I was wondering how discharging all the energy from one 18650 Lithium-Ion battery into another 18650 Lithium-Ion battery would be achieved. So far I’ve consulted using MOSFETs, DC to DC Converters and current sources/sinks without avail.

The whole circuit I’m attempting to achieve is a circuit that can fully charge battery 1, and then transfer all of the energy (less any losses) from battery 1 into battery 2.

Any direction or reference papers would be appreciated. Thank you.

Edit: Will update with more details shortly.
The end goal is to have a battery testing circuit that does not entirely dependent on the energy from a wall adapter. The circuit will also be testing for battery voltage, charging current and temperature among other things.

So far I have a discharging circuit to discharge both batteries and a charging circuit to charge one of the two the two batteries. The transferring of energy from battery 1 to battery 2 is the essential step I’m currently working on.

Thanks again for all the insight it’s has been a huge help.

  • 5
    \$\begingroup\$ Buck-boost converter. But why? \$\endgroup\$
    – winny
    Jan 11, 2020 at 21:58
  • \$\begingroup\$ Do you mean - completely discharge all ENeRGY from battery one to battery 2? That's doable but is not the ame as your stated "all current". | A DC-DC converter of suitable specifications will do the job, but knowing more about exactly what "the job" is will help us to provide better answers. \$\endgroup\$
    – Russell McMahon
    Jan 13, 2020 at 3:01
  • \$\begingroup\$ @RussellMcMahon Yes, all their energy from one battery to another within reason due to losses. That’s the main goal among measuring other parameters such as voltage, charge current, temperature etc of the battery. Essentially a battery tester/evaluated which requires battery energy transfer in tandem vs charging and discharging entirely from the wall adapter. \$\endgroup\$
    – MichaelTr7
    Jan 13, 2020 at 3:08

2 Answers 2


The way lithium-ion batteries work is by being at full charge at ~4.2V and "no charge" at ~3.0V. The limits on voltage is due to safety, because lithium-ion batteries are unstable outside of this range.

So given this information, and we know from electronics that the higher voltage-potential is the one that will control the way the energy flows, we can see that you would want the "charger battery" to stay at a higher voltage than 4.2V. Of course you aren't going to charge it to a higher voltage, but you can step up the voltage with a buck-boost converter, also called a step-up regulator. Then you could have the voltage at for example 5V, and it would charge the other battery.

If you want to charge the other way again, for whatever reason, then you can use the same step-up regulator and some switching-circuitry with mosfets to control which battery that is stepped up to 5V.

I have drawn a suggested schematic below. You want to research this more of course, but I think this would be at least close to what you want.


simulate this circuit – Schematic created using CircuitLab

My schematic does not include over-charge and under-charge protection. This is of course an important aspect you need to add to ensure you don't charge above 4.2V. That part of the design is up to you to include, but should be simple enough.

Sidenote: 3.7V stated on the batteries in the schematic is because this is the nominal voltage of lithium-ion batteries. They will charge up to 4.2V and discharge as much as you allow it.

  • 1
    \$\begingroup\$ I would add that you would need the charger circuit to be able to regulate the input at a low(ish) constant current, rather than try to regulate the output current. As the voltage of the input battery will be dropping fast after a certain point of discharge. \$\endgroup\$
    – crowie
    Jan 13, 2020 at 3:30

If you connect a fully charged (4.2V) battery to a discharged (3.2V) battery then there will be almost unlimited current flowing and maybe a fire or explosion, then if both batteries survive they will each end with 3.7V.

You [b]never[/b] boost the voltage higher than 4.2V to charge a Lithium-Cobalt battery cell. Also you did not mention using a battery charger circuit.

  • \$\begingroup\$ re " ... battery charger ..." -> He said ", DC to DC Converters " which is essentially what a battery charger circuit would be in the gven context. \$\endgroup\$
    – Russell McMahon
    Jan 12, 2020 at 9:15
  • \$\begingroup\$ A DC to DC converter is not a smart battery charger circuit. \$\endgroup\$
    – Audioguru
    Jan 13, 2020 at 0:40
  • \$\begingroup\$ @Audioguru Is there an issue by limiting the current through a resistor of high enough power rating, other than power loss? Also are you suggested the ICs that can be configured to do the 3 phases constant current (CC) and constant voltage (CV) charging cycles. Thanks again appreciate the insight. \$\endgroup\$
    – MichaelTr7
    Jan 13, 2020 at 3:28
  • \$\begingroup\$ A resistor is not a smart battery charger circuit. It does not sense the voltage of the battery before charging then attempt charging at a low current, then if the coltage does not rise normally it refuses charging. \$\endgroup\$
    – Audioguru
    Jan 14, 2020 at 14:04
  • \$\begingroup\$ A resistor also does not sense the charging current drop when the battery is fully charged then completely stops charging to prevent a damaging over-charge. \$\endgroup\$
    – Audioguru
    Jan 14, 2020 at 14:09

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