this is very interesting question to me that why batteries can be charged while being used as the same time? I designed a buck converter that charges a battery from a solar panel but and when I want to use the battery I would enable a relay that will cut the power off to the battery so I can use it. but using a battery while being charged is possible. What are the principles in this case and would the battery be damaged if being used and charged at the same time?

  • \$\begingroup\$ I have edited my answer less than a minute ago. Check the first 2 sentences! \$\endgroup\$ Apr 25, 2020 at 19:06

2 Answers 2



simulate this circuit – Schematic created using CircuitLab

Figure 1. A very simplified schematic.

How a battery is being charged and used as the same time?

It can't. Either current is flowing into the battery (it's being charged) or current is flowing out (it's being discharged). You can't have current flowing both ways in the one wire.

This is the same as a car electrical system. The alternator charges the battery even though the lights, ignition and radio are switched on. The current into the battery in either case is the difference between the charging current and the load current. The result of the subtraction can be positive or negative with, for example, positive means battery charging and negative being battery discharging.

An example:

  • V1 is supplying 5 A.
  • Load is drawing 6 A.
  • Battery current = 5 - 6 = -1 A. It's discharging.
  • \$\begingroup\$ Thank you very much! \$\endgroup\$
    – f321
    Apr 25, 2020 at 19:41
  • \$\begingroup\$ This explanation is awesome, and the diagram is very useful to see it. But why is the charger "taking control" of the flow ? Why isn't the battery saying "I will provide current and load the charger" ? \$\endgroup\$
    – fallais
    Jan 28, 2021 at 8:02
  • \$\begingroup\$ @Elwyn, "I will provide current and load the charger" It could only load the charger if it was drawing current. The battery is dumb - metals and chemicals. The charger is usually configured to output a higher voltage than the battery so that current flows "downhill" into the battery and the load. \$\endgroup\$
    – Transistor
    Jan 28, 2021 at 8:06
  • \$\begingroup\$ @Transistor : So you mean that everything is based on the fact that the charger provides more voltage ? What if I have a Lithium battery of 25.2V and a charger of 24V. It will work as well ? \$\endgroup\$
    – fallais
    Jan 28, 2021 at 8:14
  • \$\begingroup\$ @Elwyn: You will stop charging when the battery reaches 24 V. In practice it's a bit more complicated. I might have a 24 V supply but the charger may have a voltage step-up converter precisely because it is needed to charge the battery. Meanwhile the battery may have a BMS (battery management system) to control the charging (prevent overcharging) and to prevent / shut-off discharge when the voltage drops to a certain level. e.g. Your phone charger is not a charger - it's a 5 V power supply (although it's more complex in recent years). The charge controller is in the phone. \$\endgroup\$
    – Transistor
    Jan 28, 2021 at 8:26

You shouldn't cut off the charge to the battery when you're using it, because you're discharging it while your available power from the solar panel is sitting idle, doing nothing, and it would be better to use it for the load.
You can use it and charge it at the same time, it shouldn't damage it.
However, a battery's type and capacity should be taken into account along with the charging current and the load current, but generally speaking a battery without a load can get overcharged, while a battery disconnected from a charger and connected to a load gets discharged sooner and can get too low. Both of those scenarios are worse for the batter, and that's why it is better to actually keep charging it while using it.
The main issue here would be the charger's ability to limit its current so it doesn't burn under too much load.

A battery doesn't really know and care about being charged and used at the same time.
What it "cares" about is the voltage across its terminals.
When the voltage applied to it is higher than its own, it will be accepting charge.
When its own voltage is higher, it will be losing charge.
You should only worry that the current being used by a load is lower than the current coming to the battery from the charger connected to the solar panel.
It is about striking balances. How to achieve all this? Well, the devil's in the details.
It takes some time and effort (and knowledge) to figure out and calculate, but it shouldn't be too hard.


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