I am trying to understand:

1) how the following circuit works and more specifically how it turns on.

2) How to to modify it to get a hold on/hold off behavior instead of push on/hold off to prevent accidental turn on.

enter image description here

It is a circuit found and partially documented here. I have tested it in the auto off position and it works as expected.

I do understand how the circuit switches off but not how it initially turns on. How is the P-MOSFET gate pulled low when the button is pushed?

EDIT: I have designed a circuit on faltstad here for simulation. I tried Mohamad's idea but it didnt work. So the question remains: how to have a hold-on behavior, without loosing the auto-off (circuit initially off when powering it)?

  • \$\begingroup\$ Try looking just at R1, R2 and C1 on their own and that should give you a good idea of how the circuit initially turns on. Falstad simulation link \$\endgroup\$
    – user103993
    Dec 17, 2018 at 11:26
  • \$\begingroup\$ You would build a soft on - soft off toggle instead of trying to copy and paste a circuit not designed to do that, and trying to modify it. \$\endgroup\$ Apr 16, 2021 at 19:09

1 Answer 1


1)Before pushing the button When the P-Mos is off the voltage across the C1 is zero (Vc1 = 0) due to R1 and R2 pull ups. When you push the the button the left side of the C1 becomes ground. The voltage across a capacitor can't change so quickly therefor the right side of the C1 should have the same voltage to keep Vc1 = 0. When this happens the P-Mos gate goes low and turns on the circuit and after that the N-Mos turns on and keeps the gate of the P-Mos low.

2) In order to have a hold on function you need to make a time constant, an RC. Put a large capacitor between the gate of the P-Mos and ground and also a series resistor with the button like 300k which gives you the same 3s hold on time.

  • \$\begingroup\$ 1)thanks! 2) doesnt work and it would loose the auto-off behavior \$\endgroup\$
    – AF14
    Dec 18, 2018 at 12:44

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