I am trying to power a device which requires 5V and 2A through a 5.5mm / 2.1 mm barrel jack connector. SInce I need the device to be portable I am using a battery bank like this: http://www.amazon.co.uk/dp/B00OB5251W/ref=pe_385721_37986871_TE_item with a USB to barrel jack adapter like this: http://www.ebay.co.uk/itm/USB-A-to-5-5mm-2-1mm-Barrel-Jack-Male-DC-5v-2ft-power-charger-connector-cable-/281293234909?ssPageName=ADME:L:OC:IE:3160
However, I don't seem to be getting enough current. I think it is because the adapter cable is limiting the current to less than 2A. Does this sound correct?
I'm thinking of creating a new adapter cable by taking the cable from the original plug-in charger which came with the device, cutting off the plug end and soldering on a USB connector. Would this work? I think it should work if it is the wires which are limiting the current but not if it is the actual USB connector which limits the current. Does anyone know if this would work?
Best Answer
You are very likely right in your suspicion of it being the cable.
Cheap cables from who knows where often use things like AWG34 wires, or in this case it might be coaxial with up to a whopping AWG30 center wire.
What you could try is adding a LOAD to it and measuring what comes out, for example a 2.5Ohm 10W resistor.
It can be made of 40 1Ohm 1/4Watt resistors by making 4 strands of 10 resistors in series and then connecting the ends together, or by connecting all strands at all points:
simulate this circuit – Schematic created using CircuitLab
The best solution is to find a cable that's ready made for 2A with low loss.
The next best one is to find a loose connector for each end and a wire with low resistance and connect them up all fresh. (The barrel connector will be easy, 100's of factories around the world make them at 20cents retail prices)
It's presumable that the device will still work with 4.7V, so the maximum loss in the cable itself will be limited by the plugs and the wire, I'll assume the connectors waste half the maximum voltage drop, because without specifics we can never be certain anyway. So that leaves 0.15V for the wire, which comes down to:
R = V/I = 0.15V/2A = 0.075OhmThat's for the total wire, but you need a positive and a negative, so one wire can be up to 37mOhm. You can do two things: Estimate your length requirement, lookup the resistivity per meter of copper wires (there's a million tables on Google images relating AWG to Ohms per meter). Take a 25% margin and order the cable you need. But that does have the risk of the wires being too fat for the connector shell - some DIY required.
You can also order something in the range of AWG22~24, measure the voltage drop at 2A across 1m and then calculate the maximum length of that wire.
The next-next best thing is using connectors of existing cables and splicing a thicker wire in, it'll be easier, but the little ends of your cable will still be somewhat limiting, so you need to make them as short as possible.