Interesting. My motor/spindle doesn't 'surge'. It takes (and is programmed to do so) 3 seconds to reach target frequency/speed. So how does the VFD control this when I'm simply switching on the VFD's output?
I was imprecise.
I am guessing that CO has wired it up so the on/off switch provides a low voltage signal to tell the VFD to start-up, and the VFD (as you said) is set-up with a 3 sec ramp-up to get to full speed. The VFD also prevents huge surges during motor reversal by controlling ramp-up of voltage and current when it swaps two phases.
If your VFD is always running, and your on/off switch is between the VFD and the motor, than you would see a large current surge when you started the motor (if you had an ammeter on it)... and your VFD would probably generate a fault (excessive Full Load Amps) and shut-down to protect itself.
As long as the VFD is allowed to control it's on/off/reverse cycle on its own, it avoids excessive current surges. The VFD prevents startup surge by ramping up both current and voltage slowly, minimizing the huge differential voltage usually placed across the windings at startup.
Using the barrel switch to switch motor directions with the VFD energized could result in current surges in excess of 5x run current... not good for the transistors in the VFD, as you are exceeding their ratings many-fold, which could lead to early death. Your VFD will probably (again) generate a fault (excessive Full Load Amps) and shut-down to protect itself.
DISCLAIMER: I have never switched the power between a VFD and motor, not wanting to brick a $250 VFD. So it is possible that the VFD won't fault on FLA if you start it that way (maybe there is a time delay). The enemy of transistors is heat; current passing through the transistor generates heat. However, short durations starting surge might not heat it up enough to fry a transistor. I don't know, and I am not curious enough to risk $250 to find out.
