How to calculate the energy output (in Joules) from an LED

Question

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?

Answer 1

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.

Answer 2

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.