# LED parallel array wizard

Discussion in 'LEDs and Optoelectronics' started by bugzapper, Jun 23, 2014.

1. ### bugzapper

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Mar 21, 2014
I'm using this wizard to tell me what resistors to use (I am a complete amateur, btw).

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

My LEDs say the typical forward voltage is 3.6v and the max is 4.0v. Which do I input into the wizard. It says if I have 4v I use 1 ohm resistors and with 3.6v I use 68 ohm resistors. Seems like a big difference. Which should I use? I don't want the circuit to fail on me.
Thanks,
Marc

2. ### BobK

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Jan 5, 2010
It would help if you gave us all the relevant information. We need to know that voltage you are using to drive the LEDs and what forward current they will need.

Bob

3. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Yes that's a big difference and it could be a problem. It shows that there's not enough voltage across the resistor.

Imagine it like this. Your LEDs are books that you need to pack side by side into a wooden box. Each book is a slightly different thickness (different forward voltages) because of manufacturing process variations, and their thickness will also vary with temperature. So you use a piece of foam at the end, to take up the remaining space, so that the books will be held fairly tightly regardless of variations in their thickness. That foam is the series resistor, and the width of the box is your power supply voltage.

A resistance of 1 ohm corresponds to a very thin piece of foam. If the thicknesses of the books varies significantly, you'll get a big variation in the force that the foam creates (the current, in this analogy). But if you take out one or two books and use a fat piece of foam, variations in the books won't make much difference to the force. That's what you want.

Now there are two possible reasons for your problem:

(1) There is only one LED and its forward voltage is too close to the power supply voltage - in the analogy, you have only one book, and it's almost as fat as the box, so there's not much room for the foam. The only thing you can do here is to increase the power supply voltage (use a wider box) so you can use a fatter piece of foam.

(2) The calculation software has put too many LEDs in each series chain - in the analogy, there are too many books side by side and there's not enough room for a proper piece of foam. In this case, you need to remove a book - possibly more than one - so there's a wider gap for the foam. Then you can use a fatter piece of foam that will absorb the variations in book thickness (LED forward voltage) without causing large changes in current (tightness of the books inside the box).

This is NOT a perfect analogy. For one thing, using a fatter piece of foam will increase the tightness of the packing, whereas using a higher resistance will decrease the current. It's just to get across the basic idea.

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4. ### Harald KappModeratorModerator

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Nov 17, 2011
A very nice comparison, Kris!

5. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Thanks Harald

6. ### bugzapper

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Mar 21, 2014
Sorry, should have provided all the info. 12v battery, 20mA forward current on each LED, 6 LEDs total.

7. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
OK, so the calculator has used two chains, with three LEDs in each chain. You should use three chains, with two LEDs in each. This is like repacking the books into more boxes. Instead of having two boxes, each containing three books and a very thin piece of foam, you have three boxes, each containing two books and a fatter piece of foam.

This will require a higher resistance. Two LEDs in series will be typically 3.6V + 3.6V = 7.2V. That leaves 4.8V across the resistor. For 20 mA current, you can calculate R using Ohm's Law: R = V / I = 4.8 / 0.02 = 240 ohms.

8. ### BobK

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Jan 5, 2010
And at 4.0V, and 3 in a chain there is 12V - 12V or 0V across the resistor. The calculator should have said that configuration was invalid. Instead it chose 1 Ohm. (I verified the numbers in the OP, when asking for 3 LEDs in each series).

Bob

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9. ### bugzapper

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Mar 21, 2014
Thanks for the analogies and the assistance. This forum has always been great.
Marc