@Hero999 I want to know what is the roll of Q1 transistor in your circuit, thanks.
Positive feedback.
When power is first applied, C1 charges via the load, R5, D1 and Q1's base. R1 now keeps Q1 on, which holds M1's gate near 0V, via R3. The circuit will remain in this stable state until S1 is activated.
When S1 is closed, it connects M1's gate to +V, turning it on. C1's positive plate is connected to 0V, via M1's drain, which now sits at 0V. C1's negative plate now sits at the power supply voltage, minus two voltage drops, below the negative rail, which will be around -11V, with a 12V supply. Q1 will be off because no current will be flowing through its base. With Q1 off, M1's gate will be held at +12V, via R2 and R3. The circuit will remain in this state, until Q1's base starts conducting. C1 will charge via R1. When the voltage on C1 reaches two diode drops above 0V, about 1V, Q1's base will start to conduct, which will turn it on. As Q1 turns on, it will connect M1's gate to 0V, thus turning it off. When M1 turns off, Q1 will start to turn on even more, this turning M1 off more, hence positive feedback. When M1 is off, C1 will charge via D1 and R5 again, returning the circuit back to the initial state.
Because the current through Q1's base and D1 is so low, one diode drop can be considered to be 0.5V, rather than the usual 0.6V, hence why two diode drops is said to be 1V.
If the power supply voltage is below 6V or so, D1 is not needed, as it protects Q1's base from negative voltages. The tiny leakage current through D1 will do no harm to Q1's base.
If the load current is under 50mA or so, then R5 is not needed, as it protects Q1's base from the current spike when C1 charges, via the load.
The circuit is fairly insensitive to supply voltage changes, because the threshold when the cycle ends is two diode drops which is small compared to the supply voltage. The single MOSFET circuit is much more sensitive because the MOSFET threshold, is much higher. The latching action of this circuit means the MOSFET turns on and off relatively quickly, compared to charging/discharging a huge capacitor via a large resistor, connected to the gate.
The Wikipedia article describes a similar circuit, quite well. If you have any questions or want more schematics or drawings, just ask.
https://en.wikipedia.org/wiki/Monostable