Calculating optical power output

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Discussion Overview

The discussion revolves around calculating the power conversion efficiency of a blue semiconductor LED connected in series with a resistor. Participants explore the relationship between electrical input power and optical output power, focusing on the equations and concepts necessary for the calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants discuss the formula for conversion efficiency, which is defined as the ratio of optical output power to electrical input power.
  • One participant calculates the electrical input power based on the current and voltage across both the LED and the resistor.
  • There is uncertainty regarding how to calculate the optical output power, with a request for the appropriate formula.
  • One participant suggests using the equation E=hf to find the energy of a photon, linking it to the calculation of optical power.
  • Another participant proposes using E = hc/e.wavelength to derive the energy per photon, but expresses confusion about the term "e.wavelength."
  • Clarification is provided regarding the value of e (the charge of an electron) in the context of the calculations.

Areas of Agreement / Disagreement

Participants generally agree on the need to calculate the optical output power but have not reached consensus on the exact formula or approach to use. There is ongoing uncertainty about the definitions and calculations involved.

Contextual Notes

Participants express confusion about specific terms and equations, indicating a need for clarity on the relationships between photon energy, wavelength, and optical power. The discussion includes various interpretations of how to approach the calculations.

snoothie
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Homework Statement


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.


Homework Equations


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

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.
 
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snoothie said:

Homework Statement


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.


Homework Equations


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

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.

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.
 
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?
 
snoothie said:
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?

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

I was referring 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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