I need to emit 20J over 5ms onto a target area of 962mm^2 from an LED source. The source as well as the target will reside in a completely encapsulated environment with reflective internal walls such that all of the energy will inevitably make its way to the target.

If I were to use a xenon flasher, rather than the LED bank, determining the output energy is quite easy: the bulb's data sheet provides the max input energy (electrical) to apply at the base to achieve the max output energy (in the form of light). Most xenon bulbs are 50% efficient at converting electrical energy into light energy so if I wanted 20J output I would purchase a bulb rated at 40J input and I would power it using the relationship E=0.5*C*V_c^2.

Does there exist a similar process for determining the output energy of an LED?

  • \$\begingroup\$ I think you are mixing J and W. I also don't see how a capacitor has anything to do with your question. \$\endgroup\$ Commented Feb 17, 2018 at 2:49
  • \$\begingroup\$ @HarrySvensson, I think the OP is asking about a case where a capacitor is dumped into a lamp of some sort. Basically the OP needs to know the efficiency of the LED at converting input power to output power. The input and output energy will also be related by the same constant. \$\endgroup\$
    – user57037
    Commented Feb 17, 2018 at 3:03
  • \$\begingroup\$ The lumen output of the LED will be specified in the datasheet. The input power required for a certain lumen output will also be specified (lumens/Watt). Lumens are basically optical power, scaled by some sensitivity factor. I hope this comment may help you a bit. I don't have time to dig in to reference material sufficiently to write a good answer. But it is an interesting question. \$\endgroup\$
    – user57037
    Commented Feb 17, 2018 at 3:20
  • \$\begingroup\$ You say you need to provide 20 J. Over what time period? Dividing energy by time produces power, and knowing the required power level will determine the LED characteristics. Also, in your xenon example, your capacitor value will only be correct if the capacitor completely discharges - and they never do. \$\endgroup\$ Commented Feb 17, 2018 at 13:00
  • \$\begingroup\$ @WhatRoughBeast 5ms is my on time \$\endgroup\$
    – Landon
    Commented Feb 19, 2018 at 16:15

2 Answers 2


Your question is about efficiency of a LED emitter. You can't calculate the efficiency, it is a design/manufacturing parameter. Some manufacturers do specify the light output in terms of power, so it should be possible to find the answer to your conditions in easy way. The parameter varies wildly, with LED bin, and chip temperature. For example, this LiteON LTPL-C034UVH405 UV emitter has more than 50% efficiency, at 3.9V and 700 mA (2.73 W of input), it emits 1.375 W of light with spectral peak of 405 nm, according to the datasheet.

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As Fredled suggested in the comments, one solution to this question is by running a simple experiment:

Step 1) Calculate energy present on target surface after exposure time. I will assume LED is 50% efficient (i.e., to get 20J output, we need 40J input):

1J/sec = 1W so 20J/5msec = 4000J/sec = 4000W ==> Energy = 4000J after 1 second

Step 2) Calculate the mass and identify the specific heat capacity of the 962mm^2 target (which is made of 2mm thick aluminum). Multiply these two values:

(Area×Thickness×Density)×(heat capacity) = ((962mm^2)×(2mm)×(2700g/mm^3))×(0.9J/g°C) = 4.675J/°C

Step 3) Dividing Energy by the result from step 2 will result in the number of degrees Celsius that the target will rise during the 1 second exposure: Delta T = Energy/(4.675J/°C) ==> Delta T = 855.556°C after 1 second.

So after 5msec, if around a 4.278°C increase in temperature is measured then we can safely conclude that the LED(s) have done their job!

The drawback to this solution is that you have to buy the LEDs before knowing the energy output. If, after performing the experiment, you discover that it doesn't output the correct amount of energy, you have wasted your money and must repeat the process until one of them works.

  • \$\begingroup\$ It's not feasable. If you need 4000W of LED then you would need 1300 LEDs concentrated on an area 962*962mm. How are you going to build this? It will be simplier to see further up the line how to decrease the amount of energy needed. Why not using a much smaller aluminium pad and less LEDs? Or change the way the system is triggered. Why does it have to rise such big mass of aluminium by 4.3C over 5 ms? \$\endgroup\$
    – Fredled
    Commented Mar 7, 2018 at 13:01

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