a high side load switch made from n channel and p channel mosfet working princip

[oppel]

Hello, I did research on the internet to understand the working principle of the circuit you see in the image, but I could not find an article that explains exactly how the N-channel and P-channel MOSFETs used here work according to their transmission cut-off states. I would appreciate it if you could give detailed information about the working principle of this circuit.

[moffy]

When 'enable' is low then Q1 is 'off', and the gate of Q2 is pulled up to its source by R2 and it too is 'off'. When 'enable' is high then Q1 is 'on' and the gate of Q2 is pulled down to 0V turning Q2 'on'.

[BennoG]

This type of driver will only work up to about 20V, above that you need extra protection to not go above the maximum Vgs value.

Benno

[Terry Bites]

For higher voltages, you can split R2 to keep Vgs to a safe level.

[BennoG]

Yes that is a possibility, myne had to work from 12V to 50V, so that made it a little more complex.

Benno

[Doctorandus_P]

Yes that is a possibility, myne had to work from 12V to 50V, so that made it a little more complex.

Benno

Meh, replace R2a in the previous schematic with a TVS diode, or better, add a TVS diode to limit the maximum gate voltage, but leave  the resistor so it will firmly turn the FET off when it should be off.

Circuits like this switch quite slowly, because of the combination of resistors with the gate capacitance. This is OK for switching loads occasionally (such as also done with Relays), but if you want to switch loads fast (such as with PWM) then this slow switching may cause a lot of trouble and the FET is likely to let out it's magic smoke and die in a puff or a loud bang.

[temperance]

This is how I do that.

The base drive voltage together with the emitter resistor form a current sink. (Udrive-0.7V)/ R1

Select R2 such that the voltage across R2 never exceeds the maximum gate source voltage for M1.

You can adjust the turn on speed with C1 (Miller integrator) to prevent high current surges if the load is capacitive.

You can add a diode in series with C1 if only the turn on speed should be adjusted.

[HwAoRrDk]

It should also be noted that R3 in OP's schematic is of minor relevance to the switching functionality. It could be removed.

It only serves to add a small constant load to the output, and will also (slowly) discharge the output when Q2 is switched off.