Procedure to express radiation intensity of an LED in watts/cm^2

AI Thread Summary
To calculate the radiant intensity of an LED in watts/cm^2, the total radiant power should be divided by the area of the circle projected by the light cone onto the target, as the illumination is not uniform. For the UV LED with a power of 20mW, this approach assumes that the entire power is contained within the projected area. For the green LED, luminous intensity is converted to watts using the luminosity efficiency function, but care must be taken regarding the peak value and the effective spot size due to non-uniform illumination. It's also important to consider the sensor's color response when measuring intensity, as typical lux meters may not accurately reflect LED light, especially for monochromatic or mixed colors. Understanding the difference between photometric and radiometric measurements is crucial for accurate results.
Abhishek K J
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Procedure to calculate radiant intensity of a LED in watts/cm^2 given total radiant power of the LED or luminous intensity of the LED. Note that the target (on which radiation from LED is falling) is placed close to the LED source and is big enough to accommodate circle projected by the light cone from the LED. LED has narrow wavelength band, and assume almost uniform intensity distribution. Viewing angle and other details are accurate.
I am calculating Responsivity of a pn junction photodiode (a.k.a the target) by irradiating radiation from LED sources. For this purpose, i have two LEDs, one UV and another green LED. Note that LEDs are placed close to the target.UV LED : Manufacturer has given total radiant power to be 20mW. How can i calculate radiant intensity in watts/cm^2 ? should i just divide the power of LED by the total area of the target since all radiation from UV LED is reaching the target (since the LED is placed close to the target)?Green LED : Luminous intensity is given in mcd (as"x" mcd). I have converted mcd into lumens then finally to watts as follows :Firstly, i find total solid angle using the expression: Ω (solid angle)= 2(pi)(1-cos(θ)). here, θ is half the apex angle of the light cone. Apex angle is also equal to viewing angle of the LED mentioned in the datasheet.Now, Lumens = Ω*(x) mcd. After this, i use the luminosity efficiency function to convert lumens into watts.Now, given that i have placed the LED near the target, how to find the radiant intensity in watts/cm^2 ? Should i just divide the power by area of the target ? Or, should i divide the power of the LED by the area of the circle projected onto the target by the light cone?
 
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n each case you need to estimate the intensity patern because

UV: the 20mW is distributed across the entire illuminated beam somehow

Green: use the luminous efficiency as you have done (correctly I presume) and then realize that the quoted number is typically the peak value luminous flux (for the beam center)

The radial beam pattern for LEDs is published if you know the type. Make a +/-20% estimate and calculate

.
 
hutchphd said:
n each case you need to estimate the intensity patern because

UV: the 20mW is distributed across the entire illuminated beam somehow

Green: use the luminous efficiency as you have done (correctly I presume) and then realize that the quoted number is typically the peak value luminous flux (for the beam center)

The radial beam pattern for LEDs is published if you know the type. Make a +/-20% estimate and calculate

.
Okay. Yes,i have used luminosity efficiency function correctly, i presume !

Can you answer my last doubt, which is, while expressing radiant intensity (in watts/cm^2), should i divide by the area of the target or by the area of the circle projected by the light cone on the target? i am assuming that the circle projected on the target contains 20mW in it, the reason behind the assumption is that, datasheet mentions viewing angle .
 
Be careful about what sensor and meter you use. The sensor itself has a color response that must be considered in any measurement.

Also, IIRC, most commonly used meters assume that the whole aperature is illuminated.

a lux meter cannot accurately display the intensity of an LED, especially if the LED is monocolored blue or red light, and certainly not when it is a mixed purple light.
https://ledlightinginfo.com/can-a-lux-meter-measure-led-light
(https://www.google.com/search?&q=led+light+meter)

https://bioslighting.com/how-to-measure-light-intensity/architectural-lighting/
(https://www.google.com/search?&q=how+measure+optical+intensity)

Cheers,
Tom
 
Abhishek K J said:
Can you answer my last doubt, which is, while expressing radiant intensity (in watts/cm^2), should i divide by the area of the target or by the area of the circle projected by the light cone on the target?
The area of the illuminated circle (approximately). The art of the matter is that the circle is not illuminated uniformly, yet the specified intensity is usually the max (central) value and so you should make an estimate of the "effective" spot size. Your eyeball is a very nonlinear (log) detector, so this can be difficult without data.
An indeed be wary of photometric vs radiometric reporting by the light meter.
 
hutchphd said:
The area of the illuminated circle (approximately). The art of the matter is that the circle is not illuminated uniformly, yet the specified intensity is usually the max (central) value and so you should make an estimate of the "effective" spot size. Your eyeball is a very nonlinear (log) detector, so this can be difficult without data.
An indeed be wary of photometric vs radiometric reporting by the light meter.
Alright. Thank you, i think i got it.
 
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I did this for a living for a while and I had a cheatsheet on the wall to remind me of the names and units of all the various units (and per steradian or per area or per 4pi steradian or photometric or radiometric or IR filtered and sensitivity of eyeball or Si and a different name for each). I think you've got this but I always wanted my cheatsheet..and still do, but now there is Wikipedia!
 
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