Engineering Light absorption in a semiconductor

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SUMMARY

The discussion centers on the light absorption characteristics of doped amorphous silicon (a-Si:H) layers in HIT solar cells. Specifically, a 25nm thick doped a-Si:H layer absorbs 77% of light at 400nm, based on Beer’s Law, indicating that only 23% of the light passes through. The confusion arises from interpreting the percentage of light lost versus the percentage transmitted, with participants clarifying that a lower transmission percentage correlates with higher absorption loss.

PREREQUISITES
  • Understanding of Beer’s Law and its application in optical physics.
  • Familiarity with the properties of doped amorphous silicon (a-Si:H).
  • Knowledge of semiconductor physics, particularly in relation to solar cells.
  • Basic concepts of light absorption and transmission in materials.
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  • Research the optical properties of doped amorphous silicon (a-Si:H) in solar cell applications.
  • Learn about Beer’s Law and its implications in photonics and materials science.
  • Explore the impact of layer thickness on light absorption in semiconductor materials.
  • Investigate methods to optimize light absorption in solar cell designs.
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Researchers, materials scientists, and solar energy engineers focused on improving the efficiency of photovoltaic devices and understanding light-matter interactions in semiconductors.

semiconductooor1
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Homework Statement
Silicon Heterojunctions
Relevant Equations
Beer's law: I = I0 * exp (-alpha*x)
The doped a-Si: H layers in a HIT solar cell do not contribute to the photocurrent. The light they absorb (according to their absorption curve below) is lost.

For a doped a-Si: H layer at the front side of the cell that is 25nm thick, what percentage of light at 400nm will be lost due to absorption in this layer? For this question, we can ignore the effect of reflection at the front interface. Express your answer as a number between 0 and 100.

My guess:
using beer's law: exp (-alpha*x) = I/I0 = percentage of light lost due to absortion
alpha = 5.8*10^5 cm-1 (because 400 nm = 3.1 eV)
x = 25*10^-7 cm
light lost due to absorption = 23.46 %
But I'm told that this isn't the correct answer, I'm so confused can someone help me please?
 

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I think you did the arithmetic correctly, but you didn't think through what it means. If I/I0 = 0.23, what percentage of the light is lost?
 
phyzguy said:
I think you did the arithmetic correctly, but you didn't think through what it means. If I/I0 = 0.23, what percentage of the light is lost?
For me it means that I = 0.23×I0, that means 23% of the light got absorbed (lost) and 1-0.23 = 77 % didn't get absorbed.
 
semiconductooor1 said:
For me it means that I = 0.23×I0, that means 23% of the light got absorbed (lost) and 1-0.23 = 77 % didn't get absorbed.
Nvm after further thinking, it means that 23% got through the semiconductor without getting absorbed that means 77% got absorbed.
But does that mean that 77% of the light is lost? Thats seems to high.
 
It seems high to me too, but that's what the numbers say. I got the same result as you did, so either that's correct or we both made a mistake.
 
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phyzguy said:
It seems high to me too, but that's what the numbers say. I got the same result as you did, so either that's correct or we both made a mistake.
Thank you for your help, I appreciate it, if this comes wrong I will update you, but thanks again.
 
semiconductooor1 said:
Thank you for your help, I appreciate it, if this comes wrong I will update you, but thanks again.
SO, what's the final answer?
 
Welcome to PF.

ossiedissie said:
SO, what's the final answer?
What do *you* think the final answer should be? :smile:
 

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