I'm planning on building a 4s 18650 (16.8V) lithium-ion battery pack with a 4s BMS with balance function to replace an old built-in Ni-Mh battpack. I would like to charge the new lithium battpack via Micro USB (5V) with maximum 2A current.

My question is the following: I can implement the USB charging by:

A) Connecting a DC-DC step up converter in-between the female USB plug and BMS. [5V source in -> DC-DC step up booster -> BMS]

B) Add a TP4056 (or something similar) connecting it directly to the 18650 cells and charging them in parallel. [5V source in -> TP4056 -> charging output to each cell separately, bypassing BMS altogether {just for charging}]

Which option is more recommended?

The problem with option "A" is that the booster module produces tons of heat even with a heat sink attached (the battpack is installed in an enclosed plastic case) or if I go with option "B" should I use single 1S BMS on each battery cell with a separate balancing board?


2 Answers 2


When one of your suggestions is "bypass the BMS" that is a bad idea. The point of the BMS is safety. Safety matters. If you are going to break your pack up into parallel cells, you want to completely isolate them from each other. That is hard to do well. So my advice is to not do it that way.

You other option, option A leads to some more questions:

How are you asking for charging current out of the USB? The USB spec says you only need to provide 100mA (or may be 10mA if you want it to be fully backward compatible) at 4.5V for USB. Making 450mW, So at your charging voltage (4 cells, 4.5V each, 18V) that makes a current of 25mA into your pack. Even if you draw the max official standard USB current (0.5A) that only lets you have: 2.25W, at 18V that gets you... 0.125A. This pack is going to take a long time to charge up, even if you have managed to make a fantastic and impossible 100% efficient boost converter. More realistic, 85% efficient, 4.5V, 0.5A (2.25W) in, 1.9W out, 100mA into your 2000mAh cells... 20 hours to charge up the pack.

Yes there are way to get more power out of power supply designed USB sockets. Also it is possible to get USB adaptors which push out more current. But if you're doing that, why not get a proper mains adaptor? They can be bought pretty cheaply. It depends on what this pack is for, but if it's a mass market thing and you are using USB because it's everywhere, your charge times are going to be really long. Get a proper supply. If it's just for home use and you can cope with a 20hr charge time, and can design a decent constant current boost converter, then that is a possible route, but I'd advise against going that way for safety, convenience and cost.

I would advise: get a suitable constant current power supply to charge your pack up straight from mains. And 18V, 1A supply (assuming your BMS has all the required safety features in it) would to a lot better job.


A BMS is not a charger.

BMSes are for cell balancing and last-ditch over/undervoltage and overcurrent protection when something goes wrong with charging or discharging your battery pack.

You need a proper, dedicated charger to charge Li-ion batteries/battery packs. A BMS is not it, however many YouTube videos and Chinese designs may suggest otherwise.

As for option A: you could use a boost converter to get a high-enough voltage and use that as the input for a 4S Li-ion charger, which then charges your 4S, BMS-equipped battery pack. If you have 5 V, 2 A to play with, then at, say, 18 V output of the boost converter, you would have about 500 mA of charging current.

As for option B: you can't charge the cells in parallel while the battery pack has cells also connected in series. I suppose it would be possible to switch the pack's configuration from series to parallel for charging only, but that is impractical, would greatly confuse the BMS, and also your load if it is still connected, and it would also have other problems if the cells aren't well-matched. I wouldn't go there.

  • \$\begingroup\$ Thank you for your answer I greatly appreciate it. If the boost converter has 80% efficiency and at the 5V source draws 2A current, does that mean the boost converter will output 1.6A maximum at 18V? \$\endgroup\$
    – ryzen
    Sep 24, 2020 at 12:14
  • \$\begingroup\$ No; that would mean that 10W on the input generates 28.8W on the output. You have 5V * 2A = 10W of power on the input. At 100% efficiency, you get 10W / 18V = 0.555A on the output. At 80% efficiency, that would leave 0.444A on the output; at 90% efficiency, 0.5A. \$\endgroup\$
    – ocrdu
    Sep 24, 2020 at 12:22
  • \$\begingroup\$ @ryzen: You may want to vote on answers, and accept one of them, so the question doesn't remain open. \$\endgroup\$
    – ocrdu
    Oct 28, 2020 at 11:17

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