Why is my buck regulator based on LM22679 limiting current way under the set limit?
Based on the simulation, I would have three suggestions for a likely solution to the problem.
Basically, the source of the problem is overcurrent protection and the output capacitor.
Here, two things can cause overcurrent, one is a poorly chosen (small) inductance, and the other is a capacitor that is too large at the output.
I took the value of the SS capacitor smaller than the original to keep the simulation for a normal period of time.

Inductance

With an inductance of 4.7uH or less, the peak current exceeds 7.1A (5A load) and the system becomes unstable.

Thus, the value of the inductance returns to the calculated 10uH.


Current limit

The set current limit 5A is removed (6.8k) because the output current 5A has a higher switching (inductance) current.
This retains the built-in 7.1A current limit. This will be just enough for the 5A output current.
There should be no resistance at all, but the model doesn’t like that state, so it will be 5k at the limiting input (for the model only).

The 5A load is OK, now let's look at the 25 Ohm load with which the measurements were made (28V and 40V inputs).



Apparently everything works except that I use a 100 uF capacitor on the output instead of 200uF.

Output capacitor

200uF at 500kHz is completely unnecessary, 100uF is enough, only a capacitor with a small ESR value is needed (electrolyte plus ceramic).
A large output capacitor with a large time constant degrades the stability of the system.

Unfortunately, several hours are not enough to model the phenomenon with a large capacitor.
So this part is not exact, just a conclusion.
Due to the large output capacitor, each switch-on activates the internal overcurrent protection beyond the SS timing.
This puts you in a state where you can only get out when you are turned off.
So we can't turn it on. We see this on the oscilloscope. It also switches off at the moment of switching on, so the signal oscillates.
It then switches to a lower frequency as if there is a short circuit at its output.
The phenomenon is described in the data sheet.

Therefore, even if the
load is shorted with zero volts across its terminals, the inductor will still see a finite voltage. It is this value that
should be used for V x and V sc in the calculations. In order to return from foldback mode, the load must be
reduced to a value much lower than that required to initiate foldback. This load &quot;hysteresis&quot; is a normal aspect of
any type of current limit foldback associated with voltage regulators.

So what will hopefully help is 10uH inductance, no special current limit and the output capacitor is only 100uF.