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r11567
Lv 4
r11567 asked in Science & MathematicsEngineering · 1 decade ago

How can I measure and compare the light efficiency of LED against traditional devices like light bulb?

Update:

I will appreciate if you can show some technical details.

5 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    There are three figures of merit when evaluating a white light source. They are:

    1. Efficacy (lumens/watt). This is the lumens (power your eye sees) out of the source per watts inputted into the source. Obviously higher numbers denote more efficient devices.

    2. Color temperature (K). This is the "color" of white that your eye sees. White light comprises many wavelengths in the visible spectrum. White light that is missing the red is called cool white (5000-1200K) and appears blueish to the eye. Most LEDs are cool white. If a significant portion of the red is also emitted then the source can be classified as warm white (3000K). Incandescent light bulbs are typically warm white.

    3. Color rendering index (CRI=0-100). This metric tries to show how well the various colors will be illuminated by the source. The higher the CRI the better the source. If you have a source that emits no red light then you will not be able to see a red object properly under its illumination.

    Theoretically you can measure all these values in an integrating sphere with a calibrated spectrometer. There are a lot of companies that will sell you a calibrated setup that will give the above values. As you can imagine the LED systems are small, but to measure a fluorescent bulb the sphere would need to be room size. You shouldn't have to do this though because the manufacturer of the source should be able to give the values for their sources. This is all you need to compare.

    After all that I need to caution that the comparison is not really fair. This is because the CRI metric is flawed. There is some controversy in the industry because the CRI was created for bulbs and fluorescent and not LEDs. The method to measure CRI will give a source like a fluorescent, which emits discrete wavelengths, high values even though many of the colors are missingn from its emission. The white LED with a similar CRI and broad wavelength emission will actually have more emitted colors and therefore your eye will be able to perceive more colors with the white LED. Until this changes though we have to live with the current CRI measuring method.

    You can find more information on LEDs at:

    www.lumileds.com

    www.cree.com

    More about evaluating light sources:

    http://www.cie.co.at/cie/

  • gp4rts
    Lv 7
    1 decade ago

    You have to gather some data. Get some LED data sheets and see how many Lumens are produced at a given forward current. Also find the forward voltage drop. The forward voltage drop (in volts) times the current (in amperes) gives the power used by the LED in watts. Compute its efficiency in Lumens per watt by dividing the lumen output by the wattage computed.

    Get a standard light bulb, say 100w. On the carton, the number of lumens will be stated. You can now compute its efficiency and compare.

    Depending on how much you want to delve into this, there are other factors to consider:

    a) The color of the LED will not be the same as the light bulb. To be accurate, the two should have similar color appearance to the eye. I don't know of any LEDs (even "white" ones) that match the color temperature of an incandescent light. But using any "white" LED (which is really a combination of three LEDs) is a better choice than a red one which would be much more efficient. Lacking that, pick a yellow LED.

    b) While the power consumed by the LED is the forward voltage times the current, that is not the power needed to drive the LED. LEDs need series resistors to limit the current flow, and there will be power lost in that resistor that reduces the efficiently that can be obtained in practical applications. More efficient circuits use transistors as current sources, so no resistor is needed, but even there some power is lost in the driving circuit.

  • 1 decade ago

    Actually in lighting instead off efficiency we use efficacy.

    Efficiency is output power over input power. A bulb outputs 10% approx worth of power as light (visible) and 90% as heat. it's output is 100% but not all usable.

    Efficacy on the other hand is how many lumens for how many watts (lumen/watt). I would use a LUX meter at a set distance for the lumens (multiply result by the sensor area in square meters) and measure the voltsxamps of both the bulb and led using a multimeter for the power. unless you find some way to collect and collimate the light source or fully enclose it in a callbrated reflector you are only going to get an indication of comparison of efficacy. Hope this helps

  • 5 years ago

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  • Lee J
    Lv 4
    1 decade ago

    Multimeter

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