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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 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:
https://www.sciencebuddies.org/science- ... #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:
https://www.sciencebuddies.org/science- ... #procedure
Optical output power of LED
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Optical output power of LED
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norman40
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Re: Optical output power of LED
Hello Anonymouse,
In this experiment you built a light detection circuit that outputs a voltage that is proportional to the light intensity or optical power. You know that the optical power from the LED is directly related to the voltage from the detector (Pout = N X Vres) but you don’t know the value of the factor N.
Because you don’t know the value of N, you can’t calculate the actual optical power. But the voltage (Vres) is a relative measure of the optical power since N will be a constant for your circuit at the distance you set between the light bulb and detector. So if you measured 2 volts across the resistor, your optical power output would be 2N.
You do need to experiment with the distance between the photocell and light bulb. If you get the photocell and light bulb too close together, the photocell will “see” too much light (become saturated) and the voltage output won’t change for different light inputs. You correctly followed the procedure when you set your breadboard to get 2.5v across the resistor. The photocell/light bulb distance isn’t used in any of the calculations. But you should record the distance in case you want to rebuild your experimental set up later.
I hope this helps. Please post again if you have more questions.
A. Norman
In this experiment you built a light detection circuit that outputs a voltage that is proportional to the light intensity or optical power. You know that the optical power from the LED is directly related to the voltage from the detector (Pout = N X Vres) but you don’t know the value of the factor N.
Because you don’t know the value of N, you can’t calculate the actual optical power. But the voltage (Vres) is a relative measure of the optical power since N will be a constant for your circuit at the distance you set between the light bulb and detector. So if you measured 2 volts across the resistor, your optical power output would be 2N.
You do need to experiment with the distance between the photocell and light bulb. If you get the photocell and light bulb too close together, the photocell will “see” too much light (become saturated) and the voltage output won’t change for different light inputs. You correctly followed the procedure when you set your breadboard to get 2.5v across the resistor. The photocell/light bulb distance isn’t used in any of the calculations. But you should record the distance in case you want to rebuild your experimental set up later.
I hope this helps. Please post again if you have more questions.
A. Norman