# LED's and constant current

Discussion in 'LEDs and Optoelectronics' started by user4572, Jul 5, 2016.

1. ### user4572

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Jul 5, 2016
I'm playing with an array of 6x5 LEDs. I wanted to see what the total current draw would be will all of the LEDs on.
When I first fired it up my meter read .9a and then as I left them on it slowly crept up to 1.5a

Do LED's draw more current as they are left on?

I thought it was supposed to be a constant draw...

I'm going to use them as a flasher unit so the wont be left on all the time but I want to make sure I get it right the first time..

Thanks for any input..

2. ### BobK

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Jan 5, 2010
Yes, they are supposed to have a constant current flow, but they do not do that themselves, you need a circuit that does that.

It sounds like you supplied the LEDs with a constant voltage. As they heat up they will draw more and more current at the same voltage. This is called thermal runaway and it is why you cannot simply apply a voltage source to an LED.

Read our resource on driving LEDs.

Bob

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3. ### user4572

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Jul 5, 2016

Sorry I should have been more specific. I have my board setup like #4, multiple strings in parallel.. My other creations have been 2 rows of 3 LEDs, each row with a resistor and I didn't have any issues. This one used more LED's and resistors so I wanted to make sure I didn't mess it up.
I even used the LED calculator that's in the link to make the circuit.

I fired it up again and noticed that a couple LEDs are not as bright as the others and 1 is flickering. I'm going to replace those tomorrow.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
I'm not sure what you mean by #4.

can you post the schematic?

5. ### Harald KappModeratorModerator

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Nov 17, 2011
True. And applies partly to current limiting by a resistor, too. As the voltage across the diode drops, the voltage across the resistor increases, driving more current through the circuit. Nevertheless, due to the non-linear characteristic of the LED thermal runaway is avoided because a small change in LED voltage will lead to a huge change in current and thus a huge a huge change in voltage across the resistor which in turn will counteract the change in voltage across the diode.

Each string should have its own current limiting resistor.

6. ### BobK

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Jan 5, 2010
If the LEDs are properly connected with resistors, I cannot see the current slowly rising from 0.9A to 1.5A. How about a schematic?

Also, are they heat sinked? High power LEDs need heat sinking, and the star board they might be on is not a heat sink, it is used to mount them to a heat sink.

Bob

7. ### user4572

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Jul 5, 2016
I've uploaded the schematic I used from the led wizard and a picture of the board. This is the same setup I've been using for a couple years now.
My mistake..its 6x4 not 6x5...
I've never needed a heat sink before with these LEDs. they are each .5 watt LEDs
Normally these are not left on. They flash so they do build up a little heat over time but never like what I've seen and always a constant current draw.

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8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Go ask the wizard for the resistor value for 5 LEDs in series. The 6.8 ohm resistor is too small to be effective in limiting the current. This is covered in the resource on driving LEDs.

9. ### BobK

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Jan 5, 2010
I hate those LED calculators. I have seen at least one that will recommend 0Ω for the resistor if the forward voltages add up to the power supply voltage.

And BTW 2.16V x 100mA is not 0.5W.

Bob

10. ### user4572

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Jul 5, 2016
I've been using that calculator for a couple years now without issue. This is also the biggest LED flasher I've built.
My last class in electronics was back in the early 90's and then I switched to computers so this is now a hobby for me. I guess I'll have to find my old books and verify everything before I do anymore big projects.

Thanks

11. ### BobK

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Jan 5, 2010
No, you have an an issue right now. It led you to believe that a 6.8Ω resistor dropping 0.6V was enough to control the current to your LED chains. It clearly was not.

The effectiveness of a resistor as a current limiter depends on what percentage of the source voltage it is dropping. Somewhere around 90% would make it very effective, 25% would make it suitable, 3%, which is what you design does, not not effective. The calculator should have told you this.

Bob

12. ### Herschel Peeler

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Feb 21, 2016
Re: "I fired it up again and noticed that a couple LEDs are not as bright as the others"
That sounds like you over stressed the LEDs.

13. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
I agree, "should". But find me one that does, or even better, one that refuses to offer this as a solution

14. ### BobK

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Jan 5, 2010
My point exactly.

Bob

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