Hello,
We are doing a 3 PHASE BLDC water pump (sensorless control) for our Tulip fields irrigation -and we have a flower show coming up soon.
The pump spins at 8000RPM, and is controlled by an ML4425 bldc driver IC.
What happens when the pump suddenly runs dry?......is it like this...
1...Pump runs dry thus pump motor is suddenly on no load....the next coil excitation pulse it receives is obviously too high in current as the feedback loop can't suddenly "realise" that its gone "No-load".
2...Therefore the high current pulse in the unloaded motor literally slingshots the rotor round at a phenominal speed......however, as mentioned the speed feedback loop won't yet have caught up, and so the commutation of the next motor coil occurs after the usual time....and this is unlikely to be at the "correct" time...thus the motor could very likely be "braked" by this next pulse....and so basically, instead of going to overspeed, the rotor will just spin in a very unsmooth manner, and not necessarily go overspeed at all.
....after all, the rotor can only truly go overspeed if the bridge switching frequency calls for this....however,the bridge switching frequency is limited to 400Hz (...what it needs to be to get 8000rpm..) , since the pump never needs to spin faster than 8000rpm.
So how can the motor ever spin faster than 8000rpm?....even when suddenly non-loaded, surely it cannot spin faster than 8000rpm?
The next question is, ...if the motor cannot spin at more than 8000rpm, then how does the controller know that it must limit the coil current in response to a sudden load removal? (the pump suddenly going dry)