Can you break an led with too much voltage

[daviddoria]

I believe the LEDs i'm working with drop 1.9 volts... does this mean that if i put +12 to the positive through a 220 resistor then negative to ground that i will simply have 10.1 v left over? or will that break the LED? it seems to work.... will it burn out faster? the thing is that i am trying to turn on 3 LEDs and as soon as i turn on the 3rd one when i am running it on 5v the whole thing turns off. is it safe to run it on 12v?

thanks
david

 

[Chippie]

Generally speaking, leds have a forward voltage of 2.0v and emit light commensurate with the current flowing through them, this is usually around 20mA,but consult manufacturers data sheets........

so to calculate the series resistor use this formula:
(Vs-Vled)/Iled

where Vs= your supply voltage
Vled is the forward voltage of the led
Iled is the currnet of the led.

so if we use your information:

12-2/20 is 500ohms.......use either 470 or 560....

Dont connect the led direct to the 12v....it wont live very long.. :lol:

 

[AstrumPreliator]

Just something interesting, read this article.

 

[daviddoria]

thanks for the explanation. i get it now.

R=V/I

and using the idea that V is the same over all branches in a parallel setup

yay i am starting to understand!

 

[Gene]

I agree with Chippie's formula and the final solution but the example needs to be tweeked a bit.

the 2 volt, 20 ma LED in the 12 volt line:

(12 - 2) volts/0.02 Amps = 8/0.02 = 400 ohms for the dropping resistor

This is a minimum so you might consider a 470 ohm as a readily available solution.

 

[daviddoria]

ya i assumed it would have to be converted to amps

 

[Chippie]

I agree with Chippie's formula and the final solution but the example needs to be tweeked a bit.

the 2 volt, 20 ma LED in the 12 volt line:

(12 - 2) volts/0.02 Amps = 8/0.02 = 400 ohms for the dropping resistor

This is a minimum so you might consider a 470 ohm as a readily available solution.

Errr think your calculator aint working right........ 8)


12-2=8 ? more like 10.........roflmao!!!

 

[bogdanfirst]

from the formula you get that the current that goes trough the led is the thing that destroyes it. the current that passes trough it it as an affect of its resistance and of the voltage you aply to it.

 

[daviddoria]

new question

so i have 4 parallel branches with an LED and a fan hooked up to each one.

if the LED is dropping lets say 2v, and the fan lets say 5v, then the formula is (all powered on 12v)
R=(12-(2+5))/.02A

so the resistor eats up the rest of the voltage? so there is a total of 0? is that the idea? we are giving it 12, and we are using up 7, so we have to make the resistor eat 5?

 

[bogdanfirst]

it is a bit more complicated, beacuse the fan modifies its resistance when rotatin, it decreases with the turation, up to a limit.
if you consider it constant, in order that you have what you say means that when the fan is powered from 5 V it eats up 20mA, but i doubt it is so small.
thing of something in real life. try to find 2 lamps of the same voltage and different power or amps. if you connect them in series, you wil see that the one that has a smaller amps rating will make more light, because it has a larger resistance, and because they are passed by the same current then it will have a bigger voltage drop on it.
of course, lamps resistance modifies too, as they heat up. it can actually decrease 10 times from when it is not powered to when you turn it on.