

Some common challenges include differences in software or hardware used, variations in experimental conditions, and errors or discrepancies in the original calculation method. These factors can all affect the reproducibility of the results.

3. How can these challenges be addressed when replicating calculations in Silicon Nanoelectronics?

To address these challenges, it is important to carefully document all experimental procedures and conditions, use standardized software and hardware, and perform multiple replications to ensure consistency and identify any potential errors.

4. What are the implications of not being able to replicate a calculation in Silicon Nanoelectronics?

If a calculation cannot be replicated, it could indicate that there are errors or limitations in the original research, which could lead to incorrect conclusions being drawn. This could also impact the validity and reproducibility of future studies that build upon the original research.

5. How can the scientific community promote and ensure reproducibility in Silicon Nanoelectronics research?

To promote and ensure reproducibility, it is important for researchers to share their data, methods, and software openly and transparently. This allows for others to replicate the calculations and validate the results. Collaborative efforts, open access publications, and reproducibility initiatives can also help to promote reproducibility in the scientific community.

Similar threads

Replies
2
Views
933
  • Introductory Physics Homework Help
Replies
4
Views
4K
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
1K
Replies
7
Views
686
  • Sci-Fi Writing and World Building
Replies
0
Views
709
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
619
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
2K
Back
Top