how to match voltage source for 3w star led?

[alec_t]

and then,what will happen then,when i,suddenly, increase the voltage source from 5v to 5.3v,what then will happen to the led vf voltage?

It will increase slightly because of the increase in current through the LED. The variation of Vf with current is non-linear and will vary from one LED to another as well as (AFAIK) with temperature.

 

[rgbbv]

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:

It will increase slightly because of the increase in current through the LED. The variation of Vf with current is non-linear and will vary from one LED to another as well as (AFAIK) with temperature.

ok. 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.

 

[throbscottle]

Deleted

 

[alec_t]

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.

Without knowing the (unpublished) exact characteristics of the particular LED you are using you won't be able to calculate Vf. You could, however, measure it.

 

[MikeMl]

This might help you visualize what is happening as you increase the supply voltage. This simulation uses a similar LED to yours, but I do not have an exact model of yours. The simulation sweeps the supply voltage from 2 to 8V, while plotting the LED's V(f), the voltage across the resistor V(s)-V(f), and the current through the LED I(D1).

Note the non-linear relationship between V(f) and I(D1). Note that the exact shape of this relationship is a function of each LED you grab out of the bag, and also of temperature.

 

[MrAl]

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 saidk. 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.

Hi again,


You have to slow down a little and think about what is going on and use the equation i gave you:
vLED=3v-0.0025*T

and the equation:
I=(Vs-vLED)/R

In the first equation, the 3v is the initial voltage of the LED at room temperature which you can measure by powering it with a resistor and measuring the voltage quickly. You can observe that as the LED heats up, the voltage across the LED drops.

In the second equation, Vs is the source voltage. So you can plus in the new value and see what happens. If you go from 5v to 5.3v then you plug in the new 5.3v and go from there.

The actual temperature rise is a little harder to calculate so you have to estimate based on a previous measurement or two. But you can get some idea what happens even without that. As Vs rises, the current increases, and it increases more when there is a small resistance R than with a larger resistance for R. Thus the heating is more with a lower value R than with a higher value R.

 

[throbscottle]

What a very educational thread - once again I have learnt things I didn't know that I didn't know

Moral of the story: help newbies because someone more knowledgeable than you might chip in with better information!

 

[rgbbv]

MikeMI,your attached simulation was exactly what i was looking for,and it is,definitely,help me to understand,what is happening as i increase the supply voltage.
MrAl,also add an essential information,that when the resistor is larger the current increase less and the heating is less.and,now,i can understand why is so nessecery to use a resistor in series with the led,and ,also,why is more simple and safe to use constant currenr drive.
i want to thanks,again, for all your replies,you gave me very professional explanations.
best regards