0
\$\begingroup\$

Is there any problem with joining up the two USB outputs of a single 5V phone charger to increase/amalgamate the current capacity for several small varying loads? I have this device:

https://www.amazon.co.uk/gp/product/B075L2QK6R (2 ports, each rated at 2.4A with a total device rating of 4.8A)

\$\endgroup\$
  • 2
    \$\begingroup\$ In all probability you will achieve nothing, on simple USB phone chargers there is not separate limiting for each port, but a combined limit for the device. \$\endgroup\$ – Jack Creasey Aug 29 '18 at 19:30
  • \$\begingroup\$ are you sure? One port is rated at 1.0A and the other is rated at 2.1A \$\endgroup\$ – ezekiel Aug 29 '18 at 19:37
  • 1
    \$\begingroup\$ I did say 'In all probability' ….since you provided not details of your unit it's impossible to say with certainty. I personally have multiple USB chargers and all of them have a single current limiter. \$\endgroup\$ – Jack Creasey Aug 29 '18 at 19:39
  • \$\begingroup\$ fair enough! I think that this may not be the case with my devices - I have added more details to the question. \$\endgroup\$ – ezekiel Aug 29 '18 at 19:48
  • \$\begingroup\$ What is the purpose of your "amalgamating"? Keep in mind that commercial powerbanks have a nasty habit to shut the power down under light loads, which happens if they are used to power some well-power-managed (periodically sleeping) MCUs. \$\endgroup\$ – Ale..chenski Aug 29 '18 at 21:08
1
\$\begingroup\$

You should NOT connect together your USB ports on your Powerbank, you will not get the results you want.

The USB Powerbank you point to has 'Smart' outputs, that means each port has a controlling IC that sets the current limit for the port. This also allows the current limit to be set to a high value if the device asks (it does this by manipulating the D-/D+ lines on the USB cable).

While you could certainly set up each port to the high current limit, there will be slight differences between each port that will result in current differences between the ports. If either of the ports does reach a current limit threshold then the other port will shutdown immediately (since you are over the current level from either port).

You should partition the current requirements of you project such that you don't overload a port. I use a very similar Powerbank in a small robot with R'Pi and have the servos on one port (I use servos as motors too) and the R'Pi on the other port.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

If the ports have seperate power supplies then you can get more current. The second problem is the supplies will fight each other because they won't be the same voltage, the cables will not be the same resistance.

Consider this: If one of the ports is 5.19V and the other 5.12V (this wouldn't be unusual since most consumer power supplies have 1% tolerance on them), with cables that are 0.3 Ohm and 0.2 Ohm. This case would happen if you were to plug in the y-cable without plugging in a load. There would be 140mA flowing from one supply to the next, with most of that power dissipated in the supply. Not a great way to run a circuit. Also keep if your making a y-cable, 4A probably violates the spec of any USB connector.

schematic

simulate this circuit – Schematic created using CircuitLab

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ NO, the ports won't fight each other, how silly a statement. If one port is a higher voltage (internally) than the other (which is very unlikely) then all that happens is the current balances across the ports. What exactly happens when ports fight each other ….do they get cuts and bruises? \$\endgroup\$ – Jack Creasey Aug 29 '18 at 21:48
  • \$\begingroup\$ Power get dissipated in one of the supplies, as stated in the rest of the post. What I was not clear on is that this happens if there is no load (if you were to plug in the two ends to the sources before the device), but the diagram shows that. Usually dissipating power in a source is not ideal. I edited the question to be clear \$\endgroup\$ – Voltage Spike Aug 29 '18 at 22:43
  • \$\begingroup\$ Unless the two ports have different DC-DC converters the no load voltage will be identical. Even if there was two converters (which is very rare in the packs I've seen), you can't feed power back into a DC-DC converter. So you answer makes no sense at all. \$\endgroup\$ – Jack Creasey Aug 30 '18 at 0:25
  • \$\begingroup\$ I mean the case in which they are not identical (as shown in the diagram, othwise how would they be different voltages?) If they were identical there would not be a problem, so why post at all about potential issues? Many USB ports with DC DC regulators don't have droop control, and the tolerances on feedback many cheap DC DC regulators can range from 2% to 5% of the setpoint accuracy. \$\endgroup\$ – Voltage Spike Aug 30 '18 at 1:58
  • \$\begingroup\$ As I alteady pointed out, even with two DC-DC converters you can't feed power back into one from the other, so your answer makes no sense. The worst that can happen is that the current to the load will be different from the two ports. \$\endgroup\$ – Jack Creasey Aug 30 '18 at 5:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.