LED question

vskid · Post by **vskid** » Wed May 24, 2006 4:02 pm

I was looking at the LED mod for pocket pc on hackaday and it said that too much current will kill LEDs. How would that happen if current is drawn by the load, not pushed? Its like drowning yourself when drinking through a straw.

timmeh87 · Post by **timmeh87** » Wed May 24, 2006 4:49 pm

things only take the current they need, but sometimes they "need" enough current to kill themselves many times over.

in the case of an LED, it is a diode and has almost zero resistance. without a resistor it would act almost like a short circuit.

oh, also, "power = current squared times resistance". so as the current gets higher the amount of heat produced increases exponentially.

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another extreme example would be connecting a 1ohm, 1/4watt resistor across a car battery. it will want to draw about 12A of current (V=IR). do the math and you will find that that equals 144W of power in the resistor.

so you see, electronics dont have brains, and will pull suicidal amounts of current.

vskid · Post by **vskid** » Wed May 24, 2006 5:08 pm

So how can you make it so that your LEDs won't commit suicide?

joevennix · Post by **joevennix** » Wed May 24, 2006 5:39 pm

One day I will understand power, voltage and current. But today is not that day.

daguuy · Post by **daguuy** » Wed May 24, 2006 6:58 pm

vskid wrote:So how can you make it so that your LEDs won't commit suicide?

put a resistor in series with it.

vskid · Post by **vskid** » Wed May 24, 2006 9:07 pm

daguuy wrote:
vskid wrote:So how can you make it so that your LEDs won't commit suicide?
put a resistor in series with it.

Just any resistor?

daguuy · Post by **daguuy** » Wed May 24, 2006 9:12 pm

it should be a certain resistance but that depends on a lot of stuff. i can't help youi with that part

vskid · Post by **vskid** » Wed May 24, 2006 9:13 pm

daguuy wrote:it should be a certain resistance but that depends on a lot of stuff. i can't help youi with that part

Is there a website that could help me?

timmeh87 · Post by **timmeh87** » Thu May 25, 2006 7:34 am

indeed there is, but isnt learning something for yourself more fun?

diodes and LEDs will drop a specific amount of voltage, no matter what you try to give them (called the 'forward voltage'). for a normal red LED thats about 2.2v. we also know we are going to have a resistor in series with it. so we can come up with the hypothetical circuit below:

Code: Select all

   ----/\/\/\----
   |            | 
   |            __
 -----          /\
  ---           |
   |            |
   --------------

where we have a battery, resistor and LED.

so, we know that the battery will provide a set amount of volts (lets say its 7.2 for now) and that he LED will drop a set amount of volts (2.2). one of Kirchoff's laws is that the sum of the voltages around a closed loop in a ciruit is <i> always 0 </i>. The battery is usually considered a positive source of votlage, and anything that drops voltage is negative.

so we need to find the voltage across the resistor. lets call it 'x'

7.2v - 2.2v - x = 0 so x = 5v

so the voltage drop across our resistor is going to be set at 5v no matter what.

The whole purpose of this exercise is to set the current though our LED. most normal red LEDs are rated for about 50mA. so now we can juse use V = IR to find out what resistance we need.

5v = 50mA * R

5/.05 = R

R = 100

so we need a 100 ohm resistor in this situation.

now, with that said, this little website will do that exact same calculation for you automatically.

http://led.linear1.org/1led.wiz

timmeh87 · Post by **timmeh87** » Thu May 25, 2006 9:23 am

I would assume that the current will increase until the remaining 5v is lost to resistances (the battery would probably have signifigant internal resistance). Im not actually sure. Ask your teacher and see what they say.

codeman · Post by **codeman** » Thu May 25, 2006 9:34 am

do you mean if you connected the LED directly across the 7.2v battery without a resistor? If you did that then the LED would actually have the 7.2v drop across it, but it would only do that for an instant. Though it is a good approximation...the voltage across a diode is not always constant, it changes slightly with changes in current...or at least that's what the diode V-I characteristic says. If you forced 7.2v across it, the current would be near infinite and it would break down the semiconductor P-N junction.

@tim: is the "LED" in that circuit backwards or am I looking at it wrong?

timmeh87 · Post by **timmeh87** » Thu May 25, 2006 9:49 am

its definitely backwards. i was so focused on how to draw an led with ascii that i didnt really pay attention to its direction

vskid · Post by **vskid** » Thu May 25, 2006 4:10 pm

timmeh87 wrote:its definitely backwards. i was so focused on how to draw an led with ascii that i didnt really pay attention to its direction

So does the resistor go between the LED and the + on the power, or between LED and -, or does it not matter?

daguuy · Post by **daguuy** » Thu May 25, 2006 5:10 pm

it doesn't matter

vskid · Post by **vskid** » Fri May 26, 2006 2:37 pm

daguuy wrote:it doesn't matter

I didn't think so, just making sure.