# Discrepancy in AWG/current requirement tables?

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**1**Member

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Join Date: May 2014

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**Discrepancy in AWG/current requirement tables?**

Hi,

I am working on building a light fixture, and learning a bit about electricity along the way. I know a bit about electronics, so not totally new, have been learning a ton, but have come across a problem that probably amounts to looking at the wrong information, i hope someone can help.

The light fixture is actually made up of nine fixtures, all mounted to the same base. The light sources are all high power LEDs, each fixture has one 40watt LED. I am powering each with 12VDC. The LEDs are rated to 2.8amps total, but i'm only giving them 2.4 amps for each fixture.

So, the total current draw is going to be 2.8 times nine fixtures = 25.23 amps.

I have one of them working just fine. I haven't wired the rest yet.

Here is where i'm not sure what to reference.

The table at

http://www.powerstream.com/Wire_Size.htm

shows how many amps i can put through a given gauge of wire. It's got two columns, one for chassis wiring, one for power transmission, i was thinking the power transmission column would be the one to use.

But then i see at:

http://www.mjobee.com/projects&news/...t%20310.15.pdf

It indicates completely different numbers for a given gauge. How do i know which table to go by? I've been looking into that for a couple of days, one site seems to imply that i'm reading the right chart (1st link), another site seems to ignore those numbers completely and use something closer (but not exactly) what's in the second link. So far i haven't been able to find any info on why the discrepancy, but i'm sure i must be seeing something wrong. It seems like a fairly straightforward problem.

This is how i wanted to wire the power- each fixture will have two connectors for power, one for power in, one for power out. The power coming in is soldered to the power IN connector of the first fixture, and that connector is soldered directly to the fixture's power out connector, and also to the fixture's electronics to power the led. The second fixture is powered by hooking a power cable between power OUT from one fixture to power IN of the next.

So the other problem is, is it reasonable to even power these? To carry that many amps, from the table in the first link, i need 6 or 7 gauge wire to run from power supply to first fixture power IN, from power IN to power OUT, and so on for all fixtures? In series parallel circuits, i remember that it's the sum of each series leg of a parallel circuit that adds up to the total current, so each leg in this case will carry 2.4 amps, which puts the entire 25.24 amp load on the power leads?

But then, nine LEDs shouldn't take 25.24 amps should it?

Thanks for any help, i have several of these fixtures running with lower power LEDs that i built as a test, and everything works great, but coming up against the limits of how much power a wire can carry (not to mention a breakers 20amp limit) has me wondering.

I am working on building a light fixture, and learning a bit about electricity along the way. I know a bit about electronics, so not totally new, have been learning a ton, but have come across a problem that probably amounts to looking at the wrong information, i hope someone can help.

The light fixture is actually made up of nine fixtures, all mounted to the same base. The light sources are all high power LEDs, each fixture has one 40watt LED. I am powering each with 12VDC. The LEDs are rated to 2.8amps total, but i'm only giving them 2.4 amps for each fixture.

So, the total current draw is going to be 2.8 times nine fixtures = 25.23 amps.

I have one of them working just fine. I haven't wired the rest yet.

Here is where i'm not sure what to reference.

The table at

http://www.powerstream.com/Wire_Size.htm

shows how many amps i can put through a given gauge of wire. It's got two columns, one for chassis wiring, one for power transmission, i was thinking the power transmission column would be the one to use.

But then i see at:

http://www.mjobee.com/projects&news/...t%20310.15.pdf

It indicates completely different numbers for a given gauge. How do i know which table to go by? I've been looking into that for a couple of days, one site seems to imply that i'm reading the right chart (1st link), another site seems to ignore those numbers completely and use something closer (but not exactly) what's in the second link. So far i haven't been able to find any info on why the discrepancy, but i'm sure i must be seeing something wrong. It seems like a fairly straightforward problem.

This is how i wanted to wire the power- each fixture will have two connectors for power, one for power in, one for power out. The power coming in is soldered to the power IN connector of the first fixture, and that connector is soldered directly to the fixture's power out connector, and also to the fixture's electronics to power the led. The second fixture is powered by hooking a power cable between power OUT from one fixture to power IN of the next.

So the other problem is, is it reasonable to even power these? To carry that many amps, from the table in the first link, i need 6 or 7 gauge wire to run from power supply to first fixture power IN, from power IN to power OUT, and so on for all fixtures? In series parallel circuits, i remember that it's the sum of each series leg of a parallel circuit that adds up to the total current, so each leg in this case will carry 2.4 amps, which puts the entire 25.24 amp load on the power leads?

But then, nine LEDs shouldn't take 25.24 amps should it?

Thanks for any help, i have several of these fixtures running with lower power LEDs that i built as a test, and everything works great, but coming up against the limits of how much power a wire can carry (not to mention a breakers 20amp limit) has me wondering.

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**2**
This is from the National Electric Code and while it is intended for derating it also serves as a good rule of thumb for wire sizes. Houston Wire & Cable Company - NEC Table 310.16

However keep in mind that for 120 volt house wiring the NEC limits #14 to 15 amps and #12 to 20 amps. Of course you are using it for fixture wiring and it can be higher because of higher temperature rating of the insulation. Usually a bare wire in free air can handle much more current then rated for because the temperature of the wire increases with amperage and must not exceed the temperature rating of the insulation.

However keep in mind that for 120 volt house wiring the NEC limits #14 to 15 amps and #12 to 20 amps. Of course you are using it for fixture wiring and it can be higher because of higher temperature rating of the insulation. Usually a bare wire in free air can handle much more current then rated for because the temperature of the wire increases with amperage and must not exceed the temperature rating of the insulation.

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**3**
Welcome to the forums!

First off Light fixtures are required to be listed by an underwriter such as UL. It is not recommended to just "make" light fixtures to be installed in a home. For a hobby it is fine.

Quote:
Sure they can, but remember, this is at 12 volts DC. Current on the line side of the transformer/driver will be 1/10th or 2.52 amps. At 25 amps you are going to need one heck of a power supply.

Two Code sections you need to look at are Art. 402 Fixture Wires and Art. 410 Luminaires

Fixture wires have a whole new set of rules to follow. However, if each fixture is going to carry the next, you are going to need a pretty substantial wire to carry it all. Likely #10.

First off Light fixtures are required to be listed by an underwriter such as UL. It is not recommended to just "make" light fixtures to be installed in a home. For a hobby it is fine.

But then, nine LEDs shouldn't take 25.24 amps should it?

Two Code sections you need to look at are Art. 402 Fixture Wires and Art. 410 Luminaires

Fixture wires have a whole new set of rules to follow. However, if each fixture is going to carry the next, you are going to need a pretty substantial wire to carry it all. Likely #10.

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Hi Tolyn,

Since 12v is 1/10th of 120v, you take 1/10th of the current? So, i should be looking at gauges that can handle 9 fixtures at 2.5amps, not 25.24 amps, right? That sounds more in line with what i was hoping for, rather than require two separate breakers for a single light fixture.

In other words, the chart assumes you are using 120V?

That makes lots of sense, but still has me confused, according to the LED data sheets, the LEDs draw 2.4 amps each, and the data sheets assume they're run at 12vdc, since that's the max voltage they're rated to, how does (2.4amps * 9 fixtures) become 2.5 amps?

Also, it's a hobby, won't be installing these in the house, i just want to learn more about the LEDs.

I am using a computer power supply, 12 volt rail. This particular supply is i believe 500W. I think it's just average as far as computer power supplies go.. Thanks for the link, i'll take a look tonight for sure.

Ray, i'll look at that link tonight as well, thanks for sending. I didn't know that insulation can affect the power traveling through a conductor.

Thanks!

Since 12v is 1/10th of 120v, you take 1/10th of the current? So, i should be looking at gauges that can handle 9 fixtures at 2.5amps, not 25.24 amps, right? That sounds more in line with what i was hoping for, rather than require two separate breakers for a single light fixture.

In other words, the chart assumes you are using 120V?

That makes lots of sense, but still has me confused, according to the LED data sheets, the LEDs draw 2.4 amps each, and the data sheets assume they're run at 12vdc, since that's the max voltage they're rated to, how does (2.4amps * 9 fixtures) become 2.5 amps?

Also, it's a hobby, won't be installing these in the house, i just want to learn more about the LEDs.

I am using a computer power supply, 12 volt rail. This particular supply is i believe 500W. I think it's just average as far as computer power supplies go.. Thanks for the link, i'll take a look tonight for sure.

Ray, i'll look at that link tonight as well, thanks for sending. I didn't know that insulation can affect the power traveling through a conductor.

Thanks!

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Wire gauge versus amperes of current flow is primarily based on the ability of the wire to dissipate heat to the surrounding air and surrounding objects. For household wiring, for simplicity, just one rating is published for each wire size for both wiring in conduit and (Romex type) wiring in stud cavities. There are some modifications (deratings) for multiple conductors running juxtaposed.

There is a secondary calculation for voltage drop. Twelve volt circuits are more sensitive to this compared with 120 or 240 volt circuits because the voltage drop (or loss) in a given piece of wire for a given number of amps is a number of volts, not a percentage, and depends on amperes only and not supply voltage.

You perform both sets of calculations (independently of each other) and select the larger size if you come up with two wire sizes.

For the daisy chained light fixtures connected in parallel, the wire run from the power source to the first fixture will carry the heaviest load and may need to be larger than the run from the last fixture to the second to last fixture.

Quote:
No! For a given number of watts, if we start with fewer volts (potential; "pressure") then we need more amperes (current; "volume"). Remember, watts equals volts times amperes. Suppose we have ten 12 volt 2-1/2 amp lamps connected in series on a 120 volt circuit like Christmas tree lights. Each gets 12 volts and the same 2-1/2 amperes goes through each lamp in turn. If we connect them in parallel to a 12 volt supply, each still draws 2-1/2 amps and all ten together draw a total of 25 amps from the power supply. (If the power supply took 120 volt power instead of being a 12 volt battery, it would be drawing about 2-1/2 amps at 120 volts while delivering 25 amps at 12 volts.)

Given a "500 watt" computer power supply, if you don't draw any power from the 3-1/2 volt rail or the 5 volt rail that does not mean you can draw all 500 watts (41 amps) from the 12 volt rail. Check the label on the power supply to see the maximum power draw from each rail. Now you might be able to power some of the (12 volt) lights from the plus 12 volt rail and some from the minus 12 volt rail if you could not power all from the same rail.

There is a secondary calculation for voltage drop. Twelve volt circuits are more sensitive to this compared with 120 or 240 volt circuits because the voltage drop (or loss) in a given piece of wire for a given number of amps is a number of volts, not a percentage, and depends on amperes only and not supply voltage.

You perform both sets of calculations (independently of each other) and select the larger size if you come up with two wire sizes.

For the daisy chained light fixtures connected in parallel, the wire run from the power source to the first fixture will carry the heaviest load and may need to be larger than the run from the last fixture to the second to last fixture.

Since 12v is 1/10th of 120v, you take 1/10th of the current?

Given a "500 watt" computer power supply, if you don't draw any power from the 3-1/2 volt rail or the 5 volt rail that does not mean you can draw all 500 watts (41 amps) from the 12 volt rail. Check the label on the power supply to see the maximum power draw from each rail. Now you might be able to power some of the (12 volt) lights from the plus 12 volt rail and some from the minus 12 volt rail if you could not power all from the same rail.

*Last edited by AllanJ; 05-22-14 at 06:07 AM.*

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Thanks alan, i read that through a few times, that's making a lot of sense. I've done tons of ohms law calculations in the past, but obviously i need to sit with the math for awhile and spend more time understanding the relationships between P, I, E and R, instead of blindly calculating.

I guess i'm still not sure though what the proper way is to find gauge when one AWG chart says a wire needs X gauge for Y amps, and another says something completely different. I get it that at 12 volts, there's more amps available, lower voltage means higher current. At 2 amps, the gauge should be the same, whether it's 120v or 12v?

With the 10watt led system i made before (which works fantastic by the way, just not bright enough), i also wasn't able to find out how to calculate how many LEDs i could hook to a given power supply. I was calculating that since each led took 700mA, and each was hooked in parallel to a 500 watt 12 volt source. With ohms law. p = i * e, so power was at 8.5 watts per led. At that number, i could power 58 LEDs with the 500 watt supply. But someone who knows more suggested this was wrong, and with that power supply i should be able to power many more LEDs than 58.

Using the same thinking, each of the new 40watt LEDs is 2.4A, same power supply, so 2.4A * 12v = 28.8watts, and i should be able to run 500/28=17 LEDs. Less to stay away from absolute maximums.

Also, from what you said alan, i need thicker wires for the first fixture's power, thinner for the ones later in the parallel string. Each leg (or LED) is 2.4, so if i have 3 fixtures in parallel, the max load will be 2.4*3 = 7.2 amps on the first. But realistically, i'd use a gauge that can handle 7.2 amps (call it 10?). Current on the wires for the second, 4.8, the third, 2.4, right?

Still some cloud, but it's looking clearer.

Thanks again,

Rob

I guess i'm still not sure though what the proper way is to find gauge when one AWG chart says a wire needs X gauge for Y amps, and another says something completely different. I get it that at 12 volts, there's more amps available, lower voltage means higher current. At 2 amps, the gauge should be the same, whether it's 120v or 12v?

With the 10watt led system i made before (which works fantastic by the way, just not bright enough), i also wasn't able to find out how to calculate how many LEDs i could hook to a given power supply. I was calculating that since each led took 700mA, and each was hooked in parallel to a 500 watt 12 volt source. With ohms law. p = i * e, so power was at 8.5 watts per led. At that number, i could power 58 LEDs with the 500 watt supply. But someone who knows more suggested this was wrong, and with that power supply i should be able to power many more LEDs than 58.

Using the same thinking, each of the new 40watt LEDs is 2.4A, same power supply, so 2.4A * 12v = 28.8watts, and i should be able to run 500/28=17 LEDs. Less to stay away from absolute maximums.

Also, from what you said alan, i need thicker wires for the first fixture's power, thinner for the ones later in the parallel string. Each leg (or LED) is 2.4, so if i have 3 fixtures in parallel, the max load will be 2.4*3 = 7.2 amps on the first. But realistically, i'd use a gauge that can handle 7.2 amps (call it 10?). Current on the wires for the second, 4.8, the third, 2.4, right?

Still some cloud, but it's looking clearer.

Thanks again,

Rob

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**7**
Allen hit on most of the highlights, but the main point I was trying to make is you have to think of the circuits as separate circuits before, and after the power supply.

Before the power supply you have 120 volts. You mentioned having to use larger breakers or wire to feed the power supply, however on the primary side of the power supply, the current drops because the voltage goes up. (Current (amps) is inversely proportional to voltage) So for your 12 volt power supply, if you have 20 amps on the secondary (the 12 volt side) you will have only

Yes, amps is amps, and you need to base your wire size on the current that you are putting through it regardless of the voltage.

If I was building this, I would avoid running the fixtures in series, otherwise your "feeder" wire would need to be rather large as you are finding. Run one "feeder" from the power supply to each fixture, and then tap off of that "feeder" to each group of LEDs.

Based on the fixture wire table, #16 wire is good for 8 amps, and #14 is good for 17 amps.

Before the power supply you have 120 volts. You mentioned having to use larger breakers or wire to feed the power supply, however on the primary side of the power supply, the current drops because the voltage goes up. (Current (amps) is inversely proportional to voltage) So for your 12 volt power supply, if you have 20 amps on the secondary (the 12 volt side) you will have only

__about__2 amps on the 120 volt side.Yes, amps is amps, and you need to base your wire size on the current that you are putting through it regardless of the voltage.

If I was building this, I would avoid running the fixtures in series, otherwise your "feeder" wire would need to be rather large as you are finding. Run one "feeder" from the power supply to each fixture, and then tap off of that "feeder" to each group of LEDs.

Based on the fixture wire table, #16 wire is good for 8 amps, and #14 is good for 17 amps.

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If you can run two pairs of power wiring , you can solve some of your problems . Each pair will be carrying half as much current & will need to be roughly half the ampacity .

As has been said , all this depends on several things . But go by the general rule of thumb , # 14 -> 15 amps ; # 12 -> 20 amps ; # 10 -> 30 amps . Multiply these ampacities by .80 for continues duty .

But , at the end of the day , when you get it built and it has run for 3 hours , continually , check the temperature of the wires and the terminations . If none is hot , you should be OK . Warm is all right .

God bless

Wyr

As has been said , all this depends on several things . But go by the general rule of thumb , # 14 -> 15 amps ; # 12 -> 20 amps ; # 10 -> 30 amps . Multiply these ampacities by .80 for continues duty .

But , at the end of the day , when you get it built and it has run for 3 hours , continually , check the temperature of the wires and the terminations . If none is hot , you should be OK . Warm is all right .

God bless

Wyr

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Making a mountain out of a mole hill and possibly off topic.

Given nine 2.8 amp (e.g. garden light or counter light) fixtures the current requirement is 25.2 amps.

Trade school level homework assignment: Pretend that all nine fixtures are at the far end of the run. Given the resistances for common copper wire sizes (#12, #8. etc.) (you need to look it up using Google) and the round trip distance from the power source to the farthest fixture (you fill in this number) what wire size will result in no more than half a volt loss (about 4%) for 12 volts within the wire run out and back. Use Ohm's Law; the voltage drop in a piece of wire equals the current flowing at that moment times the resistance of that piece of wire.

College electrical engineering level homework assignment. With the nine fixtures in their actual or proposed locations, write down the distances: power source to first light, first light to second light, etc. Write down the amperes from power source to first light (25.2 amps), first light to second light (25.2 - 2.8 = 22.4 amps), second light to third light, etc. Given the resistances for common copper wire sizes (#10, #14, etc.) (you will have to look this up) figure out voltage drops for each segment in your circuit (source to first light, first light to second light, etc.) Select wire sizes for each segment such that the total voltage drop from power source to ninth light and back is no more than 0.5 volt.

After completing either of the above assignments, modify your answer(s) as follows: For each wire segment carrying more than 20 amps use a minimum size of 10 gauge. For each wire segment carrying more than 15 amps on up to 20 amps, use a minimum size of 12 gauge. Use a minimum size of 14 gauge for the remaining segment(s).

Given nine 2.8 amp (e.g. garden light or counter light) fixtures the current requirement is 25.2 amps.

Trade school level homework assignment: Pretend that all nine fixtures are at the far end of the run. Given the resistances for common copper wire sizes (#12, #8. etc.) (you need to look it up using Google) and the round trip distance from the power source to the farthest fixture (you fill in this number) what wire size will result in no more than half a volt loss (about 4%) for 12 volts within the wire run out and back. Use Ohm's Law; the voltage drop in a piece of wire equals the current flowing at that moment times the resistance of that piece of wire.

College electrical engineering level homework assignment. With the nine fixtures in their actual or proposed locations, write down the distances: power source to first light, first light to second light, etc. Write down the amperes from power source to first light (25.2 amps), first light to second light (25.2 - 2.8 = 22.4 amps), second light to third light, etc. Given the resistances for common copper wire sizes (#10, #14, etc.) (you will have to look this up) figure out voltage drops for each segment in your circuit (source to first light, first light to second light, etc.) Select wire sizes for each segment such that the total voltage drop from power source to ninth light and back is no more than 0.5 volt.

After completing either of the above assignments, modify your answer(s) as follows: For each wire segment carrying more than 20 amps use a minimum size of 10 gauge. For each wire segment carrying more than 15 amps on up to 20 amps, use a minimum size of 12 gauge. Use a minimum size of 14 gauge for the remaining segment(s).

*Last edited by AllanJ; 05-24-14 at 06:34 PM.*

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Are you running the lights in series? If so you need to increase your voltage so each light gets 12 volts. Nine times 12 volts = 108 volts to drive the circuit. Current is the same in a series circuit so it would draw 2.8 amps in each wire. But if one of your LEDs goes bad and somehow shorts it's circuit then your remaining LEDs will get more voltage and you will probably end up burning them all out.

A 500 watt computer power supply will not provide 500 watts on the +12 volt line. You may get 3 to 5 amps maximum on that +12 volt line. You need to find the max current your supply is rated for at the 12 volt output.

A 500 watt computer power supply will not provide 500 watts on the +12 volt line. You may get 3 to 5 amps maximum on that +12 volt line. You need to find the max current your supply is rated for at the 12 volt output.

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I am assuming the OP is connecting the lights in parallel on 12 volts DC. You could think of the circuit as a "ladder" where the sides are the positive and negative respectively and the rungs are the lights.

If the nine lights together draw 25.2 amps then the ladder sides from ground up to the first run support 25.2 amps, the first rung takes 2.8 amps, and the rest of the ladder takes 22.4 amps.

The sides of the ladder between the first and second rungs take 22.4 amps, the second rung takes 2.8 amps, and what is above takes 19.6 amps. Etc.

If the nine lights together draw 25.2 amps then the ladder sides from ground up to the first run support 25.2 amps, the first rung takes 2.8 amps, and the rest of the ladder takes 22.4 amps.

The sides of the ladder between the first and second rungs take 22.4 amps, the second rung takes 2.8 amps, and what is above takes 19.6 amps. Etc.

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Thanks Tolyn, that all makes sense. I think my problem before was that i was reading the same basic ideas, but it was fragmented, not so much applicable info particular to what i'm doing, lots of theory between useful info, which made it harder to put together.

WyrTwister, yeah, i have been testing them by monitoring continuous on time for several hours. The gauges you mention, this is where it's getting impossibly frustrating. I've posted the question about gauge inconsistincies to a few forums, and noone has an answer- the ratings you mention don't even come close to what this site says:

American Wire Gauge table and AWG Electrical Current Load Limits with skin depth frequencies and wire breaking strength

And what that site says is completely different from other sites i've seen. I'm sure i believe you, but when safety is at stake, i want to understand why there is so much variance in the numbers between the sites, or maybe i am looking at the wrong thing, either way it would be good to know for the sake of longevity (electrocution and fire prevention).

No, not running in series, i need to stay with 12VDC since i'm using a computer power supply, i think that's the best way i've found so far to keep the power source dependable and cheap.

So, with parallel power connection, each led takes what it needs, but the main legs from power supply need to be of a gauge that will support the sum of the current for all of the LED's combined right? i.e. 3 leds drawing 1 amp each (hypothetical), the power will be carrying 3 amps, but i should plan on using only 80% of the wires rating, right?

This is where my question about inconsistencies between sites regarding gauge and currents come from. If one site says 18 gauge can carry 2.3 amps and another says it'll take 9.5 amps, i'd rather use the 9.5 figure because i can get more LEDs on the same wire. But it might be a less attractive option if i know the risk could be, say, death for example.

Thanks for helping!

Rob

WyrTwister, yeah, i have been testing them by monitoring continuous on time for several hours. The gauges you mention, this is where it's getting impossibly frustrating. I've posted the question about gauge inconsistincies to a few forums, and noone has an answer- the ratings you mention don't even come close to what this site says:

American Wire Gauge table and AWG Electrical Current Load Limits with skin depth frequencies and wire breaking strength

And what that site says is completely different from other sites i've seen. I'm sure i believe you, but when safety is at stake, i want to understand why there is so much variance in the numbers between the sites, or maybe i am looking at the wrong thing, either way it would be good to know for the sake of longevity (electrocution and fire prevention).

No, not running in series, i need to stay with 12VDC since i'm using a computer power supply, i think that's the best way i've found so far to keep the power source dependable and cheap.

So, with parallel power connection, each led takes what it needs, but the main legs from power supply need to be of a gauge that will support the sum of the current for all of the LED's combined right? i.e. 3 leds drawing 1 amp each (hypothetical), the power will be carrying 3 amps, but i should plan on using only 80% of the wires rating, right?

This is where my question about inconsistencies between sites regarding gauge and currents come from. If one site says 18 gauge can carry 2.3 amps and another says it'll take 9.5 amps, i'd rather use the 9.5 figure because i can get more LEDs on the same wire. But it might be a less attractive option if i know the risk could be, say, death for example.

Thanks for helping!

Rob

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I don't understand where the figures of 9.5 amps for 12 gauge and 5.9 amps for 14 gauge come from.

For fire protection purposes, find a table where 14 gauge is rated at roughly 15 amps and 12 gauge is rated at roughly 20 amps and work from there.

For voltage drop purposes (should have less than 5% drop from source to farthest load) you may use the resistance ratings in the table you were looking at.

Ignore "skin effect" and "breaking strength" for this situation.

For fire protection purposes, find a table where 14 gauge is rated at roughly 15 amps and 12 gauge is rated at roughly 20 amps and work from there.

For voltage drop purposes (should have less than 5% drop from source to farthest load) you may use the resistance ratings in the table you were looking at.

Ignore "skin effect" and "breaking strength" for this situation.

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Thanks allan. This is where the numbers came from again. This site says 12 gauge works with 9.3 amps and 14 gauge works with 5.9 amps (max) (it has different numbers for chassis wiring, but that's irrelevant i think).

American Wire Gauge table and AWG Electrical Current Load Limits with skin depth frequencies and wire breaking strength

Wikipedia i didn't notice before, but it says 20 amps for 14 gauge and 25 amps for 12. It has higher amps ratings for higher temp rated insulation. Maybe this one is much closer, within 5 amps, but can't 5 amps kill you once in awhile if you don't take that difference into account?

American wire gauge - Wikipedia, the free encyclopedia

Another site with 9.3 and 5.9

Electrical Wires: American Wire Gauge (AWG) Sizes and Current Limits

There are other places with different numbers, i can't remember them now, i think i put them before. I'm pretty sure we couldn't have things like computers and cell phones these days if everyone is going by such wildly inconsistent information, so probably i am looking at something wrong, but I can't be the first guy in the world to notice the differences.

American Wire Gauge table and AWG Electrical Current Load Limits with skin depth frequencies and wire breaking strength

Wikipedia i didn't notice before, but it says 20 amps for 14 gauge and 25 amps for 12. It has higher amps ratings for higher temp rated insulation. Maybe this one is much closer, within 5 amps, but can't 5 amps kill you once in awhile if you don't take that difference into account?

American wire gauge - Wikipedia, the free encyclopedia

Another site with 9.3 and 5.9

Electrical Wires: American Wire Gauge (AWG) Sizes and Current Limits

There are other places with different numbers, i can't remember them now, i think i put them before. I'm pretty sure we couldn't have things like computers and cell phones these days if everyone is going by such wildly inconsistent information, so probably i am looking at something wrong, but I can't be the first guy in the world to notice the differences.

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**15**
There are differences in wire ampacity because there are many variables when it comes to determining how many amps a wire can safely handle. Even in the NEC book you will find different allowed ampacities for the vary same wire. This is the kind of expertise electricians get paid for.

For example: One of the most common wires out there in the trade is #12 THHN. If you look at the table 310.16 (2008) you will see that #12 THHN is rated for 30 amps. But hold on, there is an asterisk next to it pointing you 240.4(D) which tells you that the max overcurrent you may put on #12 is 20 amps. But wait! There is more exceptions in part 240.4 (E, F, and G) when you can put a higher overcurrent protection then the 20 amp limit for small conductors.

Now throw in other insulation types, conductor materials, and applications and things will get more confusing.

Quote:
As little as 4 milliamps (.004) can kill you. This is the level that a GFCI will trip on a normal 120 volt AC circuit.

All of this is a non-factor because you are using low voltage (12 VDC) 12 volts will not hurt you even if you grabbed both bare wires with your bare hands. Everything on the output side of the power supply is 100% safe. The only thing that can happen is the wire will get hot, and if left overloaded, could start a fire.

I suggest trying with one light and test how hot the wire that you choose will get. If it gets too hot that you can't hold on to it, increase your wire size, or change to a better wire insulation. Based on your original post, I would run from the power supply with #10 as your feeder to each "fixture" which will carry 25 amps, and then tap off the feeder with #18 to make your connections to the LEDs. That wire should only be carrying 2.8 amps. (Based on your OP)

For example: One of the most common wires out there in the trade is #12 THHN. If you look at the table 310.16 (2008) you will see that #12 THHN is rated for 30 amps. But hold on, there is an asterisk next to it pointing you 240.4(D) which tells you that the max overcurrent you may put on #12 is 20 amps. But wait! There is more exceptions in part 240.4 (E, F, and G) when you can put a higher overcurrent protection then the 20 amp limit for small conductors.

Now throw in other insulation types, conductor materials, and applications and things will get more confusing.

but can't 5 amps kill you once in awhile if you don't take that difference into account

All of this is a non-factor because you are using low voltage (12 VDC) 12 volts will not hurt you even if you grabbed both bare wires with your bare hands. Everything on the output side of the power supply is 100% safe. The only thing that can happen is the wire will get hot, and if left overloaded, could start a fire.

I suggest trying with one light and test how hot the wire that you choose will get. If it gets too hot that you can't hold on to it, increase your wire size, or change to a better wire insulation. Based on your original post, I would run from the power supply with #10 as your feeder to each "fixture" which will carry 25 amps, and then tap off the feeder with #18 to make your connections to the LEDs. That wire should only be carrying 2.8 amps. (Based on your OP)

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**16**Member

Join Date: Jan 2014

Location: USA

Posts: 699

The tables that speak of skin effect were for high frequency AC . If I recall correctly , skin effect may take place at high voltage ?

Neither much apply at 60 hertz or 480 VAC or below . Which is the range I work with .

Very seldom work with DC , except on automobiles .

But skin effect is not a factor with 12 VDC .

Stick with the NEC numbers you have been given & you should be fine .

God bless

Wyr

Neither much apply at 60 hertz or 480 VAC or below . Which is the range I work with .

Very seldom work with DC , except on automobiles .

But skin effect is not a factor with 12 VDC .

Stick with the NEC numbers you have been given & you should be fine .

God bless

Wyr

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**17**Member

Join Date: Aug 2012

Posts: 231

When working on electronics circuits with low voltage, wire gauge selection is selected with high AWG wire to reduce voltage drop. When building computer units, we often run 2 AWG or larger on 3.3 volt power supplys because we don't want to drop much more than 0.1 volts.

But your big problem is your computer power supply is not going to supply 25 amps on the +12 volt output

But your big problem is your computer power supply is not going to supply 25 amps on the +12 volt output

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