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Topic: Low power shutdown/failure/t'is quittin' time ... (Read 2 times) previous topic - next topic

KirAsh4

Ok, I'm back after thoroughly sleeping off the effects of too much wine ...  In the mean time, I received another suggestion and I decided to dive into that and see if I can figure it out.  It's all about experimenting, right?

Anyway, the suggestion was to use a switching FET to turn the regulator on/off.  So I came up with the schematic below.  By the default, the regulator must remain off.  When the MCU's pin goes HIGH, it closes the gate and Vbat flows into the ENable pin for the regulator and turning it on.

The way this ought to work is:

  • By default the regulator is off, Vbat is +6V.

  • Pressing the momentary switch will feed Vbat directly into the regulator's ENable pin, turning it on (this is fine, it can handle up to +17V)

  • With the regulator now on, it feeds power to the rest of the circuit, including the MCU.

  • MCU starts up, and brings its pin connected to the FET's gate to HIGH.

  • FET sees this and closes the gate, enabling Vbat to flow through the FET on to the regulator's ENable pin

  • When I see the MCU operational (it has a blinky LED), I can then let go of the momentary switch and all remains running.

  • When Vbat drops past the MCU's operating threshold, it shuts off, taking it's one pin LOW.

  • FET's gate goes down, opening and thus cutting power to the regulator (and thus also the rest of the circuit).



Yes?  (this is one of those moments where in my head I'm going 'please oh please oh please tell me I got it right and didn't screw up horribly')  :)

If that's all fine, I have three questions:
a) Is what I came up with actually correct for how the FET works?
b) How do I calculate the value of R3?  Again, keep in mind the gate should be open except when the MCU pin tells it to close.
c) Does this mean I can now safely put a momentary switch across Vbat and the ENable pin of the regulator to kick-start it?

KirAsh4

Nevermind, I think I have that backwards ... back to studying.

Graynomad

I don't think you can use an N-ch FET like this. When the MCU switches it on the source voltage will rise to > than the gate voltage. I'm not sure what this will do but it doesn't sound right.

______
Rob
Rob Gray aka the GRAYnomad www.robgray.com

KirAsh4

Yeah, I caught on to that. I think I need a second "stage" FET to reverse the logic ...  Still learning.

Graynomad

Yep, use a p-ch to do the high-side switching and an n-ch (or npn) to switch the p-ch.

______
Rob
Rob Gray aka the GRAYnomad www.robgray.com

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