I am trying to make a transmission wire as light as possible and wondering how much I can overload it. I have a (300W) 12 amp load and a 22V DC source and need to transmit it about 1000 feet, which round trip I guess would be 2000 feet.

Normally with 12 amps you use 11 gauge transmission wire.

However, my understanding is that I can use as thin as 20 gauge wire without it melting and the only penalty I will pay is a 4 amp loss of electricity which is meaningless for my application.

Is this reasoning correct, or will my design fail for some reason?

• Your reasoning is somewhat correct, but your calculations seem to be off quite a bit. Your question is a bit ambiguous too. Is the distance to the load 1000 feet, meaning a 2000 feet roundtrip for the current? 1000 feet of 20 AWG wire (it is American wire gauge you are describing, right?) has a resistance of 10 Ohms, 2000 feet has 20 Ohms. Apply Ohm's Law and it doesn't look like you will be loosing only "4 amp". What is the resistance of the load? (It seems to be about 2 Ohms the way you are describing it here.) Can the load work properly at a reduced voltage? Oct 26, 2016 at 19:39
• I forgot I needed two wires, I will update the question. Oct 26, 2016 at 19:41
• What's your application? Oct 27, 2016 at 19:44
• It sounds like your 22 V source is about 1000 feet away from where it should be. Is there a reason for that? Oct 27, 2016 at 19:47
• @AndrewMorton No, I just did that because I wanted to create a really hard problem for myself. Oct 27, 2016 at 19:47

Your reasoning is partly correct, and partly makes no sense at all.

AWG 11 for 12 A is very conservative, considering 10 gauge is approved for 30 A in house wiring.

Look up the resistance per unit length of wire for the different wire sizes. The power dissipation over that same length will be that resistance times the square of the current. For example, if some wire has 100 mΩ resistance per foot, then with 12 A thru it the dissipation will be (12 A)2(100 mΩ) = 14.4 W, or 1.2 W/inch. That's going to get quite toasty and probably requires high temperature insulation and making sure the wire is kept away from flammable things, like the dry wooden supports of a house.

The part of your statement that makes no sense is saying that you will get "4 amp loss of electricity". That's wrong in so many ways. First, the current thru the wire is all the same. The one thing you don't "loose" is current. Second, "electricity" isn't a quantity or something specific. It's like saying today you'll have more weather, as opposed to snow, rain, etc.

What you will lose is a voltage along the end of the wire. The amount of voltage dropped by the wire is its resistance times the current thru it.

If you really need to use thin wire and still want to transfer substantial power, use a higher voltage. Send 100 V or something, then down-convert at the receiving end. The extra switching power supply may well be cheaper than a thicker cable.

The reason higher voltage helps is because less current is required to carry the same power. You say you want to transfer 300 W. That only takes 3 A at 100 V, but 15 A at 20 V, for example. Since the losses and requirement for wire size comes from the current, not the voltage, using high voltage and low current saves on wire and losses. Of course if you go too high in voltage you have to consider safety and possibly extra insulation. Physics can be annoying like that.

OK, I just looked up 20 gauge wire. It has a resistance of 33.3 mΩ/m, which is 10.2 mΩ/foot. At 12 A it will dissipate 1.5 W/foot, or 122 mW/inch. That will get noticeably warm, but should be fine for most purposes.

Now let's look at the losses. 1000 feet distance means 2000 feet of wire, since the current has to flow out and back. (2000 ft)(10.2 mΩ/ft) = 20.3 Ω. At 12 A that will drop 244 V. That's how much you have to apply to the two ends just to get 12 A thru the wire with a short at the other end. If you want 12 V available at the other end, then you have to apply 256 V at the sending end. It should be clear that this is horribly inefficient.

Use a thicker wire, a higher voltage, or both.

• What about using AC and then converting it to DC at the load? Oct 26, 2016 at 20:02
• @TylerDurden AC doesn't let you violate Ohm's Law. The resistance will still be present and probably slightly more than the DC equivalent, due to the skin effect. Olin is correct, thicker wire or higher voltage with a step-down converter at the other end is really your only good option. Oct 26, 2016 at 20:54
• @BrendanSimpson I thought the whole point of AC was to allow efficient transmission of electricity over a distance. Oct 26, 2016 at 20:55
• @TylerDurden Long distance transmission lines use many kilovolts at a relatively low current. In fact, HVDC is better for long distances because a solid core conductor can carry more current in a given size for DC than it can for AC (skin effect). However, you still need inverters and rectifiers at each end. Voltage drop is purely a function of wire resistance and current, not the source voltage. Oct 26, 2016 at 20:59
• @TylerDurden The purpose of AC is to allow efficient transmission of electricity in a time when transformers were the only or most cost effective means to raise and lower voltage. With the advent of high power electronics, AC in some applications is no longer the more cost effective means of transmitting electricity. Oct 26, 2016 at 21:05

If your load dissipates 300 watts with 12 amperes through it, it'll require:

$$E =\frac{P}{I} = \frac {300W}{12A} = 25 \text{ volts}$$

across it, and since your driving source can only supply 22 volts, your design will fail.

But it gets worse. ;) In the graphic below I've shown your setup using 11 AWG copper wire which has a resistance of 1.26 ohms per thousand feet, and 20 AWG, which has a resistance of 10.2 ohms per thousand feet.

$$R = \frac{E}{I} = \frac{25V}{12A} \approx 2.1\text{ ohms,}$$

and since resistances in series add, your source will be looking at (for AWG 11 wire):

$$Rt = R1+R2+R3 = 1.26\Omega+2.1\Omega+1.26\Omega = 4.62\text { ohms}$$

Then, since current in a series circuit is everywhere the same, and since your source is limited to a 22 volt output, the current it can send into the string, thence into the load, is:

$$I =\frac{E}{Rt} = \frac{22V}{4.62\Omega}\approx 4.8\text{ amperes}$$

$$E = IR = 4.8A\times 2.1\Omega \approx 10.1\text{ volts}$$

and it'll dissipate:

$$P = IE = 4.8A\times10.1V \approx\text{ 48.5 watts}$$

Using 20 gauge wire makes the situation much worse, as shown below.

So what's the solution???

Using a buck converter connected to the load at the load, and driving the buck converter with a fairly high voltage from a source 1000 feet away might work, as suggested by others.

Just for grins, I've depicted, below, a buck converter running at 90% efficiency with a 300 watt output, a 100 volt DC input, and a variety of wire resistances connecting it to the source.