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Please help me to run 4 LED's in series from 12v.

Hey guys...

Can anyone please help me to sort my problem out.
I am trying to run 4 3v superbright LED's in series from a 12v source. (4.3 Amp).

These are the Led's...

I have used this led circuit calculator : http://led.linear1.org/led.wiz
See Attached file for the values i put in..

It suggests putting 4 LEDS in series followed by a 1 watt 1 ohm resistor.

I have been to the shop and bought a 1w 1ohm resistor as this suggested, but it's not doing the job. The LEDS are coming on for about 30 seconds before then starting to flash randomly. I suspect they are getting to hot.

When i measure the voltage across one of the LED's I am getting 3.3v. Not the 3v that the calc suggests.

I have tried many different other higher resistors, but then the LED's are to faint.

What am I doing wrong ?
 

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just tried 6 of those 1Ohm resistors in series and that got me down to 3.1v (measured across one LED).
The leds stayed on much longer but then started flashing. So it looks like that calc is a load of crap. I will order some 8ohm resistors.
 
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If you have 3.3V across each of 4 LEDs that is not 12V. What does your power supply read? Is in an auto? When running they are typically over 14V.

Anyway, 4 3V LEDs is too many to run off 12V with a resistor. The voltage across the resistor would be 0.
Try again with 2 strings of 2 LEDs. You will get a larger resistor and it will work correctly.

Bob
 
Hi Martin from a Martin.
Those calculators are great 'sometimes' .
It isn.t always about the input voltoage.
It also depends on vf (forward voltage drop)
Martin
 
Hi Bob. Thanks for the reply.
I am getting a reading of 12.2 across my power supply. It's a (BEC) on an RC quad.

The calculator thingy must be a load of crap as it's saying if I do what you suggest, then
it's saying "the power dissipated in your resistors is a concern".

It's then suggesting 10 Ohm resistor.

Still confused.
 
Here is how the calculation works:

R = (Vs - N * Vf)) / I

Where Vs is the supply voltage and Vf is the forward voltage, N is the number of LEDs in series and I is the current.

With Vs = 12, N = 4, and Vf = 3, you get 0 for the resistor.

If, indeed, the Vf was exactly 3, it should work without a resistor, but that is not the way real LEDs work. The Vf is in some range, 2.5 to 3 was given in the link you posted. But I don't believe it. White LEDs typically have Vf of more like 3.3 to 3.6V. I have never seen one that has a Vf as low as 2.5V.

And, your link did not give a current. Where did you get 700mA?

For LEDs of this power, a resistor supply is not practical, you need a constant current driver.

Bob
 
The LEDs take about 1 amp.
The characteristic voltage across the LED is about 3.3v to 3.5v.
You can only put 3 LEDs in series if you want the greatest efficiency.
3 LEDs in series will require three x 2watt 10 ohm resistors in parallel
2 LEDs in series will require two x 3watt 10 ohm resistors in parallel
You cannot allow the supply voltage to rise above 12.2v
 
The problem is the "spec" for the LED. While all LED's have a forward voltage drop, it is not a reliable way to regulate the power in the device. High power LEDs have an optimum operating current. When you drive them with a constant current source at the appropriate value, you will see the correct forward voltage. Search for LED power supply or LED current source to see lotsa examples of both linear and switching topographies.

ak
 
Thanks guys.
I realised in my first test that the problem was I had mixed Red LEDS and White LEDS.

I have now found that 4 white LED's run fine using a 3.3Ohm Resistor, (The voltage across each led is 2.9v).
The 4 red leds seem to run better on a lower voltage. So I have used a 10ohm resistor and they seem to run consistently at around 2.6v.

Anyone got any idea what current this lot might be drawing. I am using a 3A UBEC, so don't want to over cook it.
 
The 4 white LEDs would probably also run just fine with no resistor, because you are not supplying them with enough voltage. If you want the LEDs to be operating at full power, you need to control the current as several people here have stated.

Bob
 
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