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I am putting together a latching circuit and just want to make sure I am not missing something obvious that would prevent this from working. Circuit:Spice circuit

Note: the actual LDO in the design is XC6701B502PR (DS: https://www.mouser.co.uk/datasheet/2/760/TOSL_S_A0003615205_1-2575136.pdf), but using what is already available in SPICE was quicker, should be sufficient for the purpose.

My explanation of what I am trying to achieve:

  1. POWER_SW simulates a switch which, when closed, connects to battery ~14V.
  2. M4 just a level shifter of the switch input down to 5V.
  3. Switch turns on M2 which then turns on M1. This pulls enable to 5V turning ON the LDO.
  4. With LDO ON, the output is fed back to the EN pin via D1. The LDO is now latched and will stay ON regardless of what the power button is doing.
  5. When POWER_SW is pushed down for a number of seconds, this detected by a controller and GPIO(SHUTDOWN) is pulled HIGH by the controller. M3 is turned on, EN is pulled LOW, LDO turns OFF with the rest of the system.

Running the sim seems to do what I expect:Sim result

A few questions I have:

  1. Am I missing something obvious here or will this work?
  2. Better/easier to achieve the same result?
  3. Why is output pulled to around -1V when my switch is off?

Appreciate any help or comments.

Cheers

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  • \$\begingroup\$ The output is pulled down because you are using a current source as your load. You should just place a resistor. \$\endgroup\$
    – Edin Fifić
    Commented Apr 18, 2023 at 15:20
  • \$\begingroup\$ Looks OK to me, the -1 V comes from the curent source added to the output :) \$\endgroup\$
    – Jens
    Commented Apr 18, 2023 at 15:20
  • \$\begingroup\$ Instead of detailed explanation how each of your components function, you should just give a basic idea of what is the purpose of your circuit and what are you trying to achieve. Explaining into detail what each component does and why is not a why to state briefly what you want to accomplish, while we have no idea what will your circuit be used for. Right now it is just hard to follow your main idea/purpose of your circuit, as you lose your audience in too much detail without the most basic outline. \$\endgroup\$
    – Edin Fifić
    Commented Apr 18, 2023 at 15:25
  • \$\begingroup\$ @EdinFifić Oh, right, thanks! \$\endgroup\$
    – Zygis
    Commented Apr 18, 2023 at 15:37
  • \$\begingroup\$ @EdinFifić, sure. Battery powered device. The LDO output enables a more powerful SMPS down the line. Powers a fairly beefy processor which takes time to boot up. Want a circuit that will turn ON immediately after button press and will stay ON even if the button is released. The OFF function must be more controlled and initiated by the processor itself. \$\endgroup\$
    – Zygis
    Commented Apr 18, 2023 at 15:40

1 Answer 1

Reset to default
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I do not understand the purpose of R8 resistor. In my view, you should remove it.

When the power_sw is OFF, POWER_SW_5V can be pulled high because of R8 and this can keep EN high even after getting a SHUTDOWN pulse. It is working because M2's VTH might be higher than the voltage on POWER_SW_5V but, why take the risk.

Edit: Another minor point, you need to have a pulldown resistor on SHUTDOWN so that when the micro-processor is not powered, this signal is defined to 0V.

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  • \$\begingroup\$ I just searched "MOSFET level shifter" and that is the circuit that came out, didn't think too much. I think you are right that this only works with high enough Vth and the right resistor values. Removing resistor from the board after the fact is quite easy, I'll keep it in as a placeholder. Yes, missed the pull-down on SHUTDOWN, will fix. Thank you. \$\endgroup\$
    – Zygis
    Commented Apr 19, 2023 at 9:46

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