How efficient is an LED compared to a CFL or an incandescent bulb?
Q. Visible LEDs seem to be specified in millicandelas, while IR LEDs are specified in milliwatts. What are millicandelas (mcd) and how can I compare them to milliwatts?
Candelas, candles, foot-candles, lux and related visible light output units are all photometric units, meaning that they are measured using devices that mimic the human eye response. Watts are radiometric units, meaning that they are used to measure power without regard to the response characteristics of the sensor.. The human eye has its maximum sensitivity in the yellow green part of the spectrum, where it has easily 50 times more sensitivity than it does for deep red light. In other words, the sensitivity of the human visual system varies strongly as a function of wavelength. LEDs emit light over a range of wavelengths, and human eye sensitivity is different at each wavelength.
If you want to get an estimate of how many milliwatts are output by an LED with a given millicandela rating, there is a procedure for Using a triple convolution integral, it is possible to convert from mcd to mw, there's an easier way. Most silicon photodiodes convert light to current with an 'efficacy' of about 0.45Amps/Watt. So, procure a large area photodiode and connect it to a low impedance current metering device. Good quality digital multimeters work well, but cheapies won't do. You need to present nearly a dead short load to the photodiode. If you have any trouble getting a big photodiode, see about getting a single solar cell (same thing). You need a photodiode that is at least as big and preferably bigger than your red LED diameter. Fire up the LED, position it quite near to the photodiode and aimed squarely at it, and measure the current. Adjust the LED current (vary voltage, series resistance, etc.) until you get about 2.5 milliamps photocurrent. Since
2.5/0.45 ~ = 6
you know that about 6mW got converted to about 2.5mA! The large photodiode is monitoring total power.
The photodiode should not have any magnifying lens over it…just a flat window or no window. If you can't obtain a big one, you can estimate the amount of current you would get if it were larger by multiplying measured current by the ratio of beam 'spot' area to photodiode area. For example, suppose that the red LED makes a 2 cm diameter spot at 2 cm distance away from itself. Such a spot has about 320 sq mm area. If your photodiode is only 2mm x 2mm, you're only get about 1/80th of the power, so multiply measured current by 80 to estimate what you would measure with a big enough photodiode.
It's very important to verify that the photodiode is operating linearly, which only happens if it has very little forward voltage. If possible, connect a voltmeter across the photodiode/ammeter combination and verify that the voltage is < 200mV. Otherwise, your measurement will underestimate power.
Good quality selected AlGaAs IR emitters will put out 10mW total power at about 40mA drive. The very latest super efficient 660nm red emitters have approx. the same efficiency. So, obtain some T1-3/4 super red parts and you'll very likely get your 6mW at 25mA or so. Highest eff. Parts come form HP, Stanley,betlux, Sharp, Toshiba, and Mitsubishi Cable.