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Schematic This is a very simple (or so I thought) power switching circuit I made for an extremely low, power consumptive device.

The device normally runs on a 3.6v lithium, but occasionally a user will plug it into a computer to grab/program data. When plugging into computer, I want to kill power from the battery and solely power from USB interface. When USB is disconnected, MOSFETs make the switch back to battery.

Both MOSFETs are P Channel.

When I first built the circuit, I left Q2 out to check if Q1 was working as expected. It was. When the USB is not plugged in, Q1 gate is 0v, and source is 3.6V. Negative 3.6V fully turns on the MOSFET. When I plug in USB, interface IC turns on, and drives Q1's gate to 3.3v which turns it off just fine.

But when I added Q2 (with no USB power), Q2 doesn't want to turn off. The VGs should be +3.6v, so it should drive it hard off! Instead, it seems like it's partially on, back feeding the voltage regulator; I'm reading 2V on the reg's input. Then when I plug in USB, now Q1 wont turn off!

Here's my thought's on what should happen:

With a battery, but no USB power, Q1 Vgs should be -3.6v which turns on Q1. When Q1 is on, it feeds the gate of Q2 with +3.6v while the source is 0v because theres no USB power. So Q2 should be off.

When USB power is added, Q1's gate is driven high by the USB IC which brings the Vgs to -0.3v which is plenty to keep Q1 off (confirmed through testing). When Q1 turns off, Q2's gate is 0v and the source is +3.3v. This gives Vgs -3.3v which should be driven fully on.

Any help would be awesome. Thanks guys! MOSFETs data sheet

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  • \$\begingroup\$ Your Gate-Source shunt resistor is connected wrong and going to ground instead of Vbat. Also shown upside down. For Pch . Vgs R is gate to Vbat and for Nch Gate to Gnd \$\endgroup\$
    – Tony Stewart EE75
    Dec 12, 2019 at 19:04
  • \$\begingroup\$ Those are not shunt resistors. They are pulldowns. I want to make sure that the gate is pulled to ground when there's no voltage on either gate, they're supposed to be connected to ground. Is there something I did wrong? \$\endgroup\$
    – James Pie
    Dec 12, 2019 at 19:46
  • \$\begingroup\$ Yes read again.. Its wrong . To turn off Pch you must use a Pull Up gate. Even though the current is low, I still call it a shunt to turn off Gate with an open circuit \$\endgroup\$
    – Tony Stewart EE75
    Dec 12, 2019 at 20:07
  • \$\begingroup\$ Okay I see what you're saying, but again, I removed BOTH pulldowns and the problem still persists. What you're recommending wont help. The gates (without anyresistors) are still pulled high. With no USB power, Q2's gate is 3.6v (from battery). And with just battery power, you wouldnt want to pullup Q1, because That will bring Vgs positive when it needs to be atleast -1v. \$\endgroup\$
    – James Pie
    Dec 12, 2019 at 20:25
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    \$\begingroup\$ @JamesPie "I removed BOTH pulldowns and the problem still persists." Yeah, but you didn't remove the load, which is in parallel with the pulldown for Q2. So removing that pulldown won't affect the behavior of the circuit. \$\endgroup\$
    – WhatRoughBeast
    Dec 12, 2019 at 20:32

1 Answer 1

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Q2 leakage is probably through the body diode of the FET, which is not shown on your schematic. I think a bigger problem is that the gate and drain of Q2 are connected together, meaning you can only get within Vgs(th) of your 3.3V supply. And Tony's right, if those resistors are meant to turn off the FET in a floating gate situation, they should be connected between gate and source.

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  • \$\begingroup\$ "you can only get within Vgs of your 3.3V supply". I'm sorry, I'm confused by what you mean. Could you elaborate please? \$\endgroup\$
    – James Pie
    Dec 12, 2019 at 20:37
  • \$\begingroup\$ I meant Vgs(th)...with the gate connected to the drain, while the 100K is pulling your gate down the drain is pulling it up when the FET starts to conduct. It can't pull it up past 3.3V+Vgs(th) without turning off the FET. For this FET, the datasheet shows Vgs(th) to be -1V to -3V at 25C. \$\endgroup\$
    – Cristobol Polychronopolis
    Dec 13, 2019 at 13:54
  • \$\begingroup\$ YES the diode body! I completely forgot about that. I cant "hold" voltage against the drain pin. It will conduct through the source. That was the problem. Thank you \$\endgroup\$
    – James Pie
    Dec 13, 2019 at 15:32

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