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I'm boosting a 7000mAh Li-ion battery pack to 16V to drive a dc brushed motor that has a startup current limited to 1.4A. Due to my switching regulator efficiency a 1.4A output current means about 6.5A input current requirement from the battery.

I noticed the inrush current on the motor and the startup time do not change even when the battery reaches 3.3V

How is it possible that the battery at voltage near cutoff is still capable of supplying that much current, whereas my discharge curve does show at that voltage level the remaining charge is less than 2000mAh ?

I have to admit my knowledge of Li-Ion battery is limited.

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  • \$\begingroup\$ Vbat range is 4.2V-3.0V \$\endgroup\$
    – igorvanhelsing
    Commented Oct 22, 2016 at 22:38
  • \$\begingroup\$ Accounting for the remaining charge in a LION (or any secondary battery) is a cross between art, reading the (reputable) manufacturer's specification sheet for THAT battery and choosing & configuring a battery management chip (it.com makes most of them) to make all the calculations easy for you. In the end, the remaining charge is more an educated guess. I suspect you are getting the "safety margin" performance of a fresh pack which will eventually not be there after the pack ages. \$\endgroup\$
    – st2000
    Commented Oct 22, 2016 at 23:00
  • \$\begingroup\$ Lithium ion batteries are able to deliver high current even when they are almost fully discharged. It is all about the surface area and chemical reaction rate at the anode and cathode. For Lithium Ion (including lithium polymer) that reaction is pretty fast. \$\endgroup\$
    – user57037
    Commented Oct 22, 2016 at 23:12

2 Answers 2

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2000mAh does not mean 2 Amps. It means if you draw 2 Amps you can do it for 1 hour before the battery is empty. In reality, the slower you draw current the better the results. That is, you might get 1 hour at 2 Amps. But it is more likely you will get 2 hours at 1 Amp. Or 4 hours at 500mA.

The maximum current a battery can produce has more to do with its internal resistance. In an ideal battery there is no internal resistance. There is no such battery in reality. But if there were it would produce an infinite amount of current for an infinitely short amount of time.

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There is no contradiction in your observations.

Assuming the initial voltage of a cell is 4.2V, you boost it to 16V, which makes the boost ratio to 3.8.

You also say that it takes 4.6A to make 1.4A, which gives you the current transformation ratio of 6.5/1.4=4.6. In other words, you have a converter with 3.8/4.6=0.82 = 82% efficiency. Which is pretty normal.

Now, at 3.3V you have the same output, 1.4A, meaning that the voltage boost is 16/3.3=4.84, so the current that your converter will draw out of the battery is 1.4 * 4.84 /0.82 = 8.2A

This means that your battery can supply 8.2A even at discharged state of 3.3V.

Regarding the 2000mA remaining capacity, see the answer by st2000

Obviously you need to manage your battery to keep it from discharging too low.

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