Take some time to look at the results the simulator is giving you and think about them. The current in your motor (L6) is hitting almost 240A! This is not realistic. There is parasitic resistance and back EMF that will come into play when using a real motor.
From there, you are only simulating a single side of the drive on this. When the top side driver turns off, the bottom side usually turns on after a certain amount of blanking time (usually in the tens of nsec range). You have it fixed such that the bottom side FET never turns on, so the current in your inductor gets run through the body diodes of your bottom and top side devices, and those diodes likely have enough resistance to get to the voltages you are looking at. The current in your load will NOT change quickly when you are simulating a 10mH inductor with zero parasitic effects.
I dont find it that unrealistic. In my real circuit i am hitting somewhere above 60A without anything smoking. Only the shunt gets hot, nothing else.
You are talking 4X the current in your simulation, which is 16X the power. If your shunt is getting hot, that's 1-2W of power that is shifting to MUCH higher numbers in the simulations.
What parasitic effects do i have to set up, in order for the current in my load to change quickly? What are you considering my load?
Do you have any resistance in series with the 10mH inductor? Your original post said this is a BLDC driver, the assumption I made is that the 10mH is the DC motor as a load. It's pretty easy to look at the circuit and see that you have a high side switch and a low side switch and the simulations you are doing only uses one of them (the bottom switch has its gate tied to source, forced off). So, let's remember back to the inductor equation: V = L * dI / dt. Inductors do NOT change current easily. The larger the inductor, the harder it is to change currents quickly.
So we have a current that needs to keep flowing in this inductor as you turn off the top side FET. Where does this current come from? It runs through the body diode of the bottom side FET. When you are talking 200+A through a body diode, you get large voltages. Now, what is happening to other currents in the circuit? When that top side FET is on, you're getting 200+A running through all of those inductors and resistors in your circuit (look closely at L9). Turn off that top side FET, and that current wants to continue as well. Where does it go? Into the caps and resistors you have there. Remember that the energy stored in an inductor has to go somewhere which is to shunt into the capacitors/resistors you have. Just calculate what happens if you shunt the energy from the inductor into you caps and you'll see a massive shift up in voltage.
The physics aren't changing anywhere here. You've created a tank circuit with an extremely high Q (@iMo was right about the ESR, getting below a few milliohms is nigh impossible, let alone sub-microohm). This will ring like CRAZY with any transient on it. The thing to look at in debugging your simulation and circuit is to review not only the voltages around the circuit, but the currents as well. What happens where as you have a change happen? When M2 turns off, where does current flow happen at one time point versus the next? Put that together with the understanding that inductors can have voltage change rapidly but will fight current change and capacitors can have current change quickly but not voltage change and you will begin to understand what happens in the circuit.
I think part of the issue that a lot of people are going to have is that we also have a hard time lining up which node is which when they are unlabeled on the schematic you provide.
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