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 dont know what help is. Id appreciate help a lot, thanks.
All the values I have are experimental. Im trying to calculate the relative wall plug efficiency of an LED. heres my data right now.
I do not have a data sheet and the steps Im 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 thats 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?
This project comes from here btw:
http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p003.shtml#procedure

Edited December 27, 201510 yr by daPoseidonGuy

OK so do you understand what the first equation you posted means and where it come from?

 

 

Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)

 

I ask because in your project instructions they tell you where it comes from immediately before they quote it.

 

Why are do they have you measuring the voltage across a resistor?

What does that tell you?

Edited December 27, 201510 yr by studiot