I just want to ask, is electric power equal to radiant flux? In calculating intensity, it is the total light power emitted (radiant flux) divided by the surface unit area.


closed as unclear what you're asking by Dmitry Grigoryev, Arsenal, Voltage Spike, Daniel Grillo, pipe Sep 9 '16 at 15:20

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  • \$\begingroup\$ Peripheral" You say " ...it is ..."-> What is? You don't actually say. That would be eg lux = lumens/m^2. You seem not to have introduced any equivalent concept to area illuminated in your attempted analogy. The question can't be answered as right or wrong until you suggest some obvious analogy. || I'd suggest that some some parallel may be able to be suggested but it's liable to be artificial and situation specific. Someone may have a more satisfying analog than my brain deigns to proclaim at 2:30am :-). \$\endgroup\$ – Russell McMahon Sep 8 '16 at 14:27
  • \$\begingroup\$ The units of electric power and radiant flux are the same, but it's not light which is doing the work in electric circuits, so I'd say no, they are not equal. I'm a bit confused what you want to know exactly. \$\endgroup\$ – Arsenal Sep 8 '16 at 15:03
  • \$\begingroup\$ thank you for ur response sir , i am actually going to characterize my LED for my project, and i need to know the intensity or the power density of my blue led, according to the formula , I= P(light power emitted ) divided by the surface unit area . ive read an article ,it says that the power consume by the light source is equal to the light power emitted . and i want to know if it is right. \$\endgroup\$ – Ralph Bryan Sep 8 '16 at 15:22
  • \$\begingroup\$ No it's not. If it was, you would have a 100% efficient light source, which as far as I know is not available at the moment (even LEDs heat up). (at least if you only consider visible light) \$\endgroup\$ – Arsenal Sep 8 '16 at 15:34

It says that the power consume by the light source is equal to the light power emitted

That's probably not a true statement: at best it's intuitively misleading.

By the law of conservation of energy we know that all the electrical power consumed must go somewhere, but it doesn't need to go into light. Certainly not all of it goes into visible light: the LED also gets hot. There are other kinds of non-visible light that might be radiated: infrared light from the heat, and RF light from variations in the current in the device.

In practice the LED is probably in air, and is attached to wires. Some of the thermal energy is lost to the air by convection, and the wires by conduction. In practice this is where most of the energy that wasn't radiated as visible light will go.

Is electric power equal to radiant flux?

Both of these things are measured in watts, so it's possible for a device that emits light and nothing else. But since there's convective and conductive cooling taking energy out of the system by a mechanism that isn't light, we can deduce there will always be some degree of discrepancy between input electric power (current times voltage), and output radiant flux (power of photons leaving the LED).

  • \$\begingroup\$ sir,do you know any alternative method in measuring radiant flux and intensity(my project is to characterize the blue LED operated with 12 v DC , no datasheets given ) ? i only saw formulas and heavy device that are not available in our school .. \$\endgroup\$ – Ralph Bryan Sep 8 '16 at 15:55
  • \$\begingroup\$ LEDs of the same color tend to have similar performance characteristics. So your best bet which doesn't require specialized equipment could be to look for a table of typical performance, or datasheets of similar LEDs, and make an estimate based on that. If you must measure directly, most smartphones have a lux meter, so with the appropriate app and some calculation you might get something that's correct at least within an order of magnitude. \$\endgroup\$ – Phil Frost Sep 8 '16 at 16:01
  • \$\begingroup\$ Lux to watts calculation with area in square meters: The power P in watts (W) is equal to the illuminance Ev in lux (lx) times the surface area A in square meters (m2), divided by the luminous efficacy η in lumens per watt (lm/W): P(W) = Ev(lx) × A(m2) / η(lm/W) (i saw this formula, but still it says it is the electric power, not the power emitted by a light source) \$\endgroup\$ – Ralph Bryan Sep 8 '16 at 16:06

No, it's not. Most of the energy you put into a diode goes into heat rather than light. Heat can remain as energy within the device or it can radiate as well. So any LED will have energy going into the frequency of light you want, radiated energy in the infrared range due to thermal radiation and then you still have conductive heat that will move and be released by touching other objects (air, metal contacts, pcb, etc.)

See here for a typical table for luminous efficiency of LEDs.


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