Hi MrAl
yes i read your excellent post(#17)and i happy that you explained in details about the thermal runaway(i was not aware about that phenomenon before),the behavior of the led and the voltage drop when the temperature rise with all the calculation and the way you divided the led into two parts: the led itself and the resistor.all of that help me alot to understand the behavior of the led and thanks for that explaination with all the other helpful posts,i,now, can deal with leds and be aware for what i am doing,and by that avoiding mistakes and damage to the led.and again i apreciate any post that i recieve about the issue because every piece of information can help.
in your post you refer to the thermal runaway,you explained to me what will happens in the circuit: when the temperature of the led increase the voltage will drop on the led and the current will increase,i understood that and liked the way you explain that.but i didn't found explaination what will happens if i increase the voltage source(without involve the thermal runaway)like i said to 5.3v?
Alec_t said
k. that what i thought,but it is possible to calculate what will be,at that case,the vf of the led?i would like to understand by exact calculation,how much voltage does the resistor drop on itself ,when i increase the voltage source, relatively to the led.