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Intensity of LEDs

  1. May 25, 2010 #1
    Hi guys, first post here but I have been looking around a while and this forum has proved to be pretty helpful. I have a question about LEDs. I just did a lab in school where I cooled LEDs by placing them close to liquid nitrogen and also heated them close to hot water and measured their voltage and observed light intensity or brightness.

    I came to the conclusion that when cooled down, the LEDs have a higher voltage running through them and they appear brighter. The opposite was true for when they were heated up.

    Now, I've been told that unlike a normal light bulb, Ohms law doesn't hold true for LEDs. And I assume this is because they are semiconductors and behave in a different way. I am struggling however to think of a way to explain the behavior of voltage and intensity in a way that relates to band gap theory. Hopefully someone can explain this relationship to me. Thank you!
  2. jcsd
  3. May 25, 2010 #2


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    Staff: Mentor

    You should plot both LED voltage and current versus temperature. Were you controlling the current to keep it constant? Or did the current change because you were using a constant voltage power supply and a simple resistor for current limiting?
  4. May 25, 2010 #3
    I don't think Ohm's law is ever true, its not as simple as V = IR, there is a differential equation involved.

    Anyways, a normal light bulb relies on resistance to increase the electron's energy levels. So if you heat it up, the resistance is higher, and it gets brighter.

    An LED is much more efficient, a process that allows the electrons to move between energy levels in a less chaotic, more controlled way, and if you cool the LED the process works even more efficiently.
  5. May 25, 2010 #4


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    Staff: Mentor

    I'm not understanding what you are saying. What DE are you referring to?

    The resistance does not increase the electrons' energy levels. And increasing the resistance of the bulb filament does not necessarily increase the brightness -- it's all about the power dissipated, so if the current goes down with increased resistance and constant voltage, the brightness of the incandescent bulb would drop.

    What do you mean by "works more efficiently"? What solid state equations are you referring to?
  6. May 25, 2010 #5
    LEDs are diodes and the I-V characteristic is pretty much the same of any old diode.
    If you do some research, you´ll find the threshold voltage rises as temperature goes down. Even though is a bit old, McKelvey's book "Solid State and Semiconductor Physics" will give a clear explanations. The efficiency of conversion from current to photons is a bit more complex to explain because photon creation competes against other recombination processes
  7. May 25, 2010 #6
    Well, as the resistance increases the temp increases and that causes the resistance to increase again. So it is more than just V = IR in a real environment.

    I don't really understand how LEDs work so I might be wrong about it becoming more efficient in cooler temps. I just know they are much more efficient than just applying a voltage to a tungsten wire.
  8. May 25, 2010 #7
    For a simple resistor only circuit V=IR absolutely does hold true. Sure, things will heat up and the resistance will change but ohm's law still applies it's just that a new value for R must be used. Also, generally as metals heat up they become more resistive but as semiconductors heat up they become less resistive.

    As said above, Ohm's law does not directly apply to diodes. They are nonlinear devices as can be seen in the first link below:

    Voltage vs Current curve for a diode:

    LEDs are also nonlinear devices and a small change in voltage can give an exponential change in current. This is why they are generally run at a constant current so they stay as efficient as possible and aren't destroyed/damaged.

    As far as efficiency of an LED vs temperature I don't know enough about it. Sorry!

    Last edited: May 25, 2010
  9. May 25, 2010 #8

    Andy Resnick

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    I'm not sure your questions are getting answered.

    LEDs are simply diodes- Ohm's law does not hold in any meaningful approximation for a diode.


    The brightness of the LED is related to the current:


    Now, when the voltage changes a little (the operating voltage is temperature-dependent), the current changes *a lot*, which means the LED heats up. By cooling the LED, you allow it to safely carry a much higher current. You didn't give any information about the power supply, but LEDs are usually driven by a regulated power supply, which can limit the current draw as a safety feature.

    Does that help?
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