I have an 50Wh “Hama Pipe” powerbank with 21W output power (2×2.1A ports OR 2.4A+1.8A), which would equal 4.2A at 5V in total, minus resistance.

But the input power is only 2.5W, which equals 0.5A at 5V. (no typo, see http://www.hama.com/00137422/hama-power-pack-pipe-13000-mah-weiss-grau)

Is that a technical limitation? The powerbank does not produce any heat during charge, so it is probably an *easily bypassable** limitation of the charging IC of the powerbank.

I have a different powerbank with 10W of input power, which produces heat.

Is that 2W limitation technical? Can it be bypassed?


My smartphone supports fast charging (9V 1.7A), but when I connect it using 5V, it does not take 3A (charger has 20W maximum output), not 2A but only 1.2A. I should build a power step-up transformer out if hardware modules.

It is not because of the resistance in the long cable. Short cables aloow 5V 1.3A. The USB multimeter shows 5.26V 1.21A for the long cable.

It is because the charging controller refuses to take all available power.

How can I manually overthrottle the charging IC?

  • \$\begingroup\$ It doesn't have to "throttle" the input, it just draws what it was designed to draw. It doesn't care that more is available than it needs. \$\endgroup\$ – Finbarr Jan 15 '18 at 14:39
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    \$\begingroup\$ Yes, devices do "throttle the input", in some sense. Try to search the EE site for terms "charger signature". Think of the problem: how your device may know that your powerbank can do 4 A? \$\endgroup\$ – Ale..chenski Jan 17 '18 at 4:22

How can I manually overthrottle the charging IC?

You can't and you should not.

The charging current of the powerbank is a technical limitation due to the design choices made by the manufacturer.

Charging over USB can be made faster but more complex and/or more expensive electronics might be needed.

I should build a power step-up transformer out if hardware modules.

Why? It will not work and will very likely damage your phone.

You might think that you simply need a higher voltage to make the devices charge faster. It is not that simple. USB is 5 V unless the power adapter and the device "talk" to each other and both decide that a higher voltage is OK. That's needed to prevent a power supply to feed 12 V over USB to a phone that is only designed for charging with 5 V over USB.

  • \$\begingroup\$ My phone can charge with 9V, so I can transform 5v to 9V. \$\endgroup\$ – neverMind9 Jan 15 '18 at 14:52
  • \$\begingroup\$ Thanks, nevertheless. Unable to upvote due to insufficient reputation points. \$\endgroup\$ – neverMind9 Jan 15 '18 at 14:53
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    \$\begingroup\$ My phone can charge with 9V, so I can transform 5v to 9V You still don't get it. You cannot just feed 9 V to the phone because the phone expects a certain protocoll to be followed, example: Qualcomm quickcharge. If you "just" apply 9 V the phone might be damaged and might not even charge any faster. To charge efficiently at 9V or more a DCDC converter might be needed which might not be active in the "5 V" mode. But sure, ignore my advice and try it but don't complain to me that I didn't warn you when you break your phone. \$\endgroup\$ – Bimpelrekkie Jan 15 '18 at 14:56
  • \$\begingroup\$ I charged my phone often using a just-feed 7V charger, without any problems. Nothing happened. My old test phone refuses to accept any charge above 6.4V. The charging IC can already control everything. But yes, I will take care, and not just feed 9V. However, what happens, if that protocol is not followed? \$\endgroup\$ – neverMind9 Jan 15 '18 at 15:03
  • \$\begingroup\$ However, what happens, if that protocol is not followed? it is written in my comment above. What happens depends on the design of the charging circuit in the phone. When you apply more than 5 V without the proper charger then you rely on what the manufacturer did to handle that situation. The charging circuit could be damaged and/or overheat. It will very likely not charge your phone any faster. Why not simply fast charge in the way it is intended to be used: by using the charger which is supplied with the phone. \$\endgroup\$ – Bimpelrekkie Jan 15 '18 at 15:11

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