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

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The discussion focuses on calculating the optical output power of an LED for a science project, specifically using the formula Pout = N × Vres, where N is a linear factor that needs clarification. Participants suggest starting with a detailed LED datasheet to find necessary values and emphasize that N is not critical for comparing relative efficiencies between different light sources, as it cancels out in ratio calculations. The importance of measuring the distance from the photocell to the light source is questioned, with guidance provided on how to approach the experiment without needing to determine N. Overall, the conversation highlights the relationship between input and output power in assessing LED efficiency, while suggesting practical steps for experimentation.
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?
 
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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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I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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