FreshmanNewbie
Advanced Member level 1

Thank you very much for the detailed explanation which was easy for me to understand.Hi,
I will try and do that later.
One important point to note here is that all voltages in the circuit are DC. Please bear this in mind as we proceed.
Now let's assume that you just supply power to the circuit at time t=0 and that all capacitors in the circuit (I'm referring to gate capacitances here) are discharged. From time t=0 to some other point in time (call it t_steady), there will be some transient in the MOSFET gate capacitor voltages and the 1nF capacitors externally connected to the MOSFET gates as charges flow into them and charges them into their respective steady-state voltages. Please ignore other MOSFET capacitances for now. These capacitors will be charged to a certain level depending on the voltage driving the charges. Then no further charges will be driven into the gates. This time is t_steady and from this time onwards we are in steadystate. Charges cannot leave the gates because no one condition for the charges to leave (i.e. either actively extracting the charges like MOSFET drivers do, or providing a path to a lower potential as in the case of pulling down the gate) has been provided. Since the driving voltages are still present, and there is potential difference in the circuit, then current keeps flowing along paths (or branches) that have potential difference.
In this circuit, at time t_steady, the Low-side MOSFET is already charged to 2.195V. The entire 1.62mA steady-state current that I calculated flows from the drain through the zener, through the gate drive resistor, to the -14V.