resistors burning up my leds

I bought some L.E.D.s on fleebay with resiters. The green leds came with a resister dropping voltage to 3v. BUT I have some that are free flowing. I have hooked up three or four resisters now and I am getting readings of 11.75v on the other end.

Do they burn out? I thought when a resister goes nothing gets through. Are these cheap or am I missing something. I get a 465ohm reading with them.

I hooked one up and it was 3.15v removed my burned out led replaced it with another one, checked again and I am getting 11.74v again. Whats up with these??

Resistors resist current, not voltage.  Working with series resistors is completely different then working with a resistor in series with an LED, because LEDs are discreet components, not resistive.

If you have a single resistor connected to a battery you will read battery voltage on the other side of it, however, that voltage will be current limited.  If you are using a 470 ohm resistor hooked up to 12vdc you are limiting the current to the LED to around 25mA.  Depending on the LED, that may be too much current.  If you can find out what the max Forward Current is you can determine the proper resistor value for your application.  If in doubt wire two or three of the supplied resistors in series, that will limit the current even more.

so how do you limit voltage? the led is rated at 3-3.6v

it is being used as an idiot light on a motorcycle with 12-13.8v output range

You don't worry about the voltage, just limit the current.

Discreet components work a little bit different the resistive components.  If you limit the current the LED will take care of the rest and will work with no issues.

Emitted Colour : Green

Size: 5mm

Lens Color : Water Clear

Forward Voltage (V) : 3.0~3.6

Forward Current (mA):20

View Angle: About 25 degree.
Static Sense:Yes

Luminous Intensity: 15000mcd
Life Rating : 100,000 Hours

Free Resistors:
Supplied with free resistors for DC 12V

A 465ohm resistor is going to drop you to 25.8mA at 12 volts.  This is already 6mA too high.  Also remember that your standard automotive battery is actually 12.6 volts, and when running, produces over 13 volts.

I'd suggest a 510 ohm resistor (500 should do the trick) for a single LED.

Jeff

Thanks Jeff,

I think I am going to go with a 1k resister cuz these are super bright leds and I want to dull them down a bit. My bike system charges at 13.4v

KPierson wrote:

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If that is the case then I would suggest getting a pot and dialing in the brightness you want.  Then, measure the pot and get a resistor of that value.  I would guess you would be closer to the 4.7Kohm range for a super  bright LED to not be blinding, especially if its in your direct line of sight when you look at it.

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That sounds like a good idea? what is a pot?

Pot is short for potentiometer - they are adjustable resistors.

a 10K pot would be adjustable from 0ohm to 10,000 ohm.  You need to be careful when adjusting the pot though, because if you go below that 500 ohm mark you could damage the LED - to be "safe" you can hook the 470 ohm resistor in series with the pot - this will give you a 470 - 10,470 ohm range so you will never have to worry about putting too much current to the LED.

The pot will have three terminals on it - two outside terminals and a middle terminal.  You would only need two terminals, one of the outside terminals and the middle terminal.

So, basically to find the resistance you would hook your 470ohm resistor to battery power (while the bike is on and the battery is charging).  Wire the other side of the reisistor to one of the outside legs on the pot.  Wire the middle leg to the + terminal of the LED (the longer leg).  Wire the shorter leg to ground.  Now, adjust the knob/dial/etc on the pot to the desired brightness (don't look directly at the LED - super bright isn't just a clever name).  Once you have it dialed in you can remove the circuit from the bike and measure the resistance between the outside leg you used and the middle leg.  Add 470 to that value to accomidate for the 470 ohm resistor and that is the value of the resistor you need for that brightness.

If you want to get fancy (which seriously seems like overkill here) you could install a 5vdc regulator and run the LED off of it.  The regulator will keep the voltage at a rock solid 5vdc so the LED won't vary in brightness as the battery voltage goes up and down.  However, for an idiot light, that probably isn't needed at all!

I figured that since you did not know what a pot was, you might need a little help with the installation of said pot.

I drew it backwards, CCW will be full brightness.  CW will be the dimmest possible.

KPierson wrote:

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So, basically to find the resistance you would hook your 470ohm resistor to battery power (while the bike is on and the battery is charging).  Wire the other side of the reisistor to one of the outside legs on the pot.  Wire the middle leg to the + terminal of the LED (the longer leg).  Wire the shorter leg to ground. 

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Kevin, is the above statment the same as the diagram? you said to connect the resistor to bat power and  the outside terminal. then the long lead of the led to the middle termininal but that is I think going to power in the diagram and the anode to ground.

In my head red is positive/black is ground. Buddy at the shop told me the center terminal is a ground terminal.

The posted picture will work, too.

The pot has no polarity so there is no power or ground terminals.  As long as all the components are hooked up it should work.  The only thing you need to pay attention to is the polarity of the LED.

In fact, the resistors can be installed on the ground side of the LED and the positive side can be hooked stright to the battery and it will still work.  The resistance just needs to be somewhere between the battery and ground to limit the current flowing in the circuit.