Soft Switch Design Question

[kenny15963]

Put together a soft switch for a small project.  Circuit is similar to many I've seen all over the internet. Having a small issue with my implementation though.  When I'm applying the 9v, the circuit should be "off" but about 90% of the time it starts in the "on" mode.  Not a huge issue for this particular application but would like to know how to ensure it starts in the "off" mode.

More info on what I'm doing with this. Using the output to power 3.3v for an Attiny as well as a 9v rail to drive some LEDs.  The "off" signal comes from an output pin on the Attiny going high (current limiting resistor not shown.)

[kurt]

The instant you connect power to the circuit you will have a capacitive voltage divider made by the gate capacitance of the BS250P and the drain-source capacitance of the 2N7000. If the drain-source capacitance are to large compared to the gate capacitance you will pull the gate of the BS250P down enough to turn it on.

I guess this could be fixed by adding more gate capacitance. I.e. putting a small capacitor in parallell with the leftmost 100K resistor (component designators would make it easier to talk about the circuit BTW  ).

[kenny15963]

Thanks for the response. I tried your suggestion with no success.  It did get me thinking about the capacitance of the gates though and I think I figured out what was happening from that.  The P-MOSFET will initially be in a conductive state. Due to the capacitance of the gate, there will be a short time for the gate to charge up through the 100k resistor where the P-MOSFET will remain on.  The gate capacitance of the N-MOSFET used is smaller than the gate capacitance on the P-MOSFET. So the N-MMOSFET is winning the "gate charging race" most the time and keeping the P-MOSFET gate low on initial power-up.  I added a small cap between the N-MOSFET gate and ground which seems to have fixed the issue.

Sorry for the lack of designators.  Feeling the awkwardness of that mistake as I type this response. I'll be sure to include them in the future (I am trainable...)

[kurt]

Right. I guess I missed something. And I think you have the right solution, but that you conclusion for why it worked is a bit off 

The P-MOSFET will not initially be in conductive state, since it needs a gate charge for that to happen.

But you have another capacitive / resitive voltage divider consisting of the P-MOSFET Csd, the rightmost 100k resistor and the N-MOSFET Cgs. If Cgs charge to the threshold voltage before Csd has eaten enough of the voltage you will have the N-MOSFET turn on.

So your solution of adding more gate capacitance to the N-MOSFET should solve this issue.