I need to evaluate the concentration of photo carriers in illuminated thin film semiconductor, e.g. CdTe.
Suppose I have given:
I - irradiance of incident light (set as monochromatic for simplicity) in W/cm2
Q - quantum yield in # of generated electrons/# of incident photons
E - energy of the incident photon in J
R - reflectance
t - relaxation time of generated photo carriers in s
Can I estimate the order of photocarrier concentration with equation of such form:
n = (1-R)*(I/E)*Q*t
In the case of bulk materials (and light of some wavelength range) resulting in an integral over depth, absorption coefficient, wavelength, etc...
I'm not much into the subject and want to know how much is missing in the upper relation.
For the stationary case your score is, in my opinion, correct, except that you forgot the absorption coefficient. In the general case
∆n(t)= βαIτ/ℏω (1-e^(-t/τ) )
where I - the incident light intensity (considering reflection);
α - the absorption coefficient;
β - the quantum yield;
τ - the lifetime of free carriers;
ωħ - the photon energy of the incident radiation
Basically your equation is correct. I would make the following adjustments:
(1-R) should be replaced by absorption A of the CdTe in %. This equals (1-R) only if there is no transmission and no absorption in any other layers (e.g. substrate).
Also, dividing incident power density by energy gives you a current density (Acm-2). If you want photocarrier concentration you also need to divide by the elementary charge q~1.6E-19
ADDED 29.10.14: the above comment - dividing by q - is only necessary if the photon energy used in the equation was in eV, and its purpose is to convert eV to J. If it was in J initially division by q is unneccesary.
Thank You both for a fast response.
Taking both suggestions I take, the modified and more than less correct equation will be:
∆n(t)= βαIτ/qℏω (1-e^(-t/τ) )
so the version given by Leonid but including the division by elementary charge, to give carrier concentration.
That is if and only if the layer is included, neglecting the interaction of light with the substrate (e.g. no back reflection).
Thus, do You agree that the upper is more accurate?
Dear colleagues!
I do not agree with the above formula, if only for reasons of dimension. Charge in the final expression should not be! The charge appears in the expression for the current. According to Ohm's law, the photocurrent density is j = eμn (x) E, where E - the electric field strength and μ - mobility of the electron.
Therefore quote calculation estimating the concentration of photoexcited carriers as a distance x from the surface, and its average value over the sample. We will calculate concentration of the photoexcited carriers in the semiconductor in the form of a plate (lxbxa) at radiation it monochromatic light with a power of P.
(1-R)P/ℏω – the number of the photons entering in 1 second a plate surface (l, b);
(1-R)P/ℏω exp(-αx) – the number of photons that will remain at a depth x;
[(1-R)/ℏω] d[Pexp(-αx)]= [(1-R)P/ℏω] α exp(-αx)dx – the number of photons that are absorbed in a layer of thickness dx;
[(1-R)P/ℏω] α/lb exp(-αx) – the density of photoelectrons generated in 1 second at a depth x (divided expression 3 into layer volume of lbdx);
In the steady state rate of recombination and generation must be equal, hence:
n(x)/τ=[(1-R)P/ℏω] (1/lb) αexp(-αx),
as P = IS = Ilb, then n(x)=[(1-R)I/ℏω] ταexp(-αx) – carrier concentration at a distance x from the input surface in the volume dV = lbdx.
We find the total number of photoexcited carriers in the bulk of the plate:
N=∫n(x)dV= [(1-R)Ilb/ℏω] τη = [(1-R)P/ℏω] τη ,
where η=α∫exp(-αx)dx - value which takes into account the incomplete absorption; а – sample thickness. Easy to see that if the plate was infinitely thick, the factor η = 1. So there is every reason to call the value of the quantum efficiency η.
The average concentration of the photoexcited carriers then will be equal:
n ̅=1/V ∫n(x)dx=[1/abl] [(1-R) P/ℏω] τη=[(1-R)I/ℏω] (1/a) τη.
As can be seen, the dimension of the resulting expression is: (W c)/(W c M^2 M)=1/M^3
Dear Krzysztof,
Perhaps you confuse the factor 1 / a in the final expression for the concentration of photoexcited carriers in semiconductors.
I would like to clarify that if ħω ≥ Eg, then 1 / a = α and we obtain for the stationary conditions the same formula to estimate the concentration that was in my first answer, that is,
n ̅= (1-R) ηαIτ/ℏω.
In the case where Eg
Dear Leonid,
I very much love the derivation as it shows everything explained step by step, though I can't accept some pieces.
1. I'm not sure if I can express the quantum efficieny only in the terms of absorption coefficient. That does not take into account excitation to states not as delocalized as s states which will also affect the value of absorption coefficient. Why η doesn't serve as a transcription factor with dimension [electron/photon], when it would be necessary to eliminate the elementary charge?
2. The decay factor was ruled out, because of the steady state condition, right?
Could You clarify the upper cases?
Concerning abs. coefficient, I've multiplied by 'α' to sort out the absorbed part of light from the material of thickness 'a'.
Dear Krzysztof,
I will try to answer your questions.
1. Under the quantum efficiency in this consideration is implied fraction of photons created photocarriers, i.e. it is a dimensionless quantity.
It is assumed that the absorption probability is much higher than the probability of all other possible processes, particularly given you an example. It is especially shown, if ħω ≥ Eg.
You can enter another definition of quantum efficiency [electron / photon]. In this case it would be necessary to include the elementary charge.
2. With regard to the second point, you're right. This is evident from my first answer.
Overall, I totally agree with your view. My answer pursued its goal to show that the same result can be obtained by the simple reasoning.
Thanks for discussion.
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