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Old 09-09-2013, 02:18 AM   #16
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Originally Posted by jeffkrol View Post
Well a pleasent surprise today I found they DO sell even number "thermostat wire..


As to cat 5-6 I believe most bundle 2 wires and don't run individual strands for the power feeds but it is good to point that out..

quick question.. what would be the effective gauge of 2 Cat 6 pairs bundled into one?.. or does this not really apply.?.
I don't think you fully understand. If the Cat 6 cable is rated as 24 AWG, then the individual wires in the bundle are 24 AWG. Wire bundles(collections of other wires) are not typically rated by the pair or by the bundle. They are rated by the constituent wires.

When I was talking about 24 AWG in Cat 6, I was talking about using a single wire from a Cat 6 bundle. I wasn't talking about using the entire cat 6 cable. Also, there is no such thing as a "power feeds" in cat 6. All of the pairs are the exact same size. In fact, depending on the connecting standard, you change the wires around.
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Old 09-09-2013, 04:47 AM   #17
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Legot, why do you hate soldering guns? I use both guns and irons. Both have their place and use.

Why solder 2/0? Because it was required in the EW class back in the 70's. There were only 2 of us that got it right. And it was a soldered lineman's splice.
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Old 09-09-2013, 06:04 AM   #18
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Legot, why do you hate soldering guns? I use both guns and irons. Both have their place and use.

Why solder 2/0? Because it was required in the EW class back in the 70's. There were only 2 of us that got it right. And it was a soldered lineman's splice.
It's a personal preference thing, I can't see any circumstance where I would be better off using a gun over an iron. Most of the things I solder are really fine, and a gun wouldn't stand a chance. Anything heavier (8awg at the absolute most) I can just set the temp really high and use a wider tip. Pencil type or torch irons are the only thing I'll use until I find myself thinking "Wow, I could really use a soldering gun."

I respect your ability to do the lineman's.
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Old 09-09-2013, 02:03 PM   #19
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I don't think you fully understand. If the Cat 6 cable is rated as 24 AWG, then the individual wires in the bundle are 24 AWG. Wire bundles(collections of other wires) are not typically rated by the pair or by the bundle. They are rated by the constituent wires.

When I was talking about 24 AWG in Cat 6, I was talking about using a single wire from a Cat 6 bundle. I wasn't talking about using the entire cat 6 cable. Also, there is no such thing as a "power feeds" in cat 6. All of the pairs are the exact same size. In fact, depending on the connecting standard, you change the wires around.
guess I didn't ask well.. Cat 5/6 in LED lighting cases is used for the power feed to the LED strings... Sooo IF you solder a pair together and use it as one leg of the power does this, in effect change its power handling capability? Is such a thing handled in standards and ratings??
In another way to say it 2x 24Ga = x carrying capacity.
Quote:
Conductor Size:
The larger the circular mil area, the greater the current capacity.
You are doubling the copper
so 2x wires ganged together increase the circular mill area by 2?
http://www.seas.gwu.edu/~ecelabs/app...at/CCCofCC.pdf

I know it is complicated..
It appears "free air" capacity of 24Ga far exceeds (even derating by .7 (6-15 bundled conductors) leaving 4.2A before reaching "critical" temp) what we are usually dealing with..

BTW I have NO idea of the composition of the insulation of Cat 5-6 cable so I went w/ lowest common denominator nor power losses due to heating and resistance increases..

But the bottom line SEEMS to be 24Ga Cat5 for free air wiring of individual
LEDS is fine

Running Cat 5 as 4 pair of DC feeds is fine w/ caution..If you join a twisted pair to make a single "wire" is more than fine..BUT you only can do 2 strings..

bottom line for me (since I will usually do 3 or more individual strings) is to stick w/ "thermostat" wire for power (can get 3-4 strings w/ a thick copper wire 18-20ga per solid strand) and as to separate LEDs.. almost any wire works.. though practically starting at 24 ga makes sense..

Even looking at this simple composite chart.. (and assuming a "short" run 4ft ) 18G can handle 10A at 12V 22Ga 5A @ 12V

http://www.rowand.net/Shop/Tech/WireCapacityChart.htm

or this;
First number is wire gage, second is amp capacity when single wire near no other wires & third is amp capacity when in a wire bundle or conduit.

Wire Size Single Bundled
24 gage 3 2
22 gage 7 4
20 gage 11 6
18 gage 15 10
16 gage 22 12
http://www.vstrom.info/Smf/index.php?topic=2645.0;wap2

So, again it "appears" Cat5 can safely handle sub 2A through a single strand in the bundle.. leaving me w/ the OK w/ caution again for using a single "pair" for a LED power wire.. not too many LED strings are built for over 2A output though it CAN be done..

Last edited by jeffkrol; 09-09-2013 at 02:28 PM.. Reason: data
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Old 09-09-2013, 03:03 PM   #20
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It's a personal preference thing, I can't see any circumstance where I would be better off using a gun over an iron. Most of the things I solder are really fine, and a gun wouldn't stand a chance. Anything heavier (8awg at the absolute most) I can just set the temp really high and use a wider tip. Pencil type or torch irons are the only thing I'll use until I find myself thinking "Wow, I could really use a soldering gun."

I respect your ability to do the lineman's.
Yeah it is personal preference. I have both and although I use both the gun usually (OK almost always) wins out because I know exactly where I left it. Most of the time. And if I can't find it I know where 3 others are. I still haven't seen my irons in a while.

Splices were a requirement to pass the class. Back in 9th grade. A lifetime ago it feels. I still use a few of the splices I learned though.

Jeff, you do get more capacity. Your thinking is accurate. Not sure about how much increased capacity you get though.
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Old 09-09-2013, 03:06 PM   #21
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Current rating and heat aside, the other benefit to using heavier wire (wherever possible) is reduced voltage drop, especially if it's running a long distance (I have my wires routed up and into my conduit pipe hangers, then all the way down into the stand for instance)
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Old 09-09-2013, 03:09 PM   #22
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Current rating and heat aside, the other benefit to using heavier wire (wherever possible) is reduced voltage drop, especially if it's running a long distance (I have my wires routed up and into my conduit pipe hangers, then all the way down into the stand for instance)
Have you done the math and figured out what the drop would really be on such a short run?
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Old 09-09-2013, 03:17 PM   #23
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Have you done the math and figured out what the drop would really be on such a short run?
http://www.calculator.net/voltage-drop-calculator.html

My wire length from fixture to driver (in stand) is about 8 ft, which includes some extra wire for lowering the fixture, which gives, at 1A:

0.16V drop with 20 gauge
0.26V drop with 22 gauge
0.41V drop with 24 gauge

Edit: And this is for one direction. The actual voltage drop, accounting for the live and ground side, would be double this, plus whatever length of wire is in the fixture itself.
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Old 09-09-2013, 04:05 PM   #24
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guess I didn't ask well.. Cat 5/6 in LED lighting cases is used for the power feed to the LED strings... Sooo IF you solder a pair together and use it as one leg of the power does this, in effect change its power handling capability? Is such a thing handled in standards and ratings??
In another way to say it 2x 24Ga = x carrying capacity.
Alright. It is and it isn't. If you strip the wire completely out of the bundle, then it would double the capacity. If you left it wrapped up in plastic, you would be creating problems. Cat 5/6 is a communications standard and not a power transmission cable. There is a provision for power transmission in cables that meet the standard, but it is limited to 360mA per pair with only 2 pairs allowed.

Personally, I wouldn't recommend using the bundle for any real power transmission. I would use regular wire for this purpose. I was using ethernet cable as an example of 24 AWG wire to explain how small the wire could really be when you were running from LED to LED. Bigger is always better, but a lot of people are using ridiculously oversized wires which are unnecessary.
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Old 09-09-2013, 06:17 PM   #25
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http://www.calculator.net/voltage-drop-calculator.html


Edit: And this is for one direction. The actual voltage drop, accounting for the live and ground side, would be double this, plus whatever length of wire is in the fixture itself.
cool calculator but..err.. don't think it works that way (but it might).. the doubling that is..
But you do have to include internal wiring run.

20AWG seems pretty ideal..
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Old 09-09-2013, 07:59 PM   #26
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Originally Posted by Chronados View Post
http://www.calculator.net/voltage-drop-calculator.html

My wire length from fixture to driver (in stand) is about 8 ft, which includes some extra wire for lowering the fixture, which gives, at 1A:

0.16V drop with 20 gauge
0.26V drop with 22 gauge
0.41V drop with 24 gauge

Edit: And this is for one direction. The actual voltage drop, accounting for the live and ground side, would be double this, plus whatever length of wire is in the fixture itself.
Keying in 16 ft. I got the .41 Vdrop.

Hell you'd get more variation just in normal power fluctuation.
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Old 09-09-2013, 08:41 PM   #27
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cool calculator but..err.. don't think it works that way (but it might).. the doubling that is..
But you do have to include internal wiring run.

20AWG seems pretty ideal..
It does work that way. Current is constant. Resistance is a fixed amount per foot for a given gauge as referenced in the chart below the calculator. Therefore voltage drop goes up by a linear amount because Vdrop = I*R

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Keying in 16 ft. I got the .41 Vdrop.

Hell you'd get more variation just in normal power fluctuation.
???



I'm not saying it's a massive loss, but it's not like 20 gauge costs much more than 22 or 24 gauge wire.
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Old 09-09-2013, 08:46 PM   #28
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It does work that way. Current is constant. Resistance is a fixed amount per foot for a given gauge as referenced in the chart below the calculator. Therefore voltage drop goes up by a linear amount because Vdrop = I*R



???


I'm not saying it's a massive loss, but it's not like 20 gauge costs much more than 22 or 24 gauge wire.
Since it asks the conductors it may already take into account the return. I mean you have to get power out. And back. Why would it only account for a single conductor out?
I think it's 8 feet and includes both the hot and ground.

Voltage and AC or DC seems to be irrelevant in this calculator.
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Old 09-09-2013, 09:04 PM   #29
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Since it asks the conductors it may already take into account the return. I mean you have to get power out. And back. Why would it only account for a single conductor out?
I think it's 8 feet and includes both the hot and ground.

Voltage and AC or DC seems to be irrelevant in this calculator.
I think you are right. It does say "set" of connectors, and when I do the math by hand, I get 0.41V drop for 16 ft of 24 gauge. My mistake.

But yes, the base voltage doesn't matter, because the Vdrop isn't a function of it. I think they only included it in the calculator so you could get the % drop.

In any case, aside from aesthetic reasons, I can't really see a reason for not using heavier gauge wire wherever possible.
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Old 09-10-2013, 11:22 PM   #30
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That calculator is making some very simple assumptions. It is only calculating the resistance of the wire and then using that for all other assumptions.
In power, the real calculations for "transmission line" are more complex. This is just a handy rule of thumb.

The resistance of a piece of wire is really simple to calculate. All materials have a resistivity constant. The resistance of a piece of wire is = (resistance constant)*[(cross-sectional area)/length].
Does that sound complicated? It isn't. It is simply saying that the fatter the wire, the less resistance it has and the longer the wire the more resistance it has.

That is all that the calculator is telling you.
One serious concern is that the resistance of a material increases as the temperature increases. If you are running 3.5A across a 24AWG wire, it will get hotter and thus increase the resistance. This will mean you now have more current(3.6A). This will make it even hotter. See where this is going?
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