Miro, I would imagine that a MOSFET, with 10V or so on the gate, will behave like a low value resistor in series with a small inductance. Given the small size of the MOSFET chip, at HF the resistance shouldn't be much different from the DC value given in the datasheet (RdsON). And the inductance is that of the terminals and bonding wires. A typical TO-220 MOSFET will have roughly between 7 and 10nH. Larger encapsulations have higher inductance, while small SMDs have lower one.
The problem is how to turn a MOSFET off, for RF. Simply shorting the gate to the source is definitely not good enough. The drain-source capacitance gets very large when there is no voltage on the drain, enough so to be very far from an open circuit, at HF. And even if you could find a MOSFET with low enough capacitance even at zero drain voltage, a large RF voltage across it would turn on the parasitic drain-source diode, which turns on much faster than it turns off, so that any large enough RF voltage across the MOSFET will turn it on and keep it on.
So to use a MOSFET as RF switch, you need to bias its drain to a high voltage to turn the switch off. High enough to be comfortably above the peak RF voltage, so that even at the negative RF peak the MOSFET has enough positive drain-source voltage to keep its capacitance small enough. And even then the capacitance might still be a serious problem!
So I would say that using MOSFETs as RF switches is generally not a very good idea. It can surely be done, but it would require several MOSFETs per switch channel, a high voltage source, supply chokes or resistors and DC-blocking capacitors, and much care would need to be placed on switching slow enough so that no high voltage pulses are coupled into the signal inputs and outputs, and so on. In the end it's more complex, less good and more expensive than other solutions, such as PIN diodes (which have some of the same problems, but to a lesser extent), or even plain old relays.
If you use the NanoVNA to measure a MOSFET used as switch, keep in mind that the signal provided by the NanoVNA is small. It will probably not even begin to make the drain-source diode conduct. But a transmitter signal will!
Manfred
