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I can't change bulb wiring so I used P-channel MOSFET to PWM bulbs. To save the space I used gate driver and in order to save battery life all ICs are powered from ignition (interruptible 12V). MOSFETs' source is powered from separated fused 12V line.

Mosfet connection

Driver's output stage is tied to the ground so I added pull-up resistor to the gate to keep it off when the driver is off. enter image description here

The problem is MOSFET is fully on when the driver is powered off. Vgs is -2.41V. Should I unsolder the driver - MOSFET is off.

When the driver is powered - everything works as expected Can't get what am I doing wrong here.

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    \$\begingroup\$ That diode across the high-side NPN may be the culprit- combined with D2,D3 the pull - up on the FET gates will dribble current into V+ Of the driver chip when it’s not getting power from elsewhere. \$\endgroup\$
    – Frog
    Apr 26, 2021 at 0:04
  • \$\begingroup\$ But do you ever have power on +12V and not on IGN_KEY? \$\endgroup\$
    – Frog
    Apr 26, 2021 at 0:06
  • \$\begingroup\$ According to the datasheet, the irf9310 can have a threshold voltage as low as 1.3V. I do not like the fact that R4 is 10K, while the internal resistor of the driver is 100K. With a 12V supply, the drain through the driver will bring you close to that 1.3V. I would cut R4 approximately in half to 4.7K. The driver should be able to handle that. \$\endgroup\$
    – Math Keeps Me Busy
    Apr 26, 2021 at 1:17
  • \$\begingroup\$ @Frog yes, it is powered. I'm testing it on the table. I may consider removing turn-on diodes for the test purposes. \$\endgroup\$
    – pugnator
    Apr 26, 2021 at 6:01

2 Answers 2

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Looking at your problem in detail, I think you should rework the circuit like this:

schematic

simulate this circuit – Schematic created using CircuitLab AC coupling might be the simple solution to this.

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  • \$\begingroup\$ Would you explain it a bit? I'm almost desperate not able to find the root cause and ready to do it in a simple NPN-as-a-gate-driver way \$\endgroup\$
    – pugnator
    Apr 27, 2021 at 14:31
  • \$\begingroup\$ Your circuit is backfeeding in DC. Once it is ac coupled, it will break this connection. I don't know what kind of signal you are feeding the circuit. But if it is pwm this will definitely work. Another thing you can try, because the impedance is low on a power supply while off, is to put a diode in series with the Vcc of the predriver chip so that this low impedance is not exerting itself through the circuit. If the output device was reworked to be a n-channel enhancement mode mosfet, there wouldn't be much issues, as the negative loading effect would gate the mosfet off. \$\endgroup\$
    – David Mikeska
    Apr 27, 2021 at 15:21
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Vgs is -2.41V. Should I unsolder the driver - mosfet is off.

Cut R4 and R15 from 10K to 4.7K or even 2.2K. That will lower the absolute value of Vgs. According to the datasheet the IRF9310 can have a threshold voltage as low as 1.3V. The MC34141 should be able to handle the change in pull-up resistors.

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  • \$\begingroup\$ Just tried 4k7. With 12V I measured -2.11V on the gate and it's still on. I'l try 2k2 later but looks like I have to do it trial and error way. But if I'm correct, R4/R15 is a voltage divider together with pull-down resistor of the driver. Then decreasing pull up resistor I will decrease voltage on the gate? And I checked several gate driver - is there some low-side drivers with pull up instead of pull down? May be it's a better approach. \$\endgroup\$
    – pugnator
    Apr 26, 2021 at 5:51
  • \$\begingroup\$ I'm wondering if your ignition key is set up like a "kill switch". Instead of simply disconnecting the 12V, it actually grounds that rail, (to prevent theft via "hot-wiring"). Yes, a different driver might make the necessary difference in your case. \$\endgroup\$
    – Math Keeps Me Busy
    Apr 26, 2021 at 11:45
  • \$\begingroup\$ Due to the negative drive, the pull up resistor is backfeeding the predriver circuit, because the Vcc side of the perdriver rail is acting like a low impedance load. \$\endgroup\$
    – David Mikeska
    Apr 27, 2021 at 15:45

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