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Given a slab of Silicon and a photon flux $\Phi[\frac{\gamma}{cm^2sec}]$ incident on the slab of thickness $h$, the incident power $I$ given by $\Phi$ will be absorbed following the Beer - Lambert law: $$I(x)=I_0\exp(-kx)$$ where $x$ is the vertical axis from the surface of the semiconductor. What is the probability for a photon $\gamma$ of wavelength $\lambda$ to create an electron - hole pair inside the slab? Thanks.

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We will have to assume several things:

1) That your external Quantum Efficiency is unity (assuming you have a good AR coating).

- or conversely we are dealing only with the photons that make it into the substrate.

2) That your Si deep enough that all the photons are absorbed and that yo are concerned only with the photons absorbed within a certain thickness h.

- other wise you'll have to be concerned about reflections at the other interface too.

3) and we'll assume there is insignificant recombination

Since Si is an indirect bandgap material it's absorption coefficient does depend upon wavelength quite strongly.

In semi-conductor physics the coefficient is called alpha, here is a plot.enter image description here

From the chart you pick your wavelength, which gives you your alpha. The absorption from the surface to a depth of h is simply: $$ P=P_0(1-e^{-{\alpha(\lambda)}h}) $$. This is the cumulative absorption in that depth of material (i.e. the integral).

Using the above data and assigning a 450 nm to "blue", 556 nm to "green", 630 nm to "red" and 850 nm to "NIR" we derive the following plot of internal QE. The total absorption is the area under the curve of each color. You will notice that blue gets absorbed closer to the surface. with NIR (Near IR) penetrates much deeper.

enter image description here

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  • $\begingroup$ You are assuming the quantum efficiency is equal to one. This means the probability for a photon to create an electron - hole pair is one inside the bulk of the slab. This is a strong assumption and I don't see how to agree. $\endgroup$ – Riccardo.Alestra Oct 11 '13 at 16:45
  • $\begingroup$ This is wrong. Not every photon can create an electron-hole pair. The transition probability depends on the density of states, photon energy and matrix element for the transition. $\endgroup$ – mcodesmart Oct 11 '13 at 17:19
  • $\begingroup$ Oh boy ... I am assuming the external QE ** is one. i.e. every photon **once ** it is the Si is subject to the probability of absorption as defined above. Since no external conditions are noted one simply needs to apply this to the photons that **do make it into the Si. $\endgroup$ – placeholder Oct 11 '13 at 18:10
  • $\begingroup$ This is not wrong. Let me explain. The energy of the photon is the selected by the wavelength. The photon moving through the Si leaves a a probability trail of e/h pairs with a characteristic depth that \$ e^{-1} \$ absorption. That formulate above is a probability. To check this out look at "Donati, Silvano". "Photodetectors: Devices, Circuits and Applications". Prentice Hall PTR, 1999. The graph above is from internal data that I've collected and verified. Do a little research. $\endgroup$ – placeholder Oct 11 '13 at 18:17
  • $\begingroup$ Why are people down voting this good answer? @Riccardo.Alestra you define quantum efficiency (QE) incorrectly. QE is the fraction of electron hole pairs that decay radiatively. It has nothing to do with absorption. As pointed out above the transition probably depends on the density of states, however, it's convenient to use the macroscopic property of absorption coefficient as it encapsulates all of that information in a simple coefficient. A big +1 for your answer from me. $\endgroup$ – boyfarrell Oct 12 '13 at 1:06

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