Hmmm ... looks like I made a fool of myself ...
Not at all.
I was told this was a
free wheel diode, apparently it is not, so would you care to explain ?
Sure. When the MOSFET is 'on', current flows from the supply to drain
to source. When the MOSFET starts to turn off, the voltage at the
drain will *rise* to higher than the supply rail (due to motor
inductance), most likely limited in this case by the avalanche voltage
of the MOSFET. That dumps some energy (high current at high voltage,
so quite high instantaneous power) into the MOSFET. The absolute
maximum single pulse energy for the given part number is only 0.019J.
PM motors don't tend to be as high inductance as solenoids so you
might actually get away it, but it's not a good idea.
A freewheel diode across the load will limit the drain voltage to a
volt or so above the positive rail, and the diode itself will
dissipate very little power because the voltage is so low. In case
you're wondering, the energy stored in the inductance gets used up in
the motor, which is usually where you want it anyway.
The diode you see in the MOSFET datasheet (body diode) won't conduct
unless the voltage at the drain drops below ground. That is actually
possible under some conditions with a motor (for example if you
mechanically overdrive the motor at faster than full speed or if the
motor supply was suddenly pulled down with the motor spinning the
back-EMF could cause the body diode to conduct). The result would be
braking. But that diode is of no help dealing with the motor
inductance.
Best regards,
Spehro Pefhany