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I would like to prevent the battery in a device from discharging (initially) until the usb is plugged in (to keep charge on the battery during storage and shipping). I am using a typical LiPo protection circuit (BQ2970):

LiPo Protection

I need to disable the discharge MOSFET until a USB cable is plugged in. I'm sure this has been done before, but I don't have a common reference voltage to implement any of my ideas (have USB v-divided to drive MOSFET etc.).

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    \$\begingroup\$ This would be accomplished with some type of micro controller. Place the micro in ship mode, this keeps it asleep until USB power is detected. Then after USB power is detected switch the micro to normal operation. \$\endgroup\$ – vini_i Oct 12 '16 at 16:41
  • \$\begingroup\$ @vini_i That would work if I didn't have external circuitry (not just a uC) that could also drain the battery. \$\endgroup\$ – JBaczuk Oct 12 '16 at 17:01
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    \$\begingroup\$ Use a micro as your battery protection and with a custom firmware that holds the battery in ship mode until a pin goes high. \$\endgroup\$ – vini_i Oct 12 '16 at 17:09
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    \$\begingroup\$ You can capacitively couple VBUS to a self-latching PMOS fet switch. Will try to draw a circuit later. Basically, NPN pulls down FET gate. PMOS output turn on NPN. Micro can force off NPN to enter deep sleep. Capacitor from USB_VBUS to NPN base. \$\endgroup\$ – mkeith Oct 12 '16 at 17:21
  • \$\begingroup\$ @mkeith Sounds interesting, I'd love to see a quick schematic so I understand how it integrates with the above circuit I posted. \$\endgroup\$ – JBaczuk Oct 12 '16 at 22:05
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This is the type of thing I have done in the past. It might not be exactly what you are looking for. When USB VBUS goes high, capacitive coupling turns on M2 which turns on M1. Once M1 comes on, R2 pulls up the gate of M2, keeping it on.

A GPIO would drive the gate of M3 high when you want to cut power to the system. This will obviously cut power to the processor itself. It is kind of like "suicide" for the processor.

Some tweaking might be required. Circuits like this can be finicky, but I have gone to production with things like this.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ that is very useful, thanks. Can you explain how the capacitive coupling turns on M2? I've only seen coupling capacitors used to filter out DC and only allow AC signal through. \$\endgroup\$ – JBaczuk Oct 13 '16 at 15:28
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    \$\begingroup\$ When USB is attached, VBUS rises somewhat rapidly from 0 to 5V. This positive edge will cause a rise on the gate of M2. This will keep M2 on long enough for M1 to turn on (we hope), and then the pullup (R2) will keep the gate high after that. It is a type of latch. If you drive the gate of M3 high, it will force M2 off, and M1 will turn off again. You might want to add resistors from gates of M2 and M3 to GND, just in case. Definitely simulate it, prototype it and make sure it is good before putting it in your design. \$\endgroup\$ – mkeith Oct 13 '16 at 20:10

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