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Solar Panel - Voltage and Current from Spectral Radiance

  1. Dec 2, 2016 #1
    1. The problem statement, all variables and given/known data
    i)A beam of light at wavelength 600nm, with spectral width of 4nm and spectral radiance of 1300 Wm-2μm-1, illuminates an area of 10cm2 . What is the voltage this beam could generate if converted to energy?
    ii) What current could this beam produce?
    iii) How would the current change if the width of the beam were 5 meV?

    2. Relevant equations
    i) Voltage = hf / charge = change in energy / charge = h*c / q*λ

    ii)
    Spectral Radiance= eλ = hf φλ
    Where, Spectral Flux Density = φλ
    f=c/λ
    Current Density = q * φλ * δλ = I / A

    iii) Absolutely no idea what to use.
    Maybe, E (in eV)=1.24 / λ (in μm)

    3. The attempt at a solution

    i)
    I need max voltage available, so I ignore spectral width.
    Voltage = h*c/ (600*10-9 * q)= 2.060 Volts

    ii)
    I get flux density first
    Spectral radiance= eλ
    Spectral flux density = φλ
    Rearranging equation -- -- > Spectral Radiance= eλ = hf φλ
    φλ = eλ * λ / h*c = 1300 * (600*10-9) / h*c = 4.58*1021
    Rearranging equation, Current Density = q * φλ * δλ = I / A
    I = q * φλ * δλ * A = 3.68 mA

    iii) Not a the faintest of idea of what to do here...
     
    Last edited: Dec 2, 2016
  2. jcsd
  3. Dec 2, 2016 #2

    haruspex

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    I'm puzzled by this statement. Spectral radiance is for the case where light is coming from a surface at some distance. To get the total power we have to consider the area of the emitting surface as well as the solid angle the receiving surface subtends at each point of the emitting surface. I'm guessing this should say spectral irradiance. See e.g. https://en.wikipedia.org/wiki/Radiance.
    I assume you meant 1300 Wm-2μm-1
    No point in working those two out separately, just work with the average here.
    That's in Joules, right?
    That's not what I get, but I could be wrong. Please post your working.
     
  4. Dec 2, 2016 #3
    Thanks, I re edited my original post and I definatly have part 1 and part 2 correct as I checked with my study group who all confirmed they have that answer.

    I just dont know where to get part iii
     
  5. Dec 2, 2016 #4

    haruspex

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    Given the units, I have to guess that beam width refers to the range of wavelengths. In your original post you calculated the Joules per photon at each end of the 4nm range. The difference would be the beam width in this sense. What was that beam width in units of meV?
     
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