How can I compare the output of IR LEDs, specified in watts, to visible LEDs, specified in millicandelas? Q. It seems that the optical output IR LED's is generally given in milliwatts and phototransistor/diode sensitivity is given in flux density, lux or millicandelas. The only connection that I can find between the two is that 1 W = 683 lm at 555nm. A. First of all, you only need the efficacy of Si photodetectors to predict photocurrent from incident power. Of course, it varies as a function of wavelength, but at 880nm, it is about 0.5 Amps/Watt. Simply multiply the incident power produced by the IR LED by this value to get photocurrent. At other wavelengths, use the appropriate values from the efficacy curves published in datasheets. However, you won't see much variation from photodiode to photodiode. Silicon detectors all have pretty much the same curve, reaching a peak around 900nm and falling off roughly linearly toward the shorter wavelengths, crashing precipitously beyond 950nm or so. It can be difficult to calculate incident power from datasheet values, particularly for visible emitters that are specified in photometric units. It depends strongly on geometry. If an emitter is spec'd in milliwatts per steradian, you multiply the spec by the number of steradians covered by you detector. Steradians are a measure of solid angle. By definition, a sphere has 4*pi steradians. This is handy, since the surface area of a sphere with radius r is 4*pi*r*r. It follows that a small area A on a sphere of radius r subtends a solid angle T: T = (4 * pi) * A /(area of sphere with radius r) = (4 * pi) * A / (4 * pi * r * r) = A / (r * r) For a detector with active area A at a distance R from an emitter with steradiance S, the power P incident on the detector is approximately: P = A * S / (R * R) Obviously, the units of A and R have to be the same, e.g., if A is expressed in square mm, measure R in mm. The above approximation applies only to the case where a directional emitter is aimed at the detector with the detector facing the emitter. This is because those are the conditions under which the datasheet value S were measured. An example will make this more clear. For Siemens IR emitter SFH484-2, S = 900mw/sr at 1 amp. For Siemens photodiode BPW34, A = 7 sq mm. At R = 1000mm, we calculate the incident power: P = 7 * 900 / (1000 * 1000) = 6.3 microwatts From this figure, we predict about 3 microamps photocurrent. Voila! If you need to do the same calculation for visible LEDs that lack the steradiance spec, you can divide the luminosity (in millicandelas) by the value (in lumens per watt) of the standard observer curve at the wavelength of the LED. Note that a 1000mcd source has a total luminosity of 4*pi lumens, i.e., 1 lumen per steradian. Since there are 4*pi steradians over a sphere, this definition of lumens makes it easier to work with steradians. The standard observer curve is published in many places. Just look in the back of most any LED databook. Since LEDs output a band of wavelengths, you are making an approximation by using a single value from the curve, but it will generally give you a good estimate. Again, an example will make this more clear. Siemens red LED LS5420 is rated 100mcd (i.e., 0.1cd) at 10ma drive with dominant wavelength 628nm. Standard observer value at that wavelength is about K = 300 lumens per watt. Its steradiance at 10ma is therefore estimated by: S = [0.1cd * (4*pi lumens/cd)]/ [300 lumens/watt * 4*pi angle per sr] = 0.1 / 300 watts/sr = 333 microwatts/sr at 10ma. For comparison with the SFH484-2 IR emitter above, note that its steradiance at 10ma would be about 100x lower than the 900mw/sr used above, making the IR emitter about 25x more intense than the red emitter at the same current! However, note that the visible LED emits into a much broader angle (+/- 24 degrees) than the IR emitter (+/- 8 degrees). The red LED covers 9 times the area (at 3x the angle). We can infer that the red LED puts out about one third of the power of the IR emitter at the same current (25/9 = ~3).