Calculating LED Efficiency with Radiant Power and Temperature

[mathewmical]

i wonder if it is possible to get the efficiency of an LED just by knowing the radiant power an the outside temperatur. i also know the temperatur of the emmited light. well they say the LED has an efficiency >1 because it takes heat from the out side and turns it into rays. so these two things must add to have the radiant power... but how can i get the efficiency out of that?

 

[ZapperZ]

i wonder if it is possible to get the efficiency of an LED just by knowing the radiant power an the outside temperatur. i also know the temperatur of the emmited light. well they say the LED has an efficiency >1 because it takes heat from the out side and turns it into rays.

Wait, what?

Do you see any "rays" when the LED is not connected to a power source but just simply from the "outside temperature"?

Zz.

 

[mathewmical]

well no, but somehow it should be possible, I've read . if i cancel that thought, how could i solve the problem then?

 

[Cutter Ketch]

Well, you could have knocked me over with a feather! When I read this question I thought you must be very confused. Then I searched and found this Phys Rev Letter

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.108.097403

Apparently running an LED at very low voltage causes refrigeration. Some of the heat from the crystal is converted to light.

Regarding your question, I think the answer must be no. The efficiency of a light is defined as the light power out divided by the electrical power in. Therefore you are asking if you can infer the electrical power in from the ambient temperature alone. There is an awful lot of physics between those two quantities: the rate of cooling, the heat capacity, the thermal resistance of the conductive, convective, and radiative heat paths to ambient, the internal resistance of the diode, etc etc. It would take a heck of a model to connect ambient temperature alone to electrical power in based on first principles. Also, you say "these two things must add together" as if the ambient temperature tells you the thermal power flow. It does not. like any refrigerator how much heat is pumped depends on the the heat pumping mechanism. The heat pumping depends on the temperatures of the heat baths, but cannot be determined solely from them. On the other hand you might be able to define some theoretical bounds based only on the temperatures sort of like a Carnot efficiency for LED cooling!

On a practical note, I don't think this cooling phenomenon is a large part of the efficiency of standard LEDs and in practice can safely be ignored.

 

[Cutter Ketch]

Sorry for the double post, but I just wanted to say I am more and more impressed with the idea of using the two heat bath temperatures as you suggested to define a maximum theoretical limit of efficiency. I wouldn't have thought of it that way, but I think you may be into something there.

 

[ZapperZ]

Well, you could have knocked me over with a feather! When I read this question I thought you must be very confused. Then I searched and found this Phys Rev Letter

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.108.097403

Apparently running an LED at very low voltage causes refrigeration. Some of the heat from the crystal is converted to light.

This is not surprising to me because this is analogous to the Nottingham effect, and this is known in photocathodes and material junctions. But what the OP described appears to be something else.

Zz.

 

[Cutter Ketch]

I think it may be difficult to define the temperature of the photon bath as it is not created in a thermal process and is not Boltzmann distributed.

 

[ZapperZ]

well no, but somehow it should be possible, I've read . if i cancel that thought, how could i solve the problem then?

I actually have 2 undergraduate students who had completed measuring the efficiency of an in-vacuum UV LED light source. All they did was measured the input power to the LED (it has a range of variable operating parameters), and then they measure the power of the emitted light using a power meter and diode.

In other words, measure power in, and measure power out.

Zz.