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Optical Output power of LED

  1. Dec 27, 2015 #1
    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.
     
  2. jcsd
  3. Dec 27, 2015 #2

    berkeman

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    Staff: Mentor

    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:
     
  4. Dec 27, 2015 #3
    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 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?
     
    Last edited: Dec 27, 2015
  5. Dec 27, 2015 #4

    jim hardy

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    Science Advisor
    Gold Member
    2016 Award

    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 .
     
  6. Dec 27, 2015 #5

    meBigGuy

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    Gold Member

    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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