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I'm working on a project that powers 20 LEDs with a 9 volt battery. What gauge of wire should I use?

A couple of LEDs are in series with a 180 ohm resistor. Then each pair are in parallel.

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    \$\begingroup\$ We are going to need more information. How much current does each LED need? What is the structure of the LED connections? are they all in parallel, or are some in series strings with strings in parallel? \$\endgroup\$ – gbulmer Aug 19 '14 at 18:31
  • \$\begingroup\$ What's your total current draw from the battery? It's one thing if you're you're multiplexing them, it's another thing if running them all in parallel, and still another thing if you're running them in a series-parallel configuration or pulsing them. \$\endgroup\$ – TDHofstetter Aug 19 '14 at 18:32
  • \$\begingroup\$ Updated question with LED schema. \$\endgroup\$ – DataZombies Aug 19 '14 at 18:39
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    \$\begingroup\$ Huh? What's a "schema"? \$\endgroup\$ – Olin Lathrop Aug 19 '14 at 18:44
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    \$\begingroup\$ Even assuming LEDs with a 3V forward drop, you're going to be drawing 2/3 amp out of the poor battery. It's not going to last very long. \$\endgroup\$ – DoxyLover Aug 19 '14 at 18:52
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Most of the top results of a search for "AWG wire current" reference the Handbook of Electronic Tables and Formulas, each giving the same numbers, presumably copied from the print edition. If you use a different piece of wire from the battery to each R-LED-LED string, and if R=180 ohms and LED drops 3V each, that leaves resistor current I = (9-3-3V)/180ohm = 33.3mA per string. Ten of these strings in parallel (for 20 LEDs total) gives 333mA total from the battery.

The table offers a spec for 30 AWG of 142mA "Maximum amps for power transmission" which "uses the 700 circular mils per amp rule, which is very very conservative." Meanwhile Wikipedia gives a fusing current for 30 AWG copper wire as 10A for 30 seconds. A 9V battery will not be able to source anything remotely like 10A, so melting the wire is a non-issue. You could probably get away with 30AWG if it's all you have, or want a thin wire bundle for some reason. Anything thicker would remove doubt about current carrying capacity and 26 AWG with table spec of 361mA max would already be above your total expected current.

Whether a 9V battery could actually deliver 330mA for a reasonable length of time is another matter. There is wide variation in chemistry, construction, and thus capacity available. Wikipedia says a lithium model has typical capacity of 1200mAh, giving you perhaps 3 hours runtime. Lithium 9V will run $5-10 each at retail, which may be more expensive than you had in mind. Alkaline will cost less, $3 each at retail, but also have less than half the capacity and also less current capability. An Energizer or Duracell alkaline might only give 1 hour runtime, and a cheapo model with low current capability might be lucky to light it up at all. If you want longer runtime, you'll need a physically larger battery which can store more energy.

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  • \$\begingroup\$ Short circuit current of a 9V alkaline battery might be 5A.. or more, which could get magnet wire hot enough to cause a nasty burn. I'd suggest AWG22 or a series poly fuse if the wire has to be inconspicuous. \$\endgroup\$ – Spehro Pefhany Aug 19 '14 at 22:57
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If this is one of this consumer 9 V batteries that are sortof rectangular with two round clips on one end, then any wire you have lying around is probably good enough. Anything up to 28 guage should work fine for something of the size that fits on your bench.

A better answer requires a better spec.

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  • \$\begingroup\$ Updated question with LED schema \$\endgroup\$ – DataZombies Aug 19 '14 at 18:40
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    \$\begingroup\$ You sure didn't \$\endgroup\$ – ACD Aug 19 '14 at 19:39
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    \$\begingroup\$ @ACD: I think you meant your comment for DataZombies, not me. If so, you should make that clear by including "@" followed by the user name. That will also notify him. As it is, he may have no idea you replied to his comment. \$\endgroup\$ – Olin Lathrop Aug 20 '14 at 1:22
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It depends on how you connect the LEDs. They can't all be connected in series, because that would exceed the 9V you have to work with. You could connect them in parallel with appropriate current-limiting resistors on all of them, in which case (assuming 20mA for each LED) you would have 400mA total. A combination of series-parallel would likely be best, but the highest current draw would be if they were all in parallel.

Even a very thin 30 AWG wire can handle more than 400mA, so just about any wire you have handy will be sufficient. I sometimes use 24-28 AWG wire, found inside Cat5e network cable.

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At 9V, you should use whatever wire is available to you, and easy to work with. Based on your asking of the question, I suspect any gauge that is too small for the current is a gauge that you will probably have a hard time effectively working with.

Personally, my prototyping spool is 24 gauge solid wire, which is commonly available at RadioShack. Why? It's easy to work with, and (way) more than large enough for virtually any typical battery type application. It's thick enough to stay bent in just about any position you want (for routing), and it's easy to solder.

As you get to and through the 30ish range, I find the wire much more difficult to work with, but maybe it's just what I am used to, or the types of projects I tend to do with it.

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  • \$\begingroup\$ Milliamps is not a unit of power. \$\endgroup\$ – Olin Lathrop Aug 19 '14 at 19:48
  • \$\begingroup\$ Updated to mAh. Either way, it's more than adequate for the purpose of this question, and my recommendation of 24ish is appropriate given the nature of the question. \$\endgroup\$ – starvingmind Aug 19 '14 at 19:54
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    \$\begingroup\$ Still wrong. Milliamp-hours is not a unit of power. If you can't even get basic units right, then perhaps you shouldn't be trying to answer technical questions. We do engineering here, where sloppiness with units is not acceptable. -1 again if I could. \$\endgroup\$ – Olin Lathrop Aug 19 '14 at 19:55

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