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Now, please correct me if I'm wrong here.  I'm still trying to wrap my brain around these things ...  Specific to N-channel ones:

If the gate is 0V, the resistance between the (D)rain and (S) is pretty much nothing causing them to conduct.

If the gate is a specific voltage (which I have yet to figure out), the resistance between (D)rain and (S)ource goes high, effectively cutting off conductivity.

Yes?  No?
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Opposite way around

N-ch
Voltage on G (relative to S) = ON
0v on G = OFF

The actual voltage required depends on the FET, look at the VGS(th) parameter (the threshold voltage required to turn the FET on). However most FETs will not turn on properly with a VGS of 5v, in your case you need a "logic-level" FET. The 2N7000 doesn't actually have the magic words "logic-level" in the data sheet but it's VGS(th) is 3v so it might be OK. Looking at the graphs with a VGS of 5v it can switch about 750mA, more than enough to control the P-ch.

You use this FET to control the gate of a P-ch...

P-ch
Voltage on G < voltage on S = ON
G volts same as S volts = OFF


So when the N-ch is ON it switches the P-ch gate to 0v which is < the P-ch D volts, therefore P-ch is ON.
So when the N-ch is OFF the P-ch gate is pulled to the P-ch D volts with a resistor, therefore P-ch is OFF.

I'll see if I can come up with a schematic.

EDIT: Fixed error on the P-ch description.

______
Rob
 
 
« Last Edit: November 04, 2011, 07:01:44 pm by Graynomad » Logged

Rob Gray aka the GRAYnomad www.robgray.com

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This is how I see it.

I don't remember what your VIN was, as R1 and R2 form a divider if VIN is too low and Q1 has a high VGS(th) there could be an issue.

______
Rob



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Rob Gray aka the GRAYnomad www.robgray.com

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There are of course "high side switches" like the MIC94070 (only switches up to 5v5 but there may be higher voltage versions) to switch the main current and don't bother with the regulator enable.

It may be easier to use one of them.

______
Rob
« Last Edit: November 04, 2011, 07:07:25 pm by Graynomad » Logged

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Vbat starts at 6V.  The regulator is fixed at 3.3V, so the Atmel is running at that voltage (or at max that voltage.)  At 8MHz, the thing ran solidly for 15 hours before it finally "got stuck".  Measuring voltage then, it was measuring 3.1V coming straight off of the batteries, regulator was providing anywhere between 3.0 and 2.7V ... which is about where things start to go wonky and I'm perfectly okay with that.  At the first signs of the Atmel quitting (which it will), it just needs to shut stuff off.

So, looking at your schematic, will the above tolerances work for the FETs?  And also, where do I "kick-start" the regulator?  As in, where would the momentary switch get wired to so I can provide temporary power to the ENable pin of the regulator till the Atmel starts running and brings its pin HIGH to maintain the regulator?
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Note that this circuit is untested and designed by a non FET expert (that's me), do not bet the house on it. smiley

With a VBAT of 6v that R1/R2 divider will produce approx 550mV on the Q1 gate. That's 5v45 lower than the Q1 source so you need a P-ch FET that will be turned on with a GVS(th) of say -5v. As Q1 is only driving a high-impedance input it doesn't have to turn on very hard, I would think most small FETs would do. I don't have much data on small FETs but something like the IRLML5203 seems appropriate.

Quote
where do I "kick-start" the regulator?
As Q2 turns Q1 on you need to emulate Q2. A switch across Q2 should work.

Note that VBAT should not be higher than 20v with the IRLML5203 (max VGS is 20v).

______
Rob
« Last Edit: November 04, 2011, 08:37:33 pm by Graynomad » Logged

Rob Gray aka the GRAYnomad www.robgray.com

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