Choosing the Right Gauge: How to Efficiently Power Your Solar Devices

[eddie90]

Hello everyone, I have a question about wire gauges and choosing the right one.

Anyway this is related to my job. Long story short, we have systems that are powered by solar batteries(which are recharged through solar panels). Anyway these batteries supply 12V DC to various different devices, a couple of these devices are some strong LED illuminating lights. Now these LEDs come with about 2ft of wire to power them up(a red+ and a black- wire), and these two wires are 16gauge.

Now due to the position of these lights(they are not near the batteries) we need to add about 25 more feet of wires to supply them the 12V. My question here is: In order to be as efficient as possible - since we are running on solar energy and we want to use our power as wisely or efficient as possible - what gauge should we make those 25feet of additional cable? We are thinking of something like 14 gauge.

Im just a little confused because I know that thicker cable has less resistance and so then it should be more efficient to put the 14 gauge. But then again, those 2 feet of 16 gauge wires that come out of the LED will still be there, meaning we would have 25 feet of 14 gauge and then the 2 feet of 16 gauge. So the phrase "You are only as strong as your weakest link" comes to my mind. Is it pointless to have all that 14 gauge wire if at the end(the last 2 feet) we are going to have 16 gauge?

Thanks in advance

 

[es1]

Depends on the power that needs to be delivered to the LEDs.

Also, do you happen to know how the LED current is limited, if at all? If it is done with a LDO type circuit then it will not matter (as long as the wire can safely deliver the power).

If it is done with a series resistor then I would say as long as the cable resistance is a small fraction (<5%) of that (equivalent) resistor you are good.

 

[cmb]

You are choosing 8mΩ/m, or 13mΩ/m?

So #14 over 10m will save you 50mΩ compared with #16.

The resistance in a poor interconnection (like a screw terminal) will likely be more than that.

However, #16 is, let's say, 10A continuous max to be conservative. At a full 10A rating, you'd be saving 5W power. But if we're talking 1A, your saving is 50mW.

 

[eddie90]

Depends on the power that needs to be delivered to the LEDs.

Hi es1 thank you for your reply, by this, do you mean the current? If so, it is about 1.2 amps.
Also I have no idea about how the current is being limited, their datasheet doesn't provide this info but I contacted one of the manufacturer's technician, theyll get back to me tomorrow morning with that info. Ill post what they tell me.

 

[eddie90]

You are choosing 8mΩ/m, or 13mΩ/m?

So #14 over 10m will save you 50mΩ compared with #16.

The resistance in a poor interconnection (like a screw terminal) will likely be more than that.

However, #16 is, let's say, 10A continuous max to be conservative. At a full 10A rating, you'd be saving 5W power. But if we're talking 1A, your saving is 50mW.

Hi, I'm not an electrical engineer(yet) so I was a little lost at first about the 8mΩ/m, or 13mΩ/m but a quick google search led me to a table with a wire resistance table.

Ok so I get that part of your post, I did some calculations and in my case, using #14 would save me about 40mΩ compared to #16.

Im a little lost on the last part of your post. So you are saying that #16 would actually save me about 50mW? I say 50mW because the load of these lights is about 1.2 to 1.5 amps.

If so, how did you come up with this? I think that the formula to figure out power loss is P=V*I right? In my case V would be 12VDC and I would be 1.5amps so power loss would be 18W?
Im not sure if this is right though, I'm not an engineer but I try to read on this stuff.

Thank a lot

 

[cmb]

P=I^2.R

At 1.5A, using #16 you are going to lose [1.5^2].[0.13] = 0.3W
At 1.5A, using #14, you are going to lose [1.5^2].[0.8] = 0.18W

assuming 10 m of copper cored wire.

From a total of 18W, that'd be a ~0.67% saving

 

[mheslep]

You'll likely have an easier effort mechanically if you use a couple loops of the lighter gauge wire rather than one of the heavier.

 

[eddie90]

P=I^2.R

At 1.5A, using #16 you are going to lose [1.5^2].[0.13] = 0.3W
At 1.5A, using #14, you are going to lose [1.5^2].[0.8] = 0.18W

assuming 10 m of copper cored wire.

From a total of 18W, that'd be a ~0.67% saving

Aah I see, ok so I did the calculations using my lengths and in my case I would be loosing 0.22W using #16 and 0.14W using #14

Sorry but how did you get that percentage? And does it mean I would be saving 0.67% when using #14 rather than #16? Assuming 10m of copper wire of course.

Thank you very much for your help btw

 

[eddie90]

You'll likely have an easier effort mechanically if you use a couple loops of the lighter gauge wire rather than one of the heavier.

Sorry I'm not exactly sure what you mean by this? I am not sure what you mean by loops?
The reason why this is important to us is because there are tight spaces were we have to fit all our cables through and it would be nice if we could use a 16 gauge rather than a 14. But like I mentioned if it will compromise out power efficiency we will probably stick with the 14.

Thank you for the reply

 

[mheslep]

Sorry I'm not exactly sure what you mean by this? I am not sure what you mean by loops?
The reason why this is important to us is because there are tight spaces were we have to fit all our cables through and it would be nice if we could use a 16 gauge rather than a 14. But like I mentioned if it will compromise out power efficiency we will probably stick with the 14.

Thank you for the reply

Just that 16 gauge will be easier to manipulate and install, especially in tight space. To obtain the an ever lower resistance than the 14 gauge install two runs of 16 gauge, wire together in parallel.

 

[cmb]

Sorry but how did you get that percentage?

Power consumed = 12V x 1.5A = 18W

Power saved with #14 wire over #16 wire = 0.3W-0.18W = 0.12W

Efficiency saved with #14 wire over #16 wire = 0.12W/18W

 

[eddie90]

Power consumed = 12V x 1.5A = 18W

Power saved with #14 wire over #16 wire = 0.3W-0.18W = 0.12W

Efficiency saved with #14 wire over #16 wire = 0.12W/18W

Got it, thank you so much.

I did the math with our length of cable and I got .47% more efficiency.
Hmm this seems very very minimal to me.

We are using 4 6V batteries each rated at(according to the datasheet):
Capacity at C/100 - 220Ah
Capacity at C/20 - 190Ah

We have 2 pairs hooked up in parallel and then those two pairs are hooked up in series. Which doubles the voltage to 12V and also the amps per hour to 440Ah I believe. I'm not really 100% sure what's the difference between C/100 and C/20 but from what I read I believe our case is more like C/100 because the load of our entire systems is about 4 to 5 amps.

Anyway, is .47% almost insignificant in your opinion?
I might also consider mheslep's suggestion of using two sets of 16 gauge wires

 

[eddie90]

Just that 16 gauge will be easier to manipulate and install, especially in tight space. To obtain the an ever lower resistance than the 14 gauge install two runs of 16 gauge, wire together in parallel.

Hmm I see, would that have the equivalent resistance of an 8 gauge cable??

 

[cmb]

Hmm I see, would that have the equivalent resistance of an 8 gauge cable??

If only these types of numbers worked like that!

I think the AWG works by the number of times it used to take to roll out the wire through some wire rolling machine. So #16 went through the roller twice more than #14. You can see from the resistances that two #16 is about the same, resistance-wise, as the #14. The increased surface area will actually give two #14 a better max Ampere rating that the equivalent single cable with the same total cross-section.

 

[mheslep]

Hmm I see, would that have the equivalent resistance of an 8 gauge cable??

Two 16 AWG wires (~13 ohm / km) in parallel (so half, or 6.5 ohm / km) will have about the resistance a 13 AWG wire per km.

 

[eddie90]

If only these types of numbers worked like that!

I think the AWG works by the number of times it used to take to roll out the wire through some wire rolling machine. So #16 went through the roller twice more than #14. You can see from the resistances that two #16 is about the same, resistance-wise, as the #14. The increased surface area will actually give two #14 a better max Ampere rating that the equivalent single cable with the same total cross-section.

AAh lol sorry, got it

So, in your opinion, is .47% too little of a difference to have a noticeable effect on efficiency?

 

[eddie90]

Two 16 AWG wires (~13 ohm / km) in parallel (so half, or 6.5 ohm / km) will have about the resistance a 13 AWG wire per km.

OOh ok that makes sense, I shouldve looked back at that resistance table.
Hmm I was looking at the different AWG diameters though and when it comes to manipulate and install like you mention earlier, wouldn't two 16 AWG give us more trouble than one single#14? since two #16s would be thicker than one #14?

 

[jim hardy]

wire gages are almost logarithmic.

Rule of thumb:
every three gages roughly halves(or doubles) the resistance, and number ten is one ohm per thousand feet.
from those two data points you can estimate a lot of stuff without having to find the tables.

what i think es1 was getting at with his question is this:
Since LED's operate on less than 12 volts typically 2 to 4, the current to them has to be controlled.
there are two ways to control current to LED's:
a. resistors, which waste the extra energy as heat. That's brute force.
b. switching circuitry that lowers the average applied voltage via inductors. it's more complex but wastes less energy. That's finesse.


since you aren't sure which you have
but you're only pulling 1.8 amps
and only going 25 feet..

using rule of thumb above:
#16 would be four ohms per thousand feet (six wire gages smaller would be two doublings)
and you have 50 feet (25 feet each way in round trip)
that's 50/1000 = 0.05 ohms
your voltage drop on #16 would be V = I * R = 1.8 * 0.05 = 0.09 volts
so if you have 12.0 volts at panel you have 11.91 at end of your #16
which is not quite 1% loss...
#14 would be slightly less loss

If your LED's have 'brute force' current limiting, it's sort of a never-mind because the extra energy is being thrown away anyhow.
If they have 'finesse' limiting then you can halve the loss in your wire by increasing wire size three gages. BUt in the vicinity of #14-#16 you're only playing with 1% loss so don't spend a lot of money on that wire..
myself i'd use construction "Romex" it's cheap and easy to work. You may find a use for the third conductor, an intercom or something.

judging by size of your panels and batteries this LED is not the only thing you are supplying?