What is the formula for calculating the optical output power of an LED?

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Discussion Overview

The discussion revolves around calculating the optical output power of LEDs in relation to input electrical power, specifically for determining relative wall plug efficiency. Participants explore formulas, experimental setups, and the significance of certain variables in their calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents a formula for calculating optical output power: Pout = N × Vres, questioning the meaning of N and how to calculate optical output power.
  • Another participant suggests starting with a detailed LED datasheet to find curves for optical output versus current.
  • A participant shares experimental data and expresses confusion about the relevance of measuring the distance between the photocell and the light source in their experiment.
  • One response indicates that N cannot be easily determined as it varies with light emission angles and suggests that it may not be necessary for calculating relative efficiency.
  • Another participant explains that when comparing different light sources, the constant N cancels out, making it unnecessary for relative efficiency calculations.
  • There is mention of the need for calibration to measure absolute power of a single emitter, which requires a known source.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and determination of the variable N, with some suggesting it is not crucial for relative efficiency calculations while others seek clarity on its role.

Contextual Notes

Participants rely on experimental data and personal interpretations of procedures without consensus on the best approach to measure or calculate N. There are unresolved questions regarding the significance of certain measurements in the context of the experiment.

Who May Find This Useful

This discussion may be useful for students or researchers working on LED efficiency projects, particularly those interested in experimental methods and the complexities of measuring optical output power.

daPoseidonGuy
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Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency.
This is the only good formula i found
Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
What is N exactly? I do not know how to calculate optical output power as I don't know what help is. Id appreciate help a lot, thanks.
 
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daPoseidonGuy said:
Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency.
This is the only good formula i found
Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
What is N exactly? I do not know how to calculate optical output power as I don't know what help is. Id appreciate help a lot, thanks.
Welcome to the PF.

I'd start with a good datasheet for a typical LED. There should be curves for optical output versus current. The old Hewlett Packard LED datasheets were especially complete. Can you post a link to a detailed LED datasheet that has some of the numbers you are looking for? :smile:
 
Heres a more complete post:
Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency.
This is the only good formula i found
Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
What is N exactly? I do not know how to calculate optical output power as I don't know what help is. Id appreciate help a lot, thanks.
All the values I have are experimental. I am trying to calculate the relative wall plug efficiency of an LED. here's my data right now.
I do not have a data sheet and the steps I am following at this point are somewhat modeled of those shown here:
http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p003.shtml#procedure
in the testing and data collection section.

Voltage across resistor (V) ± .01 = 2.49

Distance from photocell to light (cm) ± .05 = 4.00

Voltage across light (V) ± .01 = 5.75

Current intensity (mA) ± .01 = 360
If N is just something I have to leave as a variable, then what is the point of measuring the distance between the photocell and the light bulb? In the experiment I moved the breadboard closer or further to get 2.5v across the resistor, as that's what I understood from the procedure. Was I supposed to do that? What do I do with the value for the distance between the photocell and light bulb?
 
Last edited:
that experiment doesn't give a value for N, only compares N for LED vs incandescent lamp.
Try this datasheet..http://www.vishay.com/docs/81011/tsal6400.pdf

and see if this helps you estimate N for that particular LED

upload_2015-12-27_14-55-57.png


good opportunity to learn about steradians .
 
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From the project link:
N cannot be determined easily, as it depends on the light emission vs. angle for each source.

The project is
1. Build a light detector. Don't worry about the detector's N since we assume it is constant across all emitters.
2. Try different sources and calculate input power vs output power (efficiency) using the detector.

You don't really need to know N to determine relative efficiency of two different sources since they assume the detector's N is the same for all sources. They also assume the detector is 100% efficient, that is, it is detecting all the light from each source (or a similar fraction).
"Since you are not collecting all of the light at the light-to-voltage converter (some of the light goes off to the side), the calculation is relative."

When you take the ratio of the two sources, the N will cancel out.

In other words, you are sort of measuring light output in units of N. N goes away when you take ratios of two measurements.

If you want to measure the absolute power of a single emitter, you need to somehow calibrate the detector, which requires a known source, etc.
 
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