How to turn on high-side MOSFET of a buck converter

I'm trying to simulate some buck converters with real MOSFETs instead of ideal switches. The problem is that I don't know exactly how to properly turn on the transistor.

The simple buck converter would be simulate this circuit – Schematic created using CircuitLab

Theory says that the MOSFET should enter the triode region when

$$v_{DS} < v_{GS} - V_{t}$$

but this seems complicate to use because how could we determine the voltage between the transistor and the inductor in a real situation?. So this solution came to my mind:

$$v_{DS} < v_{GS} - V_{t} \Rightarrow v_{GD} > V_t$$

So my questions are:

1. Is that correct? Then sould I apply $$\(9+V_t)\$$ to the gate to turn it on properly?
2. Is there any other way to make this? Maybe with a lower voltage on the gate?

To switch an N-channel FET on you need to pull the gate up to a voltage significantly higher than the input there are a few options here

1. If you are only simulating then you can put your gate drive circuit between the gate and the source of the transistor. This may not be practical in the real world.

2. Use a high side driver IC. These work by charging a capacitor from the supply when the source of the transistor is 0V. This occurs when the diode is conducting. These can be difficult, if not impossible to start if for example charging batteries as the initial voltage at the source of the FET is positive and not 0V. With a suitable resistive load however (as shown) it should be OK.

3. Use a gate drive transformer .

4. Use a P-Channel FET instead, as suggested by Andy, the source is now connected to the supply and you pull the gate low to turn it on. P-Channel FETs tend to be more expensive than N-channel FETs and few are available at higher powers.

If you are using an N channel device as shown in your diagram then yes, to fully turn on the device you need to have a gate voltage that is several volts higher than your incoming supply voltage. However, if you used a P channel device with source connected to incoming supply, the gate being pulled down to 0 volts will turn the P channel MOSFET on. Clearly your input supply voltage needs to have a minimum value to ensure the source-gate voltage for the P channel device is sufficient for decent MOSFET activation.

If your output voltage is higher than your gate threshold, you can use a boost cap between the transistor source and gate. You'll have to put some resistance between the gate drive and the gate, the RC constant will depend on your switching speed. When the FET is off, the source voltage is zero and it's easy to turn on; once you switch the transistor on, the source-gate cap pulls the gate even higher to improve efficiency.