Is it correct to say that 'the current' drawn by a BLDC motor is the same as the current from any single phase at a given time (assuming they are all balanced/equal in magnitude)?

I came across this diagram describing the characteristic shape of a BLDC current waveform:

BLDC Phase Current, Brushless DC (BLDC) Motor Fundamentals by Microchip Technology, 2003

From this diagram, it seems that the current is roughly flowing through one phase at a time, with only a tiny amount of overlap. What this doesn't look like is three sinusoidal waveforms 120° out of phase - a situation where the total current is more easily understood and calculated.

What is convention here?

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    \$\begingroup\$ if you invert all the negative currents, and add up the three phases, you will see that the total current is a sawtooth waveform that alternates between the + value and 2x the + value \$\endgroup\$ – jsotola Feb 6 '18 at 18:21

Motor line current is just the peak of any individual phase's current. Notice that in any of the six commutation steps, one phase always has no current, while the other phases have equal and opposite current (well, they should, the rounded leading edges in your diagram are wrong). This makes sense, as the phases are Y-connected so the current must always flow into one phase and out of another (in BLDC motors, one phase is always idle).

  • \$\begingroup\$ I thought that diagram seemed a little off. Thanks \$\endgroup\$ – C Jones Feb 7 '18 at 22:21

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