I’m currently designing a Li-Ion powered ESP32 circuit.

The ESP32 has a minimum operating voltage of 3 V. So, my instinct is to use a 3.1 V supervisor to keep the ESP32 within an acceptable range.

After the supervisor is triggered, I’d like to keep it active until the user turns the whole circuit off / on again. Without this, I’m afraid that the voltage of the Li-Ion cell might rise to the open circuit voltage, which would disable the supervisor and cause an endless loop of un-resetting / resetting.

I found this article from TI that describes how to “latch” a supervisor. At a high level, I think it makes sense. But it seems like a relatively complex approach to, at the end of the day, connect RST to MR.

So now I’m wondering — could I directly connect RST to MR, without any additional circuitry? My instinct is that, if the RST pin pulls low, it would trigger MR (which would trigger RST again, and so on), effectively keeping the supervisor in a reset state.

I’m very new to electronics, so I apologize if this is a silly question or if I’m misunderstanding any of these concepts. I think I’m missing a key piece of insight.


1 Answer 1


No, there is no simpler way. Think about what happens when you want to turn your circuit on: The supervisor pulls RST low for a short amount of time until the supply voltage has stabilized. If RST and MR were connected, the IC would immediately assert its own MR input and latch off. That's why you need the external circuitry: It connects RST to MR only after RST went high for the first time. Otherwise you wouldn't be able to turn your circuit on at all.

TL;DR: The external transistors make sure you can turn the circuit on in the first place.

  • \$\begingroup\$ Aha! That makes sense. Thanks for the clarification. \$\endgroup\$
    – Miles
    Dec 27, 2021 at 15:21

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