Hi,
Just for my understanding and curiosity is it working by the zener diode "bleeding" off any voltage it "see's" above its set value to ground (in this case 10v)
Yes, in fact that is a very good description as to what exactly a Zener diode does.
What I don't understand is how this does not create a short circuit and overload the amp capacity of the VFD 12v output?
Also can you explain what the resistor does for this setup please.
The resistor is between the VFD 12V output and the Zener. In absence of the resistor the Zener would 'short out'
the output and the VFD might not like it.
You are correct, but the 'short' is not quite the normal sense of it. Normally you would short a circuit and force the output
voltage to zero. In this instance you would 'short' it to not 0V but 9.1V (or whatever value Zener you used).
In engineering parlance that is called a 'differential short circuit'.
The resistor is to limit the current in the differential short and drop some voltage, in this case 13.2V-9.1V=4.1V.
It is entirely likely that the manufacturer of the VFD put some current limit resistor or other circuit within the VFD
to protect the output in case of a short circuit. Quality designed and built US, Japanese and European brands
will almost certainly have protection built in. Who knows with Chinese brands? Some are actually very good indeed,
anything made by Delta for instance I rate as good or better than any US, Japanese, European stuff but other Chinese
stuff I wouldn't 'cross the road to piss on'.
Craig