Light absorption in a semiconductor

In summary, 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.
  • #1
semiconductooor1
4
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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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  • #6
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.
 
  • #7
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?
 
  • #8
Welcome to PF.

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

1. How does light absorption occur in a semiconductor?

Light absorption in a semiconductor occurs when photons from incident light are absorbed by electrons in the material, causing them to move to higher energy levels. This process creates electron-hole pairs, which are essential for the generation of electrical current in devices like solar cells.

2. What factors affect the efficiency of light absorption in a semiconductor?

The efficiency of light absorption in a semiconductor is influenced by factors such as the bandgap energy of the material, the thickness of the semiconductor layer, and the wavelength of the incident light. Materials with smaller bandgaps are more efficient at absorbing light, while thicker layers can absorb more photons. Additionally, matching the wavelength of the incident light to the bandgap of the semiconductor can maximize absorption efficiency.

3. How does the bandgap energy of a semiconductor affect light absorption?

The bandgap energy of a semiconductor determines the energy of photons that can be absorbed by the material. Semiconductors with smaller bandgaps can absorb photons with lower energies, allowing them to absorb a wider range of wavelengths and therefore more light. This property is crucial for optimizing the efficiency of devices like photovoltaic cells.

4. Can light absorption in a semiconductor be controlled?

Yes, light absorption in a semiconductor can be controlled through various methods such as doping the material with impurities, changing the material's thickness, or manipulating the incident light's wavelength. By carefully engineering the semiconductor's properties, researchers can tailor its light absorption characteristics to suit specific applications, such as in photodetectors or solar cells.

5. What are some practical applications of understanding light absorption in semiconductors?

Understanding light absorption in semiconductors is crucial for the development of various technologies, including solar cells, photodetectors, and light-emitting diodes (LEDs). By optimizing the absorption of light in semiconductors, researchers can enhance the efficiency and performance of these devices, leading to advancements in renewable energy, communications, and lighting technologies.

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