LED uses p-n junction semiconductor together to produce light.
Normal diode also has p-n junction semiconductor and it doesn't heats up quickly. But why LED heats up quickly. What is science behind it? And they uses same kind of material. then, how LED produces light whereas diode cannot?
Electronic – Why LED’s heats up
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Better is coming.
Better costs money.
Present white LEDs usually use a single phosphor so you get two broad spectrum peaks - one from the direct light from the blue LED itself and the other from the yellowish phosphor.
In their recent PAR60 alternative LED bulbs (60 Watt mains replacements) people like Philips are placing the phosphor on a "globe" outside the LED and are using multiple phosphors to get several overlapping emission peaks and a better overall spectrum.
The standard measure of the "goodness" of the white from a light source is "CRI" = colour rendering index. This is essentially a measure of how well a range of test colours appear when illuminated with the test light compared to illumination with "white" light.
A CRI of 100 is "perfect".
CRI of >= 90 is very good.
CRI of 80 - 90 is bearable for many purposes.
CRI of 70 - 80 is definitely getting poor in most cases*.
CRI < 70 - they usually won't tell you.
- I say "in most cases" as I have some LEds with a CRI in the 70-80 range but most people (and me) find the light very nice indeed
Spectral peaks in multiphosphor LED. Usually you get blue plus one other.
Multiphosphor LED chromaticity diagram.
Excellent multiphosphor overview paper
If you used an ordinary LED for your your first test it's probably gone to LED heaven: you only have 1\$\Omega\$ current limiting resistor, that's good for 12A (if your power supply can deliver that). Now if the LED broke shorting that's also the current through the FET. It has a very low \$R_{DS(ON)}\$, so it shouldn't get hot, even at 12A. That probably means the gate voltage isn't high enough. Increase the supply voltage of your driver.
Try again with a higher value for R1, like 470\$\Omega\$. That should get you 25mA. Good enough for an ordinary LED.
BTW, D1 is not necessary. Get rid of it.
edit
On second thought it seems unlikely to me that an indicator LED, when broken, will conduct 12A. The thin bonding wire from the anode post to the die will break and leave an interruption. Then the next suspect is D1. If you mounted it wrong it would explain the high current, and why the LED doesn't light. One more reason to remove D1.
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Best Answer
The amount of power going into a device such as an LED or a diode is the voltage times the current.
LEDs tend to have higher voltage drop than rectifier diodes. For example, a 1N4004 might drop 0.7V at 0.35A (245mW) but a blue LED would drop more like 3V, so it would be consuming more than 1W at the same current, and if the packages were identical it would get much hotter.
The LED will actually run somewhat cooler than you might expect because some of the power is carried off as light rather than as heat.
If the LED was 100% efficient, then it would not heat up at all- all the incoming power would be converted to light. Most LEDs are probably in the 5%-20% range, so most of the incoming power still leaves as heat.