Why are the voltages of the gate, source and drain of the M2 mosfet equal even though they are on the cut-off? Could you help.

 

I think you have supplied the wrong polarity of 60V. The gate and source are approximately different due to the 1MΩ resistor. The voltage at the drain is almost equal because the source-drain parasitic diode operates in forward voltage and the diode leakage current flows through it. This current is very small and therefore little voltage drops across the parasitic diode.

 

I think you have supplied the wrong polarity of 60V. The gate and source are approximately different due to the 1MΩ resistor. The voltage at the drain is almost equal because the source-drain parasitic diode operates in forward voltage and the diode leakage current flows through it. This current is very small and therefore little voltage drops across the parasitic diode.

What I don't understand is this: Why is there voltage on the drain leg when the mosfet is on? Can you explain this more clearly?
Because according to my simple theoretical knowledge, if sufficient Vgs voltage is not obtained, the Mosfet will remain in the cut-off and will not conduct.

 

Why is there voltage on the drain leg when the mosfet is on?

In your circuit the MOSFET is not on. But whether it's fully on or fully off there will always be some voltage on the drain relative to the source.

 

In your circuit the MOSFET is not on. But whether it's fully on or fully off there will always be some voltage on the drain relative to the source.

So how do you explain the fact that the drain is 107V as seen in the picture?

 

You have V2 set to -60V. Is this what you want?

 

So how do you explain the fact that the drain is 107V as seen in the picture?

You have 107V on the source, D2 blocks reverse voltage, so there will only be a tiny leakage current through D2 and the MOSFET body diode, so the drain voltage will be almost the same as the source voltage.