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I know it's not a good practice to use a voltage divider as a power supply, but I'm considering it for an unusual circuit. Is this a good use of a voltage divider or is there a better way to do this I'm not considering?

This is a latching power shutdown circuit. The purpose is to power down the device after it is idle for some time to avoid battery drain. It is triggered via the IDLE_SHDN signal, 3.3 logic level coming from an MCU. The MCU itself and all voltage rails power down when this is triggered. The mosfet pulls down on the SHDN pin of a LT4356CMS-1, which is upstream of all power supplies, resulting in a total powerdown of the entire board.

If I read the datasheets correctly, both the latch and mosfet gate current needs are very low, so I can get away with just a voltage divider to power them.

The other main consideration is the supply line is really unstable:

The input line VIN is coming from a battery that is connected to an automotive alternator. When the engine is not running the power is 12V (though I allow as low as 9 for my circuits) - this is the scenario when the idle shutdown is needed, when it is running it's very noisy around 15.5 volts. I'm using a TVS diode (SMBJ15CA-C78410) with 16.7V breakdown and 24.4 clamping to protect VIN from transients.

My main concern is whether the voltage divider will have enough power to drive the mosfet while staying in the range needed so the latch doesn't get killed by transients when the engine is running. One question is do I correctly read the NOR gate datasheet that the output voltage may be lower than the input?

enter image description here

Update Based on feedback, here is the design using an automotive LDO with latching enable. I added in the reverse battery/transient circuit it works with for reference since there were some questions in that area. I ultimately decided to have the TVS kick in at a much higher level based on brhans' concern. I feel much better that this design will be safe in a much wider range of conditions.

enter image description here

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    \$\begingroup\$ Why is your '2A' NOR gate input floating? If you don't need it then tie it to '2B' or GND. \$\endgroup\$
    – brhans
    Commented Dec 4, 2021 at 14:46
  • \$\begingroup\$ And are you sure a 15V TVS diode will protect your circuit in an automotive environment? I suggest you read up on 'load dump' - it's possible for your supply voltage to briefly spike up to 50V or more, with the full 'force' of the alternator behind it to drive lots of current through your little TVS, turning it into a smoking cinder - or maybe just blowing the fuse of the circuit you've got it connected to ;). \$\endgroup\$
    – brhans
    Commented Dec 4, 2021 at 14:51
  • \$\begingroup\$ @brhans From wikipedia's load dump page "Special protection devices, such as TVS diodes, varistors which can withstand and absorb the energy of these spikes may be added to protect such semiconductor devices." I'm using the TVS as a first line of defense so I can use a power mosfet with a lower drain-source breakdown voltage with the LT4356CMS-1 as my main transient protection circuit \$\endgroup\$ Commented Dec 5, 2021 at 3:39

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No matter what you're doing, you want a capacitor on your VCC. This would already go far as to stabiliing the supply. Otherwise, you'd need to source the full gate charge current through your 90 kΩ resistor, and that current can be very high, even if just for very shortly.

The rest of your arguments are all actually against a simple voltage divider, not quite sure why you think saying "my input voltage fluctuates so that a linear divider fluctuates, too" suggests that said divider is a good idea.

So, honestly, use sufficiently much capacity on VCC to begin with.

Then, calculate how much energy charging Q5's gate might take - and compare that to the energy in your capacitor. Rule of thumb: if that is two orders of magnitude larger, you're fine; you can ignore the current through R72 for the duration of charging Q5's gate, it will be negligible in comparison. Calculate how long it will take to charge your capacitor through your R72. I bet it's longer than you actually want to wait.

Most likely, you'll end up realizing that a 10 or 100 nF capacitor and a cheap linear regulator is simpler, and more space / work / partcount efficient.

Most linear regulators come with very low standby current consumption (hint: try to not use regulators that start with LM31x or LM8xxx, these are from the sixties). Many even come with an (inverted) enable pin that you can use to even further reduce power consumption. This might even completely obsolete your whole U31/Q5!

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  • \$\begingroup\$ I see your point, maybe the voltage divider could work but the LDO is more reliable for nearly the same part count and cost. Good call on standby current. I don’t think I can get rid of U31/q5 unless there’s one with latching enable \$\endgroup\$ Commented Dec 4, 2021 at 13:17
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    \$\begingroup\$ you can build one with latching enable by feeding the output to the input through a high-ohmic resistor (if non-inverted enable) or through a pulling-down transistor if inverting \$\endgroup\$ Commented Dec 4, 2021 at 14:10

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