Can anyone help with this?

With a PFET, the source-drain diode conducts when the source is more positive. You need to exchange drain and source.

Cheers

Phil Hobbs

Belay that—I shouldn’t post after two G&Ts.

"TTman" snipped-for-privacy@gmail.com wrote in message news:10vnl0a$38eli$ snipped-for-privacy@dont-email.me...

A 3.3v variation on the circuit optimised by piglet some time ago may work better for that if pushbutton on/off is what you want.

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SYMATTR InstName V3 SYMBOL sw -304 1168 R0 SYMATTR InstName S3 SYMATTR Value SWCH SYMBOL voltage -448 1264 M0 WINDOW 3 59 141 Left 2 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value PULSE(0 2 {.501/F} 1u 1u 1m 2m 5) SYMATTR InstName V4 SYMBOL npn 128 432 M0 SYMATTR InstName Q3 SYMATTR Value BC846B SYMBOL res 48 -112 R0 SYMATTR InstName R11 SYMATTR Value 150k SYMBOL npn 544 432 R0 SYMATTR InstName Q4 SYMATTR Value BC846B SYMBOL res 208 368 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R15 SYMATTR Value 470k SYMBOL res 576 368 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R16 SYMATTR Value 470k SYMBOL cap 320 144 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C1 SYMATTR Value 100n SYMBOL cap 416 176 M270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName C2 SYMATTR Value 100n SYMBOL res 592 -112 R0 SYMATTR InstName R17 SYMATTR Value 100k SYMBOL res 592 32 R0 SYMATTR InstName R18 SYMATTR Value 10k SYMBOL diode 208 368 M180 WINDOW 0 42 50 Left 2 WINDOW 3 -97 51 Left 2 SYMATTR InstName D5 SYMATTR Value 1N4148 SYMBOL res 576 144 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R20 SYMATTR Value 470k SYMBOL diode 464 368 R180 WINDOW 0 42 50 Left 2 WINDOW 3 -96 52 Left 2 SYMATTR InstName D6 SYMATTR Value 1N4148 SYMBOL res 96 144 M90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R21 SYMATTR Value 470k SYMBOL diode 208 592 M180 WINDOW 0 42 34 Left 2 WINDOW 3 -96 35 Left 2 SYMATTR InstName D8 SYMATTR Value 1N4148 SYMBOL diode 464 592 R180 WINDOW 0 43 30 Left 2 WINDOW 3 -98 34 Left 2 SYMATTR InstName D9 SYMATTR Value 1N4148 SYMBOL res -112 -112 M0 SYMATTR InstName R13 SYMATTR Value 470k SYMBOL res -112 480 M0 SYMATTR InstName R22 SYMATTR Value 470k SYMBOL res 32 64 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R23 SYMATTR Value 220 SYMBOL pmos 848 -80 M270 WINDOW 0 -19 56 VLeft 2 WINDOW 3 76 99 VLeft 2 SYMATTR InstName M3 SYMATTR Value AO3423 SYMBOL res 928 176 R0 SYMATTR InstName RLOAD SYMATTR Value 1k TEXT -912 584 Left 2 !.tran 10 startup TEXT -912 704 Left 2 !.param F=0.5 TEXT -912 744 Left 2 !.MODEL SWCH SW(Ron=.1 Roff=100Meg Vt=1) TEXT 696 816 Left 2 ;EPW SED OCT 2025 TEXT 576 880 Left 2 ;INSPIRED BY EDWARD RAWDE TEXT 544 944 Left 2 ;TOGGLE POWER ON PUSH BUTTON TEXT -280 936 Left 2 ;BOUNCE ON MAKE TEXT -240 1216 Left 2 ;BOUNCE ON BREAK

You were right the first time.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Might be tricky. Most FETs are symmetrical, and for a P-FET the source is whichever of the source and drain connections that happens to be more positive.

"TTman" snipped-for-privacy@gmail.com wrote in message news:10vnl0a$38eli$ snipped-for-privacy@dont-email.me...

Phil was right first time and it won't work without a load (R4 in the simulation below).

This site has various variations of it.

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If you are going to drive a microprocessor then let that do the on-off thing: Like this one:

press once -> on press 3 times: off If you program those PICs right then off-current is in the nano nano amp region. Still working 100% on same Lipo after more than 10 years.

Circuit diagram:

Code for the PIC 18F14k22: see line 1000: asm is fun :-)

Yes, I already did that years ago with a tiny 25. But this is a 'discrete' design. No micro

But my 'load' side can never be more positive than the battery side...

So I need to flip source and drain? I'll try that...

Yes but you forgot the parasitic body diode.

piglet

As Phil pointed out the parasitic body diode in the PFET was letting current flow.

Also beware that to switch off reliably the R3 * cap 1uF time constanr needs to be longer than the load capacitance TC takes to decay - so either increase that cap or decrease the amount of capacitance across the load.

piglet

Thanks, I have no recollection of that! I think unsuitable for battery use as the standby current is high. Looks like an excercise in Eccles-Jordan maximal complexity!

pigley

Thanks, that mosaic industries website provoked some ideas, here is my take on the "press-on, hold-off" circuit

piglet

Don't bother. The FET decides which terminal is source or drain on the basis of the voltage at each connection.

Even asymmetrical FETs work that way, but the performance can vary a bit if the areas playing the role are bigger or smaller.

In that case please explain why reversing the source and drain in the simulation I posted results in a circuit which no longer works.

Yes I know the explanation is obvious but you seem to be implying that the FET will work either way in that circuit. This is incorrect so telling the OP not to bother doing that is nonsense.

Phil gave the correct solution . The only other thing I would note is the way the load characteristics may affect operation but this is explained on the web page I posted.

He forgot about substrate diodes.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

You want to use a microprocessor for a push on/off power switch? I'd have to admit I know people who might. And if there's already a processor in the system then maybe sometimes it makes sense. The usual process is:

1. Connect the switch to an input port with a pull up resistor if necessary.
2. Write software which logically will work perfectly.
3. Have trouble understanding what is going wrong.
4. Discover switch bounce.

The simulation uses a defective model for the FET

Why do you think that it is incorrect? The simulation isn't evidence of what the actual part will do.