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 LED suppliers

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Re: LED suppliers
« Reply #15 on: Mar 05, 2015, 08:10 AM »
Vraeden -

Voltage is how fast the flow is, amperage is how wide the pipe is. Does that make sense? So if you have 3 volts in a battery supplying 1 3v LED it will last for twice as long as the same battery supplying two 3v leds in paralell, and 3x as long as three 3v leds in paralell. With me so far?

So the way I did my gauntlets was the other way. With a 9v battery I put three LEDs in series, so one after the other - meaning that their voltage adds up. 3v + 3v + 3v = 9v - so no resistor required, and in series my battery will last longer. Like, accidentally left on in my bag for a week longer.

You SHOULD always use a resistor. But if your values are close, with that low voltage and amperage, there's no real need.

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Vraeden


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Re: LED suppliers
« Reply #16 on: Mar 05, 2015, 09:47 PM »
Does anyone know of a resistance calculator for different color LEDs?

I'm thinking of putting together a circuit powered by 4 3.7 volt Trustfire batteries, So I'd have 14.8 volts to work with.

My red and yellow LEDs draw 2 volts, and my purple, green and clear LEDs draw 3 volts each. If I wire them in series, I can't get to 14.8 volts, so I need to put a resistor(s) on the circuit somewhere.

Is that possible?  Or not advised?

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Re: LED suppliers
« Reply #17 on: Mar 06, 2015, 05:32 AM »
If you sum up the required voltages for all the LEDs you are having in series in your circuit and subtract the result from your supply voltage of 14,8 V and devide the result by the required current (usually 20mA=0,02A) you get the value of your resistor.
For example you want one of each kind you have in your circuit, so you have:

2V (yellow) + 2V (red) + 3V (purple) + 3V (green) + 3V (clear) = 13V

14,8V - 13V = 1,8V

1,8V / 0,02A = 90 Ohm

That should work fine.
I hope this was what you wanted to know.

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Re: LED suppliers
« Reply #18 on: Mar 06, 2015, 08:20 AM »
If you sum up the required voltages for all the LEDs you are having in series in your circuit and subtract the result from your supply voltage of 14,8 V and devide the result by the required current (usually 20mA=0,02A) you get the value of your resistor.
For example you want one of each kind you have in your circuit, so you have:

2V (yellow) + 2V (red) + 3V (purple) + 3V (green) + 3V (clear) = 13V

14,8V - 13V = 1,8V

1,8V / 0,02A = 90 Ohm

That should work fine.
I hope this was what you wanted to know.

That's exactly what I wanted to know, thanks!

With a circuit wired in series, can the resistor go anywhere in the curcuit?

Also, if I use a resistor rated at a 200 ohms when I only need 90, what effect will that have?  Will it just make the LEDs a little dimmer?

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Re: LED suppliers
« Reply #19 on: Mar 06, 2015, 02:25 PM »
I am glad I could help, the resistor can be placed anywhere in the row. If you alter the resistor you can calculate the opposite way.

U=R*I

R is your resistor with a value of 200 Ohm
I is the current of the LEDs, presumably 20mA

U=2000,02 V=4V

So you will lose 4 Volt over the resistor and will have 10,8V left for the LEDs. This will indeed result in a loss in light intensity.

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Re: LED suppliers
« Reply #20 on: Mar 08, 2015, 12:22 AM »
So after taking the advice of everyone in this thread, I spent the last couple of nights burning my fingers with the soldering iron.

The end result is this:



There are three LEDs in the circuit.  It's driven by 4 AA batteries (6V) with 3 LEDs (2 red, 1 yellow) that draw 2V each.  I did not put a resistor in.  There is also a push-switch that turns the lights on and off.

The two red LEDs are in the barrels of the flamethrower/blaster and the yellow LED is on the inside of the gauntlet and is a visual indicator to me that the lights are on.  It seems I did something right since the LEDs didn't immediately blow up when I ran juice through the circuit.

Can you guess what in the picture is an indicator that I am old?

Now that I've wired up a couple of circuits, I'm just dangerous enough to short something out and make a battery explode while it's on my person.

Thanks to everyone for their help!

« Last Edit: Mar 08, 2015, 12:48 AM by Vraeden » Logged
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