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I recently bought 5 12V ultra bright LEDs (all identical), and want to hook them up to a battery pack of 8 AA batteries. So I figured I could just make five parallel circuits with one LED in each, and hook them all up to the 12V DC. Since I'm quite new to hobby-electronics, I'd like to hear your input.

1) First of all, I've read in a related question that it's a good idea to always have a resistor in such circuits, even if the voltage matches the LED exactly. How many ohm should this resistor be?

2) It's ok to put the resistor "before/after" the the 5x parallel circuit, right? Or should I have 1/5 the resistance in five resistors, one in each parallel circuit?

3) Since the circuit now has a resistor, and it works by generating heat by consuming power, won't the battery be drained quicker with such a resistor in the circuit?

4) Finally, are all "regular" LEDs 20mA?

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  • \$\begingroup\$ You will have to give details on what these "12V LED's" actually are. Part #'s etc. the reason being is that there are no native 12V LED's so that tells me that these may already have current limiting resistors in circuit so that the LED's can be easy wired to a 12V source. \$\endgroup\$ – placeholder Oct 28 '12 at 20:28
  • \$\begingroup\$ Never assume all LED's are the same, Although 5mm parts are limited to 20mA continuous, SMD are not the same. If designed for 12V without R, then adding R will increase overall string resistance (ESR) and reduce overall current drawn. Can you specify LED type? eg. 5050 SMD or 5mm radial? If these are AC/DC string, then it has a series bridge on the back, otherwise it may have another R added. You can run these from PC 12Vdc. FLashlights are so cheap these days. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 29 '12 at 17:40
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If these are 12V LEDs then they most likely have some sort of regulation built in, so you don't have to worry about putting in a resistor. If you have the documentation for them, I would put it in your post.

2) If you wanted to put resistors within each parallel circuit, it would also need to have the same resistance as the one you would put at the beginning or end node, not 1/5. The advantage of doing this is that you could have smaller resistors rated for a lower power than one big resistor rated for a high power.

3) Short answer, no, the resistor is generating heat that would have been made at the LED if the resistor wasn't there.

4) Those don't sound like regular LEDs, but if you mean LEDs like the through hole 3mm or 5mm most of them are indeed designed for 20mA current draw at a characteristic forward voltage.

LEDs are non-linear semiconductor devices, their current draw is non-linear with the voltage across it. Look at this graph to see what I mean. http://en.wikipedia.org/wiki/File:Diode-IV-Curve.svg

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  • \$\begingroup\$ Thanks for the answer. I don't have any data about the LEDs except the 12V written on the package. They do look like "regular" LEDs with a small bulb and two pins. Prior to reading here I'd assumed I could simply wire up 8 AA batteries and everything would work according to plan, but now I have the impression there is more to it. I also found similar-looking LEDs on Ebay, and even though they were rated for 12V, the data sheet also said they'd work in the range 9-12V. I guess this supports your statement. Not sure about the voltage drop. \$\endgroup\$ – Pedery Oct 29 '12 at 9:29
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    \$\begingroup\$ ...so just create the parallel circuit with the LEDs, add the battery pack, and done, right? \$\endgroup\$ – Pedery Oct 29 '12 at 9:30
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Alright starting out yes you can do parallel set ups provided you know your power consumption will be pretty intensive. Resistors get hot from current but its actually quite the opposite of what you are thinking. Since V= IR and V is held constant when you add a higher resistance(R) in series it means current(I) will drop. This means your circuit is consuming less power.

The datasheet should tell you the drop across the led for the voltage. I don't think its 12V, normal LEDs are all around 2V give or take a couple hundred mV in either direction. Lets assume your LED has a 2V drop and your source is 12V this means the drop across your resistor is 10V and you want your current to be no greater than 20mA so plug these into the ohms law eqn 10V/(20*10^-3)A = 500 Ohms this means you want R to be at least 500 ohms for each parallel branch.

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if they are labelled 12 volt they will be the type with a built in resistor . so they are fine to put in parallel with no extra resistors needed. Its not unknown for standard 2v or 3.5 v LEDs to get into the wrong packet so I would sacrifice just one first on 12v to prove they are correctly labelled .

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