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As title, here's a schematics:

schematic

simulate this circuit – Schematic created using CircuitLab

Is this a correct way to create a power switch or can this schematic be improved?

Board will sink at max. 1A , P-MOS and N-MOS will me chosen to be rated +/-20Vgs.

I've added two voltage dividers due +24V power source to keep Vgs at 16V. There should be a very low dissipation through resistors.

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  • \$\begingroup\$ R1 and R2 should be swapped if you want Vgs to be 16V. Right now Vg is at 24V * 10K / 14.7K = 16.3V. But Vgs is at 24V - 16.3V = 7.7V. \$\endgroup\$
    – user4574
    May 21 '20 at 16:13
  • \$\begingroup\$ Commercially designed high-side switches are available because they meet quite rigid requirements specifications. You haven't given any extensive requirements specifications so who knows if the 0.3 volt drop when the P channel device is feeding 1 amp to the load is acceptable or not. Switch on and switch off times have't been considered nor fusing requirements should the PMOS fail short. Over current protection is usually a requirement on a lot of designs as well as under-voltage lock-out to prevent the PMOS burning if the supply voltages start to fall excessively. \$\endgroup\$
    – Andy aka
    May 21 '20 at 16:34
  • \$\begingroup\$ So, on that basis, your question can't really be answered as it stands. \$\endgroup\$
    – Andy aka
    May 21 '20 at 16:34
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This circuit is probably close to correct, but I can't tell without knowing what kind of load you are driving.

Most loads will cause the switching transistor to dissipate a pulse of energy during turn on. The transistor you selected can handle a pulse of 400mJ which makes it appropriate for use as a load switch. It has an on resistance of 0.3 ohms, which may be appropriate for a 1A load (although lower resistance wouldn't hurt).

Some things you should consider are...

1) How much capacitance does your load have? For example if you use this to turn on another power supply or some circuit card then most likely there will be capacitance on the front end of that setup.

a) During turn on M1 will dissipate 0.5 * C * (24V)^2 of energy. So 288uJ for every 1uF of load capacitance. Look at the safe operating area graph of the IRF9530, if the turn on time is several ms then you can dissipate up to 400mJ, which limits you to a load of 1.38mF max.

2) Your circuit doesn't control the rise rate at turn on. If your load has any capacitance whatsoever then its possible that you could exceed the 48A pulsed drain current rating.

You can fix this by adding a capacitor between the drain and gate of M1. The turn on rise rate then becomes...

dv/dt = ((24V - Vgs_th) / R2 -Vgs_th / R1) / C.

So for example C = 100nF Vgs_th=3.0V, R1=4.7K, R2=10K gives dv/dt = 14.6kv/s.

24V / (14.6kv/s) gives a rise time of 1.6ms (going from 0V to 24V).

3) The switch doesn't have any kind of short circuit protection. This may be OK if the 24V source is current limited. But if not, its worth considering.

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In addition to some of the comments/answers, you can do the same thing without M2, provided you just need a mechanical way (swtich) to enable the 24V source.

schematic

simulate this circuit – Schematic created using CircuitLab

Choose R1 and R2 such that you don't over-voltage Vgs (like you have thought). R1 keeps the source and gate at the same potential (pulled up) when SW1 is not engaged. When SW1 is engaged, then the voltage between source and gate is enough to turn on the M1.

M2 would be needed, however, if you also wanted to control the high side switch electronically. You could then add M2 back in, like this:

schematic

simulate this circuit

This way, by keeping the mechanical switch as I have shown, you don't have the high input voltage feeding back to a low voltage pin (e.g from a microcontroller). Either M2 or SW can enable M2 and therefore connect the 24V source to the load.

M1 has to be rated for the load current, and M2 is just a low profile NMOS with no need for high current capability (2N7002 for example).

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