Electronic – How to switch between two voltages of a capacitive divider

Here's a better way to drive a high-side P-channel MOSFET:

^{simulate this circuit – Schematic created using CircuitLab}

Note that the BJT Q1 is a current sink; the current value is the control voltage divided by R2. If your control circuit puts out 12V, the current will be 12 mA.

This current will create a voltage drop across R1 as well, and if you happen to make the two resistors equal, this voltage drop will be essentially equal to the control voltage supply (minus the V_BE drop of Q1). Or you can make the resistors different in order to get a particular relationship.

You need to be aware of two things: The current level that you set will have some bearing on how fast the MOSFET M1 switches; higher current means faster switching. However, keep in mind that Q1 has a lot of voltage across it. When off, it has the full high voltage supply across it. And even when switched on, it has the high voltage minus 2× the control voltage across it. This means that it will be dissipating power equal to that voltage drop times the current when on.

Refer to my previous answer: Is voltage the speed of electrons?

Voltage is best thought of as a field. We're used to thinking of gravitational fields being entirely uniform, but magnetic fields are not. If you attach a piece of ferrous metal to a pole of a magnet, it extends the field into it. Similarly the electric field between the two poles of a voltage source can be extended with electrical conductors.

This extension of the field extends all the way to the field inside the field-effect transistor.

(Knowing about fields clears up a lot of misconceptions brought on by thinking about electrons. Ignore the electrons.)

Edit: so the explanation is almost exactly your (1) with one detail different. Dirac16's hint is important. The gate-to-channel insulation has a capacitance. So the circuit looks like a capacitor. These two circuits are equivalent (ignore the values):

^{simulate this circuit – Schematic created using CircuitLab}

So there is no steady-state DC current flow, but at the point of connecting up the circuit the capacitor charges, and while it is charging a current flows. This is actually quite important when designing power MOSFET systems: you need to be able to supply adequate current for a very short time.