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Homework Statement:
 Silicon Heterojunctions
Relevant Equations:
 Beer's law: I = I0 * exp (alpha*x)
The doped aSi: 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 aSi: 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 cm1 (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?
For a doped aSi: 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 cm1 (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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