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I recently got a device that uses a USB cable (USB on one end and a power connector on the other end) as a charge cable. The device draws 16.5V, 3.65A Max. I want to extend the charge cable by using a USB extension cable. I tried using a 26AWG USB cable and it started to get warm so I unplugged it. My question is, will a 20AWG USB extension cable handle that power? Here's a link to the cable:

http://belkinbusiness.com/products/f3u134b06-usb-aa-extension-cable-mfdstp-6

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    \$\begingroup\$ You're getting 16.5V @ 3.65A from a USB cable now? What device is using that? I thought rapid charge USB only went to 12V. \$\endgroup\$ – Samuel Dec 2 '15 at 20:30
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    \$\begingroup\$ @Samuel non-standard use or the ports. \$\endgroup\$ – Passerby Dec 2 '15 at 20:31
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    \$\begingroup\$ Ah, there is PoweredUSB, up to 24V. \$\endgroup\$ – Samuel Dec 2 '15 at 20:32
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    \$\begingroup\$ @Samuel The existing cable uses a USB connector on one end and a power connector on the other end. It's not actually a USB cable it's just using a USB connector. \$\endgroup\$ – Dev01 Dec 2 '15 at 20:33
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    \$\begingroup\$ @TomKrones Re-purposing a very common household connectors, such as USB, for 16.5V is a bad idea. Some unwitting person may try to charge their iPhone with it. \$\endgroup\$ – Nick Alexeev Dec 2 '15 at 20:37
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2m (6-ish feet) of AWG20 should have a resistance of about 0.1 Ohm, assuming some inefficiency in cable make up and "wear and tear" over time. To and from would be 0.2 Ohm.

Which means that extension cable, in its wire, will create a voltage drop of up to:

V = I*R = 3.65A * 0.2 Ohm = 0.73V

That's probably acceptable on a 16.5V budget, as it's slightly less than the normal 5% minimum margin taken on external voltages. Most devices would/should work with a 10% margin on the external voltage.

The power dissipated in AWG20 would be:

P = I^2 * R = 3.65A*3.65A * 0.2 Ohm = 2.7W

Over a 2m cable that may just get a little warm, but should be fine, as long as the cable is made properly. That's where using USB extension cables for such manners of current may fall down a bit, because the main application 5V/2A would not account for 1/4th being lost in the cable. But, they'd have to be very old fashioned in their making of the cable for the slight warming to be seriously problematic.

One question does come up though: Why don't you just extend the power plug side with a proper coaxial power wire. Those come in current carrying capabilities up to 5A with limited voltage drop.

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    \$\begingroup\$ Good Math, but keep in mind that the cable loss should be 2X the length. i.e. the current goes there and back. Use 4M or 12 ft. \$\endgroup\$ – MadHatter Dec 3 '15 at 18:07
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    \$\begingroup\$ @madhatter And I quote: "2m of AWG20 should have a resistance of about 0.1 Ohm, .... To and from would be 0.2 Ohm." That makes 4m \$\endgroup\$ – Asmyldof Dec 3 '15 at 22:17
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It should be okay, up to some reasonable ambient temperature. Often the cable will give a temperature rating on it, if that temperature is 90°C then it will work to a fairly high ambient temperature, probably more than 50°C. This reference does not explain their assumptions (such as insulation rating) but they indicate up to 3.55A for 2-3 core. at 45°C. I think they are assuming a low insulation temperature rating.

The voltage drop will be maybe 0.5V for a 2m cable at 3.65A.

I agree with the others that this is a horrible idea- someone plugs a $1,000 smartphone into it and there may be many unkind sentiments expressed towards the designer.

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26 AWG has a resistance of 133.9 mohm/m. 20 AWG has 33.31 mohm/m.

3.65A through 1m of 26 AWG would give 1.77W of heat (I^2R). Through 20AWG 1/4 of that... so 0.44W or so. (For 6ft=2m, multiply those by 2.)

To calculate the temperature rise, you need to know more things about the cable and ambient conditions... How thick the insulation is and what is it made of etc. Also note that temperature rise is per unit of length; unless you coil the cable or something like that, it doesn't matter if it's 1m or 10m for the purpose of calculating its temperature.

However more total (IR) losses do mean less voltage getting to your device. I have no idea what that is, but since you said it still worked with the 26 AWG cable... it still will with the 20AWG one (which has fewer losses).

The Power-over-Ethernet people have some data for Cat5e (24AWG usually) and Cat6A (23AWG usually). The say to limit current to 720mA through Cat5e and to 845mA through Cat6A. At 720mA a Cat5e experiences 15C temp rise according to their data, so no surprises your 26AWG got hot (at 4x the current). Alas, they don't have any data for thicker wires as Ethernet hasn't gotten to those yet.

I found some thicker, older 10BASE5 wire spec'd as:

20 awg (10C Temperature Rise) 2.5 amps per conductor @ 25C ambient

So I think a 20AWG USB cable will still get warm (at 3.65A) based on that.

On the other hand a purpose-made 20AWG power cable is rated at 8.7A with a 30C temp rise. Which is probably acceptable.

Perhaps you can find some more similar cable to compare, but I don't usually shop for USB cables, so I don't really know what exact cable types are used for that. Insulation matters in this temp rise affair.

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  • \$\begingroup\$ Getting from power dissipation to temperature is actually a difficult problem in practice. See electronics.stackexchange.com/questions/22334/… for more, but no concrete answer there alas. \$\endgroup\$ – Fizz Dec 2 '15 at 21:04
  • \$\begingroup\$ It's actually possible to do that but it takes like two dozen parameters... it's probably far easier to try find this tabled somewhere. \$\endgroup\$ – Fizz Dec 2 '15 at 21:17
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If your widget draws 3.65 amperes with 16.5 volts across it, then it has a resistance of:

$$R = \frac{E}{I} = \frac{16.5V}{3.65A} \approx 4.5 \text{ ohms.} $$

AWG 20 copper wire has a resistance of 10.15 milliohms per foot, and with 3.65 amperes through it it'll dissipate

$$P = I^2R = 3.65 A^2 \times 10.15\cdot10^{-3}\text{ohms}\approx \text {135 milliwatts per foot.} $$

That's nothing to be worried about unless that heat has no way to escape, so your major concern should then shift to how much voltage will be dropped by the length of the cable connecting your widget to the power supply and how long that cable can be until it drops enough voltage to make your widget unhappy.

UPDATE:

Continuing in that same vein, if 20AWG wire exhibits a resistance of 10.15 milliohms per foot, with 3.65 amperes through it it'll drop

$$ E=IR = 3.65A \times 10.15\cdot10^{-3}\Omega \approx \text{ 37 millivolts per foot.}$$

Then, if your widget could tolerate a 5% drop in supply voltage without going belly-up, that's 825 millivolts, and since the 20 gauge wire drops 37 millivolts per foot, 825 millivolts divided by 37 millivolts per foot is about 22.3 feet.

That means that since there's a 20 gauge wire from the power supply to the widget, and another one from the widget back to the supply, the widget can't be located farther away than about 11 feet (3.4 meters) from the supply.

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It depends on the insulation and manufacturing materials. 20 AWG is used for 100 Amp fuse wire, but that is unsheathed.

Cable ratings are more around the insulation used than the physical amount of copper. The copper will melt through the insulation and short long before the copper fails.

A 70 Deg C cable is rated at 70 Deg C conductor temperature NOT external ambient, as once the conductor reaches that temperature you start melting through the insulation.

So the question really is 'What is the thermal propoerty of the insulation layers of this cable and how much current can I run before hitting conductor temperature of xx deg C?'. And the answer will rely on the materials and layup of the cable and insulation / jacket layers. also of the connector.

So in theory 20 AWG is good for prolonged use at well over 100 Amps (and if we look at wire fuse ratings tables could run 630 A for 5 seconds before conductor failure), but the insulation will fail long before that.

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