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I implemented this on/off circuit that I use with a LiPo battery and a HT7833 voltage regulator. Since my MCU (ESP8266) can only handle 3.3v, I added a voltage divider for the MCU sense pin, just like the original post recommended.
Here is my circuit:
enter image description here

The pin I am using for the BOOT/POWER_HOLD is GPIO0 which has an external 10k pullup resistor (it is a strap pin).
POWER_ENABLE goes straight to the CE pin of the voltage regulator. The voltage regulator requires at least 2V on the CE pin to turn itself on and a maximum of 0.6V to turn itself off. The Battery is also connected to the voltage regulator input pin.

The problem I am facing is that when I hold the power button to shutdown the circuit, I pull the POWER_HOLD pin low but it remains at ~0.06v. The circuit then remains in a limbo state where I have ~0.9V at the voltage regulator output and ~0.06v on the POWER_HOLD pin.
If I manually pull POWER_HOLD pin to ground with a wire, the voltage regulator shuts off completely (0v VDD).
What can I do (if anything) to be able to completely shutdown the regulator using this kind of circuit?

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    \$\begingroup\$ Do you have the same voltage on both sides of D3? \$\endgroup\$
    – MiNiMe
    Oct 26, 2023 at 20:59
  • \$\begingroup\$ @MiNiMe, I tried to measure the voltage at POWER_ENABLE but when I do it reads 0v and it also shutsdown the circuit completely (0v VDD). The multimeter internal resistance must be doing something... \$\endgroup\$
    – ruben
    Oct 27, 2023 at 7:41
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    \$\begingroup\$ You could put a high resistor from PE to GND. What does the anode of D4 read when your MCU signals power off? \$\endgroup\$
    – MiNiMe
    Oct 27, 2023 at 9:13
  • \$\begingroup\$ @MiNiMe BATT_SW is 0v as soon as I release the power button. I tried to manually connect a 50k resistor as you suggested and it does power off the circuit completely! Do you think 50k is a good value? The current on POWER_HOLD pin would be very small (~66uA) so I guess it would be ok. Many thanks in pointing me in the right direction. If you want to post an answe I will accept it. \$\endgroup\$
    – ruben
    Oct 27, 2023 at 12:44

2 Answers 2

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You commented that measuring voltage between PE and GND was enough to shutdown the regulator, a resistor at 1 M Ohm will do the same job to drain the last tiny current flow.

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The behavior you described, where the POWER_HOLD pin remains at ~0.06V, suggests that there may be some leakage current preventing it from going to 0V. You could try adding a pull-down resistor to the POWER_HOLD pin to help it discharge to ground more effectively when you want to turn off the regulator.

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  • \$\begingroup\$ Unfortunately, as I pointed out, POWER_HOLD already has a pullup resistor (it is a strap pin), so I cannot add a pulldown resistor. \$\endgroup\$
    – ruben
    Oct 27, 2023 at 14:26
  • \$\begingroup\$ ah I must've missed that, thanks for pointing it out again. then I believe adding the 1M Ohm resistor will be the choice as @MiNiMe suggested. \$\endgroup\$
    – msi
    Oct 28, 2023 at 18:05

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