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I am wondering how I could put two USB battery packs in parallel together in order to increase their amperage/current.

I do not want to tamper with the Battery packs, as breaking them apart and soldering a new connector is out of the picture for the scope of the project, so I was wondering if I could do the following to accomplish the task:

Attach USB cables to each of the two exactly same battery packs, cut off the positive and negative ends of the USB exposed cable and solder positive to positive, negative to negative. I would then solder the input red/black wires of the item I want to power to the now negative-negative positive-positive ends of their respective USB wires.

Would this work? Can a quality USB cable be able to transfer 4.8 amps?

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  • \$\begingroup\$ Remember: Current is not generated. Current is drawn. So your answer is no, the current will stay the same in series. Change your wiring if you want to change the current or find a different resistance. \$\endgroup\$ – user103380 Jan 24 '18 at 22:41
  • \$\begingroup\$ @KingDuken I thought it was obvious he meant increasing the maximum current. \$\endgroup\$ – user253751 Jan 24 '18 at 22:44
  • \$\begingroup\$ Possible duplicate of Running 2 5V USB battery power packs, Can I use them in parallel \$\endgroup\$ – Omar Sumadi Jan 24 '18 at 23:07
  • \$\begingroup\$ Searching before posting is good practise on here... \$\endgroup\$ – Solar Mike Jan 24 '18 at 23:29
  • \$\begingroup\$ Possible duplicate of Can you combine the outputs of two ports on a USB Battery Pack? \$\endgroup\$ – winny Jan 26 '18 at 21:05
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No, you cannot just parallel two USB outputs and expect to double the current capability. The USB battery packs work using a DC-DC invertor to provide the 5V/2A output, and if you measure the 5 V produced you will find differences between the units (just from the variations in components used). Since one unit will have a higher voltage than the other, it will supply more current. How bad the mismatch will be is unknown.

Since both units will have a negative output voltage slope ....as load increases the 5 V will drop, it is possible that they could balance at a point. That does give you more current than a single unit could supply, but there is no way to know when any given USB pack may shutdown, so you are unlikely to be able to produce consistent results.

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  • \$\begingroup\$ Nevertheless, a USB cable wouldn't even be able to carry 4.8A of current, correct? \$\endgroup\$ – Omar Sumadi Jan 24 '18 at 23:09
  • \$\begingroup\$ @OmarSumadi Depends on the cable. I have some (cheap) cables with very thin wires and some with decently sized wires. But you're not talking about USB cables carrying that current anyway, because it sounds like you're making your own cable to carry it. \$\endgroup\$ – user253751 Jan 24 '18 at 23:15
  • \$\begingroup\$ No I am trying to use the USB cable to carry it \$\endgroup\$ – Omar Sumadi Jan 24 '18 at 23:18
  • \$\begingroup\$ @omarSumadi "I would then solder the input red/black wires of the item I want to power to the now negative-negative positive-positive ends of their respective USB wires." - so you're connecting your cable (carrying 4.8 amps) to two other cables each hopefully carrying 2.4 amps. \$\endgroup\$ – user253751 Jan 24 '18 at 23:20
  • \$\begingroup\$ I am connecting the cable to a Peltier element. \$\endgroup\$ – Omar Sumadi Jan 24 '18 at 23:32
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Would this work? Can a quality USB cable be able to transfer 4.8 amps?

Yes, you'll get a voltage drop and heat up the cable. USB wires are typically 28AWG that is 213mOhm/m

Consider a 5V source connected to a shorted 1m usb cable. The resistance will be 0.213*2Ω (you have to return through the ground, the resistance is double). With that 0.426 of cable resistance connected to a 5V source, you will get 11A of current (and ~50W of power dissipated in the cable, which will probably ignite it, or at least melt insulation). This is your best case current transfer scenario. But don't try it with 28AWG.

At 4.8Amps there would be 4.8^2/0.426=10W dissipated in the cable, which is still quite high.

Lets say you find a USB cable with 24AWG wire (I don't know if these exist) in it which has 84.22mOhms/m of resistance.

At 4.8Amps there would be 4.8^2/0.16=3.68W dissipated in the cable and at 5V

They do actually make 20AWG usb cables, which only would be 4.8^2/0.06=1.38W dissipated in the cable at 4.8A.

Another way to avoid cable losses is to up the voltage, but if someone were to accidentally plug in a device to a USB port with a higher non standard voltage, there would be smoke.

So choose a bigger cable, there will be less power loss and more for the peltier. You will also get voltage drops from the resistance in the cable, but you are probably more interested in power transfer. One problem with USB is you can't force people to use bigger cables, someone could accidentally plug in a smaller cable.

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  • \$\begingroup\$ Note that OP has stated that he will not join the two cables until they reach their end point, thus avoiding so much loss in the cables. \$\endgroup\$ – user105652 Jan 25 '18 at 1:06
  • \$\begingroup\$ Yeah, I blatantly Ignored that statement and just answered the question. \$\endgroup\$ – Voltage Spike♦ Jan 25 '18 at 4:25
  • \$\begingroup\$ Thanks so much for your efforts though guys! \$\endgroup\$ – Omar Sumadi Jan 26 '18 at 16:26
  • \$\begingroup\$ @OmarSumadi You can say thanks by upvoting and marking your questions as answered, thanks \$\endgroup\$ – Voltage Spike♦ Jan 27 '18 at 0:12
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This is a no, as I have read onward that a single usb wire will not be able to accomplish the task.

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  • \$\begingroup\$ It was assumed that you would not parallel the wires until they reached their point of use. Two USB cables can carry 4.8 amps, but that is about their limit. \$\endgroup\$ – user105652 Jan 24 '18 at 23:20
  • \$\begingroup\$ Yes, I would not parallel the two wires until they reach their point of use, which their point of use is a Peltier element. \$\endgroup\$ – Omar Sumadi Jan 24 '18 at 23:30
  • \$\begingroup\$ It depends on the size of the wire, if you make the wire big enough it has enough carrying capacity. AWG 20 usb cables should work for your application \$\endgroup\$ – Voltage Spike♦ Jan 26 '18 at 20:51

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