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I'm looking to supply an MCU (3.3 V) from battery via voltage regulator: a pushbutton should enable the supply, the MCU then should keep the supply enabled until it wants to shut itself down. But the pushbutton should be dual-purpose, i.e. it should also work as a normal button readable by the MCU when it is powered.

I found STs STNS01 which integrates an LDO (3.1 V), a Lithium battery charger (4.2 V), and power path switches (supply or battery) and features a shutdown pin (active-high, but <4 V, with internal 500k pull-down) that disconnects the battery resulting in microamp quiescent current.

I don't get how the shutdown would be used without resorting to a voltage divider referenced to the battery voltage (~600 k pull-up on SD to 4.2 V should result in 1.6 V). For one, this would result in current (~ 4.3 uA), but also I can't think of a way to decouple the MCU/pushbutton from the battery voltage, i.e. when the regulator is shut down, the high-level on SD would cause currento to flow into the MCU supply via the protection diodes.

Using separate IO to control SD and read the pushbutton would be OK. Also, I'd think this could be done with a few transistors... but I don't see it :-/

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  • \$\begingroup\$ I'm not really sure what you're asking and I can only assume what you mean by a "normal button". Would you clarify please? \$\endgroup\$ – DiBosco Feb 14 '18 at 14:38
  • \$\begingroup\$ Pressing the pushbutton should A) enable supply directly to power the MCU and B) be connected to MCU so button presses can be detected by MCU without affecting supply directly. \$\endgroup\$ – handle Feb 14 '18 at 15:47
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Using the shutdown pin you would need to add some diodes and a MOSFET to do what you want. However, as you have already guessed, the pull-down in the LDO adds to your quiescent current when it is off.

schematic

simulate this circuit – Schematic created using CircuitLab

A better solution would be to add in a P-MOSFET and shut off the regulator entirely.

schematic

simulate this circuit

How that ties into the charging part though I am not sure.

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  • \$\begingroup\$ Thanks. By what means would M2's gate be charged in order to turn off? \$\endgroup\$ – handle Feb 14 '18 at 16:08
  • \$\begingroup\$ @handle when you press the button the LDO and MCU will power up. The latter should then drive the POWER_ON line high to keep the power on when you let go the switch. When the MCU is done either drive POWER_ON low, or shut off the output. \$\endgroup\$ – Trevor_G Feb 14 '18 at 16:14
  • \$\begingroup\$ I understand that, but would M2 not stay on as the gate-source voltage does not approach zero (unless through an additional pull-up)? \$\endgroup\$ – handle Feb 14 '18 at 16:27
  • \$\begingroup\$ @handle oic..yes I missed the pullup... lemme fix \$\endgroup\$ – Trevor_G Feb 14 '18 at 16:31

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