Greetings mrai,
MrAl wrote:The thing is though, the graph is misleading
because that negative current is not one of the main points
of the circuit. It happens because of the MOSFET alone,
and without that the current wont go negative.
That is counter-intuitive. The inductor is in a
magnetic field that is rapidly changing, thus
inducing a voltage across it. The coil current will indeed
oppose the initial current that built the field.
Had this been a capacitor circuit, charging would
result in stored energy with fixed voltage and no
current flow (except for leakage).
The inductor would also remain charged with fixed
current and no voltage, but such a circuit is not
practical.
When the inductor charging voltage was removed,
the collapsing field would reverse the polarity of
the terminals and the voltage would rise (to infinity)
until current flows, and the coil energy is dissipated.
The magnitude of the current is identical but the
polarity (sign) is opposite.
In the simulation circuit there is no path for
current except to charge the capacitance,
which it is agreed is parasitic to both the coil
and the FET (and strays).
This is a lossy resonant circuit. With optimization
we would have a shock-excited-resonator. The
energy would swing back and forth from inductor
to capacitor, until all is used up overcoming
circuit losses.
As you may know, the discovery of this was the
life's work of Nikola Tesla.
MrAl wrote:The more important points are that the coil energy dissipates
faster when allowing higher coil voltages, and that the higher
the faster, although high voltages can kill other parts. The high
voltage could easily kill a 50v MOSFET even though 50v is more
than 4 times the Vcc supply voltage of 12vdc.
I would think the goal is to manage the release of the
stored energy to not stress the other components.
For the OP's relay driver this is a simple back-emf
diode, which protects the transistor and limits the
radiated energy that may otherwise interfere with
the car radio or upset other circuitry.
MrAl wrote:Also, that 500v peak is not something
you can count on, it can be 100v, 500v, or 5000v
depending on the other parts of the circuit that
just happen to be there.
Actually, the 503V peak is very real for the models
used in the simulation. In practice the voltage may
be quite different, and without careful design it
would threaten the other components.
As shown earlier, with no snubber and practically
zero coil resistance the back-emf is over 70,000 volts.
The use of an inductor to store energy, and
release it into a different circuit impedance, thus
changing the voltage and current available is the
basis for a SMPS (switch mode power supply).
Output voltages higher than input are quite
practical in a flyback converter with only one
inductor (and two switches, one of which could be
a diode).
I think you and I are on the same page, and the
topic has been thoroughly reviewed, well beyond the
OPs inquiry. The take away for me is that in
simulation the magnitude of the circulating currents
and pulse voltages can be quantified, and neither
is trivial.
Comments Welcome!