I run a loaded BLDC motor. Current meter shows 30A. How much current at average will flow through each phase? In BLDC 6-step commutation current flows to a phase and comes out of another, then it changes to next phases. Do it means that each phase pases 2/3 of the full load on average, as it passes full current 2/3 of the time?
\$\begingroup\$
\$\endgroup\$
2
-
\$\begingroup\$ how are you measuring 30A with a current meter? before your motor controller? the current of the phases is more dependent on the load of the motor than the current going into the motor controller, which can increase the current that goes into the coils vs what it draws (at a reduced voltage) \$\endgroup\$– user2813274Nov 13, 2016 at 15:15
-
\$\begingroup\$ I measure it before the ESC. \$\endgroup\$– Roman SimonyanNov 13, 2016 at 15:35
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
Even though this was asked long ago, it is important to have answers recorded here for all those who will come here with similar search.
If assumption can be made that your motor controller (ESC) will switch full battery voltage in PWM pulses then your measured battery current will be equal to RMS (about 0.7 times amplitude for sinusoidal or function of a duty cycle in PWM)) of the current flowing through 2 phases cables (2 out of 3 motor feed cables) plus loss in your ESC (controller). Calculation is better done as follows:
your battery power taken P bat = Vbat * Ibat
adjust for the loss in ESC Padj = Pbat * k where K is your ESC efficiency (0.9 for 90% efficiency of ESC)
Measure amplitude of PWM pulses Vpwm (should be close to Vbat less resistive loss in the wiring), if you want only to get estimate you can assume that PWM voltage amplitude is the same as Vbat
Your motor current through 2 phases (if star wiring) will be Imotor = Padj / Vpwm , this Imotor is RMS value.
Some digital oscilloscopes show this value calculated for an observed waveform for any duty cycle of waveform.
