Consider the following schematic (and realize that it isn't perfect because I'm new to circuit lab- I'll explain in words what I'm trying to represent below). It's a highly simplified version of a circuit I'm designing, so let's please try not to get into design critique, since there's a lot more to the circuit than what is shown here. This is just meant to illustrate my specific question. Also, if you're going to tell me not to use a CC power supply to charge a battery you have already lost me. I have done it in the past and gotten great results.


simulate this circuit – Schematic created using CircuitLab

So, imagine you are using a 350ma constant current power supply with a max voltage of 257v to balance charge a high voltage hybrid vehicle battery. This battery will reach a maximum voltage somewhere between 235v and 240v and then will fluctuate slightly in that region as individual cells balance, while a lot of surplus power gets dissipated as heat. The battery has 16 modules in series with an internal resistance of about 20-25 mOHMs each, for a total resistance of about .3 or .4 ohms. This is obviously pretty straight forward stuff I think.

Now imagine you connect some resistance in parallel. What happens?

So, can I think of this as a simple current divider? If I do, I see that the battery will receive the lion's share of the current from the power supply because of it's low resistance, with just a small fraction going through the parallel resistor.

My confusion stems from the fact that a battery can act both as a current sink and a current source. Obviously the minute I turn off the power supply the battery will become a current source for the resistor. The battery is capable of supplying much higher current than the power supply, but it isn't capable for supplying higher voltage. Does the higher voltage source win out?

Realistically, I can pretty easily use a mosfet on that resistance path to open it while the battery charges, and then close the circuit when I want to discharge the battery, making all of this moot. However, if I'm correct in thinking of this as a current divider then I won't necessarily have to do that, which could have some advantages for my purposes. Thanks!

  • \$\begingroup\$ It’s not quite a simple current divider as your second battery resistor would be in series with the battery voltage source. Redraw your schematic with the battery replaced by an ideal voltage source and a series 0.4 ohm resistor and redo the analysis \$\endgroup\$
    – Bryan
    Commented Jul 6, 2022 at 1:45

1 Answer 1


Does the higher voltage source win out?

The voltage on the wire is the voltage on the wire. It's not the higher voltage source that wins, it's the stiffer one. Within the normal operating region, that's the constant-current power supply. It will set the voltage such that it's sourcing 350mA; some of that (100mA, give or take) will go into heating the resistor, and what's left will charge the battery.

If we forgot about the voltage limits on the power supply and the battery chemistry for a moment, then this would continue charging at an ever-decreasing rate until 700V, at which point the current through the resistor would equal 350mA and the net current into the battery would equal zero. But since we can't forget about those factors, the charging will continue until 257V, at which point the battery will effectively be CV top-up charging. Once the battery equalizes to a Voc of 257V, the current it takes will decrease to near zero (whatever its self-discharge rate is) and the power supply will be delivering 128mA (@33 W) into the resistor.

  • \$\begingroup\$ So long story short I should probably plan to use a mosfet on that parallel resistor? \$\endgroup\$ Commented Jul 6, 2022 at 2:17
  • \$\begingroup\$ @DerekHershman if it's there to be a discharge load... yeah. \$\endgroup\$
    – hobbs
    Commented Jul 6, 2022 at 2:21

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