What would be the best approach to constantly having an 8V low-noise output assuming a potential input voltage swinging between 7V and 12V? My initial guess would be to use a flyback or buck-boost switching transformer
It's a good approach. A buck-boost regulator can be implemented in a small area with a minimum component count (no isolation).
both fixed-output and variable-output regulators both require external components (diode, capacitor, inductor). Why do both require external components?
The reason is that components such as capacitors and inductors occupy too much volume when they are integrated into an IC. For this reason, they are left as external components.
Diodes are very complex things, made up of Forward Voltage, Forward Current, Reverse Current, Reverse Voltage, Reverse Current leak and Recovery Times. And then all voltages and currents have steady-state values, repetitive peak values and non-repetitive peak values.
Everything always has influence.
The reason diodes often are only high current or high voltage is because a lot of the features of a diode are a trade-off.
If you want a diode with huge current capability and a very good reverse voltage specification you need much more silicon material and many more controls during the process than when you choose only one to optimise.
Now, I assume your 3-phase signal is somewhere in the 1 to 100Hz, since most 3-phase power applications are.
That's a pretty low frequency to a diode, so you can pretty much skip "reverse recovery time" and all those parameters. They mean how quickly the diode will start blocking current after it previously conducted, but to 100Hz power any recovery out there is fast.
You will want to make sure the diode can handle the voltage even if it isn't exactly what you expect. One thing, for example, you didn't specify if whether the 40V is AC or expected DC. I'll assume AC. In that case, with 3-phase, you will get an approximate DC voltage of 1.8 times (rounded up) that, which is 72VDC.
So your diode must at least have a reverse voltage of 80V, preferably over 100V.
Then, the forward voltage and current are linked.
On page 4, top left, of your second datasheet (the Microsemi diode) you can see that at 25 degrees junction temperature at 40A it will only have a forward voltage of 0.8V
That forward voltage is per one diode, yes.
The difference between Steady State forward current and peak non-repetitive forward current is that a very high current will make the diode drop a higher voltage and the total peak power for a 200A spike becomes well beyond 200W, even in your first diode.
For a very short duration, and only once, the diode can handle that amount of energy, but if you keep the current constant the energy dissipated will build up. That's why the first one can only handle 12A continuous, anything higher will make it heat up more than its internal design can get rid off.
Now, many diodes have a Repetitive Peak Current, based on a 2phase 60Hz or 50Hz rectification, which is a little higher than their steady state current, that's because a diode in a rectifier will only be used part of the time. Half in a 2-phase and one third in a 3-phase.
So if you can find a diode that has only 35A steady state, but allows for 50A or such (or preferably higher of course) of Repetitive Peak current you should be reasonably safe with your 40A specification, if your 3-phase signal isn't below 35Hz.
Best Answer
You are right to be confused.
What is the meaning of "flyback diode"?
For a "flyback" diode in parallel with an inductance, and meant to dissipate inductor energy when current through the inductor is interrupted, larger Vf for fixed current dissipates this energy more quickly. (By the way, you should not depend only on maximum current rating to judge if the diode is a good fit. You MAY ALSO have to calculate power dissipated in the junction, the junction temperature rise over ambient, know the max ambient temperature, and ensure the max junction temperature is less than the data sheet. If the frequency of operation gets high enough, the diode will burn out even if the current is less than the maximum peak current on the data sheet.)
Now, for a "flyback diode" used as a diode in a "flyback" circuit which boosts voltage, the efficiency of the circuit will be higher for a diode with low Vf at a fixed current. Reverse recovery figures prominently into the efficiency calculation so reverse recovery is important too.
So you can see there is some confusion because there are two different answers.
In general, it is much better to discuss circuit questions with a schematic. Otherwise, improper generalizations are inevitable.