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High-side switch for battery protection followed by another for latching

I'm using the attached circuit to control the power supply of my battery-powered product. This is the 8th version; the newest addition is Q1 which I'm using for reverse supply protection. All previous versions work just fine.

The issue is no matter what I do to the latch node, the second FET will not turn off. I've removed C7 to rule it out, and tried to swap R7 for a 1k resistor. I've also cut the trace so that LATCH is floating and then directly applied both VBAT and GND to the latch node, neither has any effect. Measuring the supply pin of Q4 shows 4.62V, at LATCH it's 4.58. I've also removed the diodes.

VIN is going directly into an LDO and nowhere else.

Is there something obvious I'm missing here? I've swapped out Q4 for another MOSFET (still an AO3401 but from a different reel).

How do I get Q4 to turn off?

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    \$\begingroup\$ "I've swapped out Q4 for another MOSFET (still an AO3400 but from a different reel)" AO3400 is N-channel. \$\endgroup\$ Commented Jun 10, 2020 at 2:04
  • \$\begingroup\$ You measured 4.58V at GATE of P channel FET without it turning off? \$\endgroup\$ Commented Jun 10, 2020 at 14:09
  • \$\begingroup\$ Ah I wrote the wrong part number, it is an AO3401. Yes, I measured 4.58V there, even shorted it directly to VBAT and it does not turn off. \$\endgroup\$ Commented Jun 10, 2020 at 14:18

4 Answers 4

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Although you have resolved the problem, but since the actual cause is still unknown, I will suggest the following;

Assuming a defective FET is ruled out, it is very possible that the problem was coming from the LATCH circuit.

Considering that your pull-up resistor for Q4 is 100K, any leakage current in the LATCH circuit will result in a Vgs that can keep the FET on.

For example: The datasheet of AO3401 FET says it has a maximum of 1uA leakage current, therefore using 100K as pull-up resistor with Q4 Gate floating, expect Vgs of Q4 to be about -0.1V at 25 degree Centigrade. This can easily reach -3V if you add-up imperfections in the board and component material.

Therefore, the solution in this case is to reduce the 100K to a much lower value say 10K or less, for reliable Q4 turnoff.

Datasheet info

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On reverse polarity protection:

In the schematic in the question, Q1 still conducts in case of negative input voltages (= reverse polarity) and D1 creates a shortcut. I assume, that this was not intended.

Q1 must be connected the other way round like this:

reverse polarity protection

The Z-Diode is only necessary, if the maximum input voltage (+BATT) can be greater than the maximum Gate-Source-Voltage of the P-FET (usually 20V). [The 47k in my schematic can also be 100k. This should not matter.]

On positive input voltages the internal body diode charges the capacitor. The voltage over the capacitor turns the P-FET on (Vgs must be negative; gate must have lower voltage than source). On negative input voltages, the internal body diode is blocking, thus preventing reverse polarity from doing harm.

On LATCH:

If C7 charges up, Q4 is turned off. So, R7 and C7 must be interchanged. The polarity of Q4 in the schematic in the question is right.

These points should fix your problems.

Do not worry about defective FETs. There are no defective FETs shipped. All of them are right, when delivered to you from the original manufacturer. If you find one, that is defective, assume you broke it and look for the reason.

Good luck with your circuit and have fun!

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  • \$\begingroup\$ I just checked the data sheet of the AO3401. Nice choice! Low On-Resistance even at low voltages, very low Gate Threshold Voltage, small footprint. The only drawback is 30V Vds max, which is a little bit to small for 24V applications. For 5V or 12V supplies it is really great! Assuming that VBAT ist +5V, insert your 5V6 Z-Diode in my schematic and everything should be fine. \$\endgroup\$
    – user242011
    Commented Jun 11, 2020 at 0:02
  • \$\begingroup\$ Where do you buy the AO3401s? DigiKey sells it in quantities of 6000+. Other vendors do not have it in their line card. So, where can I buy this puppy? \$\endgroup\$
    – user242011
    Commented Jun 11, 2020 at 0:11
  • \$\begingroup\$ A bit late here but you can get these at LCSC.com (a chinese supplier). I get most of my prototype runs done through JLCPCB which uses them for turnkey. \$\endgroup\$ Commented Jul 21, 2020 at 1:01
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This may be wrong, but if I am not mistaken a P-MOSFET requires negative (+ thresholds..) voltage on GATE relative to source to turn on. Since your voltage measured on source is almost equal to the gate voltage, it may not have "enough" to turn off.

Also once you connected GND to GATE, Vgs was negative (0 - 4.5 V = -4.5 V) thus the FET was kept in the "ON" state. After connecting the VBAT (since you measured 4.62 V on the source of FET, I expect the VBAT voltage to be something like that) Vgs voltage was (4.58 - 4.62 = -0.04 ), which is still negative.

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I tried a third reel of AO3401's and the board now works. I reworked it several times so it's unclear whether the issue was a defective FET (makes sense - they fail to a short) or the issue that Tomas called out where there is some voltage drop causing Vgs to be slightly negative. In the later case, I may have removed whatever unintended path to ground was present, but I find this unlikely since shorting Vg directly to VBAT did not work.

TLDR for anyone exploring this same problem: make sure the FET is not defective or in a failure mode and make sure there are no unintended paths to ground.

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