I thought of an idea recently, and I wanted to know if electrical engineers agreed with my theory.

I have a smartphone. Recently I bought a power bank because I found my phone's battery not holding up long enough.

Now here is what I thought about while charging my power bank and my phone.

If both batteries (of my phone and powerbank) could in theory fill up 1 mAh/minute and both my phone and and the power bank are connected to the outlet, and are actively charging, I'm basically charging the two batteries at an effective 2 mAh/minute.

If both my phone and my battery pack could hold 1000 mAh, I could charge both half way to 500 mAh and still have an effective 1000 mAh at half the time it would take me to charge the phone alone.

So if I bought three banks then together with my phone I could get the same effective 1000 mAh in just a quarter of the time!

So if this works in theory: Why not divide the phone's battery into hundreds of separate banks to get hundreds of times the times the charge rate? What prevents this in practice?

small diagram

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    \$\begingroup\$ "Why not create hundreds of batteries inside of the phone"-What about the available space?The phones would be too big. \$\endgroup\$ Dec 20, 2015 at 15:21
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    \$\begingroup\$ the batteries would be much smaller of course. but does that math hold up? i would worry efficiency at smaller sizes goes down. but even if we use instead of 1 battery, 2. dont we get double the charging rate? \$\endgroup\$ Dec 20, 2015 at 15:26
  • \$\begingroup\$ Actually that might be true,but it feels that there is a waste of capacity that way.What would you do with the rest of the 50% battery capacity?It is simpler to just increase the voltage and current of the charger to make it faster. \$\endgroup\$ Dec 20, 2015 at 15:37
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    \$\begingroup\$ Why do you think that if you split a battery in half, you can still charge the halves at the same rate as the original battery? \$\endgroup\$
    – Dave Tweed
    Dec 20, 2015 at 15:44
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    \$\begingroup\$ so do smaller batteries take more time to charge?, my phone battery is much smaller obviously that my power pack (in physical size) \$\endgroup\$ Dec 20, 2015 at 15:46

2 Answers 2


It doesn't work that way. One of these two issues will get you:

  1. You "spilt" the original batteries into smaller batteries, maintaining the overall battery capacity. The smaller batteries now have lower capacity and lower maximum allowed charging current. If a original battery can be charged at 1 mAh/minute, then a battery resulting from a 1:2 split could only be charged at ½ mAh/minute.

    With the batteries split to conserve overall capacity, a larger fraction of the whole volume would be taken up by interrconnects, so the overall battery pack would now be larger and a little heavier.

  2. You use more of the same size batteries. Each can now be charged at the original rate, but the battery pack has become proportionally heavier and larger.

    The cell phone maker has already picked a tradeoff between size and weight on one side and longevity of charge on the other. For the same battery technology, you can't cheat this tradeoff. By adding more batteries, you increase longevity of charge, but you can't get away from the proportionally larger size and weight of the resulting battery pack.

If it were really so simple, they'd be doing it.

  • \$\begingroup\$ why cant i use the same current on a smaller sized battery? if this is truely linear, this theory is pointless, if its only slightly worse it could still be achieved.. "If it were really so simple, they'd be doing it." .. thats what i thought \$\endgroup\$ Dec 20, 2015 at 16:09
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    \$\begingroup\$ @down: The smaller volume of the chemical soup can only absorb charge at a smaller rate. Think of these batteries as being able to take some mAh/minute per cubic centimeter. \$\endgroup\$ Dec 20, 2015 at 16:12
  • \$\begingroup\$ so they are basically by our techonlogic standart as "perfectly" efficient on the atmoic scale as they can get. so the very idea of unchanged charging rate is fundamentally wrong? \$\endgroup\$ Dec 20, 2015 at 16:14
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    \$\begingroup\$ batteryuniversity.com/learn/article/ultra_fast_chargers "The goal is to avoid lithium-plating on the anode and to keep the temperature under control. A thin anode with high porosity and small graphite particles enables ultra-fast-charging because of the large surface area" : constructing better anodes is an ongoing project. There are plenty of technologies which work better on the lab bench but are hard to manufacture. \$\endgroup\$
    – pjc50
    Dec 20, 2015 at 21:16
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    \$\begingroup\$ @downrep_nation "the very idea of unchanged charging rate is fundamentally wrong" - Yes! Different sized batteries have different ideal charging rates. \$\endgroup\$
    – user253751
    Dec 21, 2015 at 2:20

Acceptable charge current somewhat scales with cell capacity, all other details being equal. This may not be entirely true (if you are generating heat, smaller cells can potentially better dissipate it due to their lower volume-to-surface ratio) however it is true enough that we generally speak of charge rates not as an absolute current, but rather as a fraction of the cell hourly capacity, or "C". So for example a ".2C" charge rate would be charging the cell at 20% of its rated capacity per hour. Given that there are inefficiencies and you have to slow down as a delicate lithium chemistry nears its voltage limit, this .2C rate might take at least six, maybe seven or eight hours. In practice, based on external measurements I think my phone will charge at more like .3C, and I have small cells for RC flying machines that are engineered for high discharge rates and can be charged at over 1C. Whatever the "C" rating, two small cells would charge no faster at it than one larger cell of the same C rating having the same total capacity.

But there's another issue in many phones: the USB spec limits supplied current to 500 mA. This is more limiting than what the cell can tolerate as a charge in most cases today, so it is common to build chargers with sense resistors or other tricks on the data lines so that the phone can identify that a special charger is present and charge at a higher rate. Unfortunately, the way they do this hasn't fully been standard - though there are attempts to make it so today, it is still quite possible you might get a mismatch between charger and phone forcing a fallback to a 500 mA or 2.5 Watt charging rate. In the past I often found an outlet-plug wattmeter useful to tell if a charger was charging a device at high rate, or only at USB rate - the reading would include losses in the charger, but the different between seeing a 2.x watts vs seeing 6 or 7 display is significant. Today you can get inline USB current meters which would also work, at least if you get one that connects the data lines through.

Your needs may ultimately be met by the new USB fast-charging standard which is intended to get you a useful fraction of capacity in a short period of time - until then make sure that you are charging your devices at their designed maximum rate, and not through mismatched chargers or bad cables at the fallback 500 mA one.

  • \$\begingroup\$ Most devices charge at 1-1.5A (as most chargers can supply). Bigger tables are often sold bundled with 2-2.5A chargers. USB charging spec is old and mature - even Apple devices can take advantage of generic charger with more than 500mA). Trick is very simple - data lines soldered together mean "as much as you can draw over 5V". The USB gadget you're describing is often known as "charger doctor" - works like a charm - I use it to identify poor quality USB cables (eg ones that won't let a device reach it's maximum charging current due to voltage drop on thin wires and poor connectors). \$\endgroup\$
    – Agent_L
    Dec 21, 2015 at 11:38

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