Calculating optical power output

  1. 1. The problem statement, all variables and given/known data
    A blue semicon LED is connected in series with a 10ohm series resistor to a voltage source. The LED emits 4.03 x 10^16 photons/s at a wavelength of 400nm. The current flowing through the LED is 40mA and the voltage across the LED is 3.5V. Calculate the power conversion efficiency of the diode.


    2. Relevant equations
    Given,
    Conversion efficiency = (optical output power / electrical input power) x 100%
    electrical input power = IV (LED) + IV(resistor)

    3. The attempt at a solution

    Calculated IV for LED = 40x10^-3 x 3.5 = 0.14W
    IV for resistor = (40x10^-3)^2 x 10 = 0.016W
    so, electrical input power = 0.156W

    Having trouble with finding the optical output power. I guess that it got to do with the LED emission and wavelength?

    Does anyone know what is the formula for calculating the optical output power? I could not find the formula in any resources on hand. Thanks a million.
     
    Last edited: Nov 10, 2009
  2. jcsd
  3. berkeman

    Staff: Mentor

    Not sure they want you to inculde the power lost in the series limiting diode in the LED efficiency calculation, but that's up to you.

    On the optical power, I'd start with the equation E=hf, where f is the frequency of the photon (often written as the greek character "nu"). Are you familiar with this equation? From that, you calculate the total power as the flux of however many photons per second, each having that energy E=hf.
     
  4. Ok. thanks for the leads.

    we can use:
    E = hc/e.wavelength

    then, the flux is simply: Eph x photons/s, which also is the optical output power in watts?
     
  5. berkeman

    Staff: Mentor

    Don't know what e.wavelength is, but it looks like you are on the right track.
     
  6. oh sorry. My bad...

    I was refering it as:

    Eph = (hc/e) x wavelength

    e = 1.6 x 10^-19 eV

    Thank you very much for your help.
     
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