question about irf9530

i read about the data sheet of irf9530 and it says that its drain-source breakdown voltage is 100 volts, its threshold voltage is 2 to 4 volts, but its gate- source voltage is can only handle ±20 volts. So how can I use the mosfet above 25 volts if it will only switch off at (25 - 4) volts gate-source voltage ?

You picked a p-mosfet so the numbers are negative.

-10 volts from S to G will turn on the fet. Note gate to source.

Shown a light bulb for a load. -75 volts power supply.

I don't know if this answers your questions.

P channel are normally ON(closed) ( 0 volts or grounded) , my question is how much voltage does it require to turn it OFF, for irf9530 based on my experience if the voltage is 20 volts connected to source lead, it would require at least 16 volts to turn it off, based on this experience if the voltage is 75 connected to source lead, it would require 71 volts to turn it off, but based on the data sheet the gate could only handle 20 volts, why is it then rated to -100 volts, if I could only use it up to -24 volts?

sorry i'm still new to this sort of things

Almost all N-MOSFETs and P-MOSFETs are enhancement which means the gate-source voltage must be less than the (absolute) threshold voltage (Vth) to turn off both devices. The only difference is the polarity of the terminal voltages.

You are, perhaps, confused in the terminology by using the P-MOSFET as a high-side driver with the transistor source connected to V+ and the drain load connected to ground. Turn-off in that configuration requires that the gate voltage be higher than Vs-Vth (to achieve gate-source voltage <Vth).

To keep Vgs less than 20V when the control signal is low, you can use a resistor divider between Vs, the gate, and the control voltage. Alternately connect a 10-15V zener between the transistor gate and source, and a resistor from the gate to the control voltage.

You could change to N-MOSFET, turn the circuit upside down. This is better to understand understand.