Understanding inductive ringing
For my own understanding I tried modelling switching off an inductive load without a freewheeling diode to observe ringing. This is the model I used:
1.) Are the two main components of the resonant circuit L1 and C_windings, where C_windings is the inherent capacitance between the two ends of L1, or are there more elements contributing to this effect that I have not placed in the circuit?
2.) Can ringing on digital signal lines also be represented in the same way, where L1 is the total inductance of the signal line and C_windings is the capacitance between the two ends of this signal line?
3.)Is adding a termination resistor on data lines essentially providing a path for the collapsing magnetic field of the line inductance that dampens the LC oscillation? (Path in red)
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are there more elements contributing to this effect that I have not placed in the circuit?
Definitely, but it's hard to say how relevant they would be. The real world is messy.
There is parasitic capacitance between every two conductors. Depending on what the dielectric is, some of those capacitors may be lossy.
Conductors are also generally not super-conducting, so are actually low-value resistors. Those resistors of course have parasitic capacitors across them. Whether super-conducting or not, they have some inductance too. This continues infinitely.
The real trick is to decide which of these unintended capacitances, resistance and inductances actually matter. In the case of shutting off an inductor suddenly as in your circuit, its parasitic capacitance certainly will matter. I'd also assume some capacitance thru the transistor. I'd probably just model all that as a small parasitic capacitance to ground. There will certainly be some of that anyway.
Can ringing on digital signal lines also be represented in the same way
Usually not, since there are transmission line effects going on. Your circuit assumes a nicely lumped system, which is yet another approximation. Series inductance and parallel capacitance can matter to digital signals, but they usually aren't too significant until the line gets long enough that you can't ignore transmission line effects.
Is adding a termination resistor on data lines essentially providing a path for the collapsing magnetic field of the line inductance
No, "terminating resistor" is a specific use of resistance that applies to a transmission line. You can't model transmission line effects with a lumped circuit as you show. There isn't a single relatively neat inductance that is releasing stored energy.
One way to think of a transmission line is that each little bit of the line is a series inductance followed by a shunt capacitance. One or a few of these by themselves doesn't make a transmission line. In a real transmission line the series inductance and shunt capacitance is distributed. You can approach this by modeling many L-C discrete pieces in series, but that's only an approximation. The approximation becomes better as you make each piece smaller and therefore use more of them. The Ls and Cs each get lower as the discrete pieces get smaller, but their ratio stays the same. The total L and C per unit length stays the same. A real transmission line is an infinite number of infinitely small of these L-C pieces in series.
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are there more elements contributing to this effect that I have not placed in the circuit?
Well, you have a simulator and it should be really easy to remove C_windings and observe if the ringing frequency alters. If it does then there is another capacitor hidden away in your circuit (aka drain-source capacitance) that sets the new ringing frequency.
Can ringing on digital signal lines also be represented in the same way, where L1 is the total inductance of the signal line and C_windings is the capacitance between the two ends of this signal line?
No, because usually on digital signal lines the problem is one of reflections rather than resonance. It may look like resonance but, it's something else tied up in transmission-line theory.
Is adding a termination resistor on data lines essentially providing a path for the collapsing magnetic field of the line inductance that dampens the LC oscillation? (Path in red)
No, the resistor is there to prevent reflections returning back down the line. It's a transmission-line effect that we are normally dealing with here on digital signal lines.
Unfortunately if you haven't studied transmission lines (a big and difficult subject) you can easily mistake t-line effects for LC oscillations.
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