Misunderstanding Ohm's law

ThePanMan

Joined Mar 13, 2020
918
I/V curve slope is often also affected by heating of the die which reduces the current (increases the die's resistance) and also reduces the light output per unit of current.
Thought it was the other way around; the warmer it gets the more current it draws. The more current it draws the warmer it gets. Isn't that the description of thermal runaway?
 

DickCappels

Joined Aug 21, 2008
10,661
Grab the nearest LED and adjustable power supply and see for yourself. The bottom line is that there is a maximum amount of light you can get out of an LED as you increase current before the curve passes the peak.
 

DickCappels

Joined Aug 21, 2008
10,661
Then you're out of luck! Just kidding. This is easier to detect when turning a pot than when switching between current "steps", of variable resistor in series with a power supply make a variable power supply with a pot an emitter follower with a resistor on the emitter and the LED feeding the collector, if you are interested in seeing it.
 

k1ng 1337

Joined Sep 11, 2020
1,038
Grab the nearest LED and adjustable power supply and see for yourself. The bottom line is that there is a maximum amount of light you can get out of an LED as you increase current before the curve passes the peak.
I've had LEDs change colours, I believe it was yellow to red on a few occasions as a result of too much current where the die emitted a different wavelength from damage. It was because I connected standard LEDs directly to 5V. Given the fixed wavelength at production I'm curious to why this happens.
 

MisterBill2

Joined Jan 23, 2018
27,508
It seems to me that there is a serious problem with the circuit if with a 5 volt supply and a 100K series resistor the LED was too bright.Either the resistor is not connected to limit the current or the voltage source is not really five volts. Sort of a rather fundamental problem.
 
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