Today, I decided to overdrive a star white LED I have with my adjustable power supply.
I extracted this LED from a little cheap flashlight, And it is a 1 watt.
It has no heat sink, But the metal plate is enough.

Here is the order of things happening:

1. Start at low power: 2.7 V .02 A, LED just glows.
2. Normal power: 3.7 V .33 A, LED is working normally, And is pretty bright.
3. Higher power: 4.0 V .6 A, LED is slightly brighter, No bad effects likely due to the little heat sinking.
4. Even more higher power: 6.0 V ? A(.8-1?): LED goes a light green! Would likely fail soon.
5. Very high power: 10 V 1 A, LED goes blue!(Phosphor likely changed, Or the blue LED chip is overpowering the phosphor, Causing a blue tint) Failure is coming.
6. Extremely high power: ? V 1 A, LED quickly burns up, Will never emit light again.
7. Highest power: 30 V 0 A, Power supply finishes off LED, exposing it to 8 times the normal voltage, Does not light up, Or have any current draw.

Goes to show how picky LEDs can be about power and heat sinking.
The LED has lots of pops, Blotches, Burns, And more.  

"Higher power: 4.0 V .6 A, LED is slightly brighter, No bad effects likely due to the little heat sinking"

That is double the power from the previous test, and 2.4 times the power rating of the LED. No bad effects likely ? I dont think so

The light output from the LED is one wavelength - Blue at around 450nm. That corresponds to the Energy gap distance the Electrons fall to put out that Blue Photon. At around that voltage (3...3.4V for Blue LED), small change in voltage makes big, exponentially growing change in how many Electrons will flow, so current and emitted light. See your experiment - The difference between 3.7V to 4V is fairly small (under 1/10th) but the current is near double

LED (or any diode) I/V behavior is exponential around its intended working point. That is why proper LED drivers are stabilized to current and not to voltage

At really excessie curent, the diode is allready as conductive as it gets, all the rest of the voltage is I*R that falls on resistances of the materials that make the LED and not on overcoming the Energy gap. So indeed more power and more heat, without as much more light. The I/V behavior in this stage behaves respectively - More like that of a resistor in series to a diode, and not the nearly pure exponential behavior of the diode itself

Finally, the heat generated damages the diode..

The color change is most likely from burning up the Phosphor. In the Green stage some Phosphor burned off so it was like lot of Blue with little Orange, then as all the Phosphor burned out only the Blue light remained

"Higher power: 4.0 V .6 A, LED is slightly brighter, No bad effects likely due to the little heat sinking"

That is double the power from the previous test, and 2.4 times the power rating of the LED. No bad effects likely ? I dont think so

The light output from the LED is one wavelength - Blue at around 450nm. That corresponds to the Energy gap distance the Electrons fall to put out that Blue Photon. At around that voltage (3...3.4V for Blue LED), small change in voltage makes big, exponentially growing change in how many Electrons will flow, so current and emitted light. See your experiment - The difference between 3.7V to 4V is fairly small (under 1/10th) but the current is near double

LED (or any diode) I/V behavior is exponential around its intended working point. That is why proper LED drivers are stabilized to current and not to voltage

At really excessie curent, the diode is allready as conductive as it gets, all the rest of the voltage is I*R that falls on resistances of the materials that make the LED and not on overcoming the Energy gap. So indeed more power and more heat, without as much more light. The I/V behavior in this stage behaves respectively - More like that of a resistor in series to a diode, and not the nearly pure exponential behavior of the diode itself

Finally, the heat generated damages the diode..

The color change is most likely from burning up the Phosphor. In the Green stage some Phosphor burned off so it was like lot of Blue with little Orange, then as all the Phosphor burned out only the Blue light remained

Interesting, I still see some phosphor intact, But good point.
If I had good heat sinking, I am sure that the phosphor would not burn off.

What do you mean by "intact" ? It may still appear white, but maybe it lost its Phosphor properties. Try ligthing on it with another Blue LED and see what happens

What do you mean by "intact" ? It may still appear white, but maybe it lost its Phosphor properties. Try ligthing on it with another Blue LED and see what happens

It lights up like a phosphor, It still works.

Maybe on the outside layer, but the ones deeper to the chip dont ? (Then the quantity of Phosphor remaining intact is insufficient to convert most of the light)

Well, The phosphor closest to the chip has a nice brown burnt color.

That might be burn products of the chip itself melted into the Phosphor, or some material in which the Phosphor is suspended (some resin ?) , or the Phosphor itself

I also think that the tiny leads burned like a fuse.