I am designing a circuit board for a small mobile robot and there is a problem with the power supply. The circuit is powered by a 1-cell Li-ion battery (VBAT) and includes a charger (MCP73831/2), a pMOS high side switch (SI2369DS), a booster (TPS61232) and a simple dual Schottky voltage selector because it can also be powered through USB (5VUSB). The relevant part of the schematics with all component part numbers is below.

The problem The pMOS high side switch does not close properly. When open, everything works as expected (green voltages). When closed (red voltages), the drain voltage is approx. 0.3V and the whole circuit is draining about 80mA. In all cases, the charger is off (jumper JP1 not present).


Things that I tried so far, without success:

  • Changed the pMOS to SI2399DS which has a lower Vgs, same behaviour.
  • Held a 330R resistor in parallel to the R14 pull-up, same behaviour.
  • Tried with both, a battery, and lab power supply, same behaviour.
  • Tried on 5 boards, behaviour is the same on all of them, so probably not a soldering problem.
  • Measured VON on oscilloscope and it is steady.
  • The pMOS footprint was checked and it is fine.

What could be the source of the problem?

  • 1
    \$\begingroup\$ You did a lot of work before asking the question. Good job. There is no reason to think there is anything wrong with the PMOS. I bet you could remove it from the circuit and all would be the same (feel free to try it). I think there is another pathway for VBAT to get to VON. I suspect it is not shown in the question. I do observe that boost regulator Vout pin is 1.67V. It could be conducting to Von through internal diodes. So why is "5VBOOST" at 1.67V? Try to figure that out. Also, how is "+5V" at 3.04V? That may be the biggest clue. Find the path from VBAT to "+5V" and or "5VBOOST" \$\endgroup\$
    – user57037
    Dec 16, 2017 at 22:06
  • \$\begingroup\$ mkeith, you were exactly right. The only other place where VBAT was routed was a multiplexer, and then to the microcontroller's ADC pin. The multiplexer select pins were grounded, so VBAT went straight to the microcontroller, and then through the microcontroller's internal diode to 5V. Thanks! (I am not sure what the proper etiquette is, should I edit the question, or do you want to write this as the answer, so I can accept?) \$\endgroup\$
    – supermedo
    Dec 18, 2017 at 15:37
  • 1
    \$\begingroup\$ You can answer your own question and accept the answer yourself. That is not bad form. I don't care about attribution, use whatever part of my comment you want. The basic idea is that whenever an IC is powered down, steps must be taken to prevent any of its pins, whether input, output or power pins, from be energized externally. Otherwise there is a path through protection diodes to VCC. \$\endgroup\$
    – user57037
    Dec 18, 2017 at 17:14

1 Answer 1


As mkeith pointed out, the problem was not in the schematics posted in the question but elsewhere: the battery supply VBAT is routed to a multiplexer and then directly to an analog input of the microcontroller where it went through the protection diode straight to 5V. This also explains why the +5V voltage was at 3.04V, which is the VBAT (4V) minus the voltage drop over the diode. enter image description here


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