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I am attempting to use a prototype 2 cell Li-ion battery charger, but the charger appears to be overheating. The charger uses a monolithic power MP2672GD IC as its basis battery charger IC. Circuit was taken from datasheet.

When I activate it the STAT and ACOK lights are solid and the battery appears to be charging. But after about 1min the 5V LDO supplying the battery management IC heats up quite a bit and so does the battery management IC. Then the STAT and ACOK light start blinking in together.

This seems like excessive current draw, but the main power supply at 14V is reading pretty steady at about 1.1A and without this portion of the circuit, it normally draws about 0.15A. The LDO IC101 has a huge ground pad and the ground plane is around it although the spokes are pretty thin, but why is it overheating so fast? Also this board is overkill and has 4 layers all of which have ground planes where no traces are. The layer directly underneath the top layer is an unbroken ground plane.

The circuit is below, the only thing I have changed is removed the P MOSFETs Q103 and Q104. With these it seemed like the IC was having difficulty detecting the battery. The battery is located at the top left of the PCB (the 3 prong connection) and the very right of the schematic drawing.

Does anyone know how my charger could be pulling only about 1A from 14V supply and still overheating?

circuit

footprint1 footprint with ground plane

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    \$\begingroup\$ I'm having trouble detecting the battery too. Where is it connected? \$\endgroup\$
    – Finbarr
    Commented Dec 7, 2022 at 12:14
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    \$\begingroup\$ Anyway, that aside, I'm struggling to understand why you're using an expensive ultra-low dropout regulator to lower 14V to 5V to then step it up again using a boost charger. At 1A the LDO will dissipate 9W of power and get hot very quickly without sufficient heatsinking. \$\endgroup\$
    – Finbarr
    Commented Dec 7, 2022 at 12:41
  • \$\begingroup\$ The charger battery is connected in the top left of the PCB and the very right of the schematic (J1). Since there are 9W to disipate will I likely need a fan? Or should I add more vias and improve the connection to the ground plane? \$\endgroup\$
    – Feynman137
    Commented Dec 7, 2022 at 15:39
  • \$\begingroup\$ Also the battery management IC heats up quite a bit, I have already limited the charging current to 0.5A but is there another way to lower the thermal stress on the battery management IC? \$\endgroup\$
    – Feynman137
    Commented Dec 7, 2022 at 16:13
  • \$\begingroup\$ 1.1 A x (14-5) V = 9.9 W of dissipation. No wonder it gets hot. Why are you stepping down to 5 V? \$\endgroup\$
    – winny
    Commented Dec 7, 2022 at 17:04

1 Answer 1

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You have the charge current set to 0.5 A. Therefore at 8.4 V out the boost charger IC must draw at least (8.4/5)*0.5/0.95 = 0.88 A at 5 V.

The LDO dropping from 14 V to 5 V will have to dissipate (14-5)*0.88 = 7.92 W. That's a lot of heat to get rid of. You should replace the linear regulator with a switching type.

With the FETs strapped out the battery will become unbalanced via R114 and R115. Do not leave it this way! When the FETs are turned on ('MotorON' signal inactive?) the battery will become unbalanced via R115. Even when turned 'off' the FETS don't completely disconnect the battery because their body diodes can conduct. It might be better to pull the VLIN pin low to disable charging (when motor is on?) instead.

For correct battery detection and accurate cell voltage measurement the battery should be connected directly to the charger IC (with MID going through R117 to limit balancing current).

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  • \$\begingroup\$ The battery charges when the motor is on (motorON line is high). Do you mean pull Vlim low instead of Vlin? If so this is actually already occuring because EN is low when motor on is low. I just was trying to protect the battery management IC from voltages on its pins when it had no voltage supply. Thanks for reminding me of R114 and R115 I will remove these now that the MOSFETs are gone. \$\endgroup\$
    – Feynman137
    Commented Dec 7, 2022 at 18:46
  • \$\begingroup\$ Also I think I will replace the LDO with MAX20077ATCA/VY+ \$\endgroup\$
    – Feynman137
    Commented Dec 7, 2022 at 19:11
  • \$\begingroup\$ Why would you want to run two switching regulators in series? Better than LDO + switcher, but why not a different li-ion battery charger IC in the first place? \$\endgroup\$
    – winny
    Commented Dec 7, 2022 at 19:50
  • \$\begingroup\$ I see your point but I cannot find many ICs that are designed for 2 cell batteries, have cell balancing, and can tolerate 15V input. \$\endgroup\$
    – Feynman137
    Commented Dec 7, 2022 at 19:58
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    \$\begingroup\$ ...and running the charger from 5V means you can use a standard USB power supply etc. if desired. The charger IC should be fine connected to the battery with no power input, since this can easily happen in typical applications. Internally it has back-to-back MOSFETs that isolate the input from the output when turned off. I can't find a spec for leakage, hopefully its low enough that the battery won't drain significantly when there is no power. \$\endgroup\$
    – Bruce Abbott
    Commented Dec 7, 2022 at 20:40

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