0
\$\begingroup\$

I'm using a circuit to charge a 2S lipo battery, and bypass the battery while it's charging, and still supply energy the circuit.

It works. But for the new charger module I'll use now, when the battery is fully charged to 8.4V, the charger starts drawing energy from the battery.

So If I leave my circuit disconnected from the charger supply, even if the circuit's load is turned off, the charger's circuit will draw a tiny bit of current from the battery non-stop.

So I wanted to isolate physically the battery from the charger when the charger is not charging.

I came with this first circuit in falstad, but I have some doubts:

Version with just a DIODE D1

Simulation of this circuit in falstad: Circuit 1

If I simply put a diode D1 to prevent energy from the Battery going back to the charger, will the charger work normally? Will the diode's voltage drop not allow the battery to be fully charged to 8.4V? The battery will not get to the full voltage output of the charger?

I did then this other version with two mosfets and removed the diode D1, and in simulation it works. But, still do the smaller voltage drop of the P mosfet will also not allow the battery to be fully charged? Or it's so small it doesn't matter? Also, will it charge the battery normally, as if the battery was directly connected to the charger?

Version with Mosfets

Simulation of this circuit in falstad: Circuit 2

Is this second circuit a good approach?

Edit: You can see the live simulation of both circuits and change it on the fly if needed by clicking in the Circuit 1 and Circuit 2 links.

Thanks!

\$\endgroup\$
8
  • \$\begingroup\$ 99.9% of discrete MOSFETs have a body diode which is anti-parallel to the source drain. That means a MOSFET can only block current in one direction. And your battery is quite a bit higher than most of the output voltages of your boost converter. I am not sure why you did not notice that in Falstad. Does Falstad not simulate the body diode? \$\endgroup\$
    – DKNguyen
    Feb 8 at 14:09
  • \$\begingroup\$ @DKNguyen the "simulate body diode option is turned on in Falstad... The boost converter is at 9V. The Battery is at max 8.4V, when the charging stops charging. \$\endgroup\$
    – Rodrigo
    Feb 8 at 14:24
  • \$\begingroup\$ @DKNguyen, did you ran the sim on the falstad link? It seems to work, but I’d like to be sure… \$\endgroup\$
    – Rodrigo
    Feb 9 at 16:10
  • \$\begingroup\$ I did not........ \$\endgroup\$
    – DKNguyen
    Feb 9 at 16:13
  • \$\begingroup\$ Your circuit is missing an important battery charger IC. Then you are risking an explosion and fire. A battery charger IC prevents trying to charge a Lithium battery that has its voltage discharged too low, it limits the charging voltage to 8.4V and it disconnects charging when the charging current drops to a low amount. You also need a circuit that disconnects the battery when it has discharged to no less than 6V. \$\endgroup\$
    – Audioguru
    Mar 1 at 16:31

1 Answer 1

Reset to default
0
\$\begingroup\$

Ok, after testing in real life the circuit indeed works. I'll post it here for future reference if someone needs it. It charges the 2S Battery and bypass it when connected to the 5V DC supply, and when the load is powered just by the battery, it blocks current flowing from the battery to the charger, as some chargers circuits keep drawing a little current forever when connected to the battery, even if the charger module is disconnected from power supply.

You can see it working by using this falstad link: Circuit Simulation

Bypass battery circuit

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.