Voltage Drop Across LEDs

It is safe enough to assume 1.7 volts for non-high-brightness red, 1.9 volts for high-brightness, high-efficiency and low-current red, and 2 volts for orange and yellow, and 2.1 volts for green. Assume 3.4 volts for bright white, bright non-yellowish green, and most blue types. Assume 4.6 volts for 430 nM bright blue types such as Everbright and Radio Shack. Design for 12 milliamps for the 3.4 volt types and 10 milliamps for the 430 NM blue.

(google is wonderful)

Yes - the answer is "it depends".

Thanks a lot.

It was a great help. Still could not understand how and why though.

The values have been discussed so I will only comment on why. Diodes are made of of 2 types semi-conductor material. For simplicity I will say positive and negative. This material is sandwiched together. When a positive voltage is applied to one side and a negative voltage is applied to the other (forward bias) there is a certain amount of potential difference (voltage) at which the materials become saturated and current begins to flow. For a silicone diode this voltage is .7 volts. Get the picture? Nothing below .7 volts will get through so you always have a voltage drop of .7 volts as long as your supply voltage exceeds .7 volts. That is of course unless you exceed the wattage of the diode by not limiting the current in your circuit to less than the wattage rating of the diode. When that happens, the diode burns up.

In the case of light emitting diodes, they glow when forward biased. The material is diffent in composition for different colors. The saturation point is also different which accounts for different voltage drops and if you apply too much voltage (current, wattage or whatever you want to call it) you blow up the diode. Actually, they usually burn out kind of slowly with just a little over-voltage and then won't burn brightly any more. I know. I found out the hard way.

LEDs are doped with different materials to change the emitted color but it also changes the properties of the Led itself hence the different forward voltages Not_so_smart mentioned

Sidban,

One key point that several people here have missed is that the voltage drop for the various LED listed here is what determines the color or wavelength of the light produced. The specifics about why this happens gets into the quantum mechanics of semiconductor material, doping and energy released per electron for jumping the band gap. It's my latter comment that determines the color of the light. Blue light at 468 nm wavelength has a higher energy per photon than red light at 645 nm wavelength. This is why LED light is monochromatic. This does imply that the "typical" 0.7 volt drop of a diode does produce light. But the energy per photo is so low (deep deep infra-red) that it is easily re-absorbed and just converted to random thermal heat. (Not quite accurate but valid.) The near exception are the white LED available today. White of course is a spectrum of colors of a variety of different wavelengths. But white light is not the light produced directly by the LED. The LED produces ultra-violet light that sprays a fluorescent phosphor coating similar to the coating found inside fluorescent bulbs. The fluorescent coating then produces the multiple spectrum that appears as white light.