How does current build up quickly in the phase coils of a 3 phase BLDC?

Hello,
In a three phase BLDC motor, (driven by a standard voltage source inverter) the three coils are obviously switched in one following the other. When the motor is spinning at nominal speed, and the motor is also full loaded, how does the current build up in a coil just after the instant that it is switched in?

The point is that if the motor is spinning, then there will be a back EMF in the coil at the instant that it is switched in. Thus there will actually be very little actual driving voltage across the coil inductance. Therefore, by Lenz’s law, (di/dt = V/L), the di/dt in the coil will be very low. So in this case, how does the current in the coil manage to quickly build up to the nominal current level in the coil?

Motor dynamics is among the toughest fields to get a grasp of.

Here is about as much as I've been able to grasp.

Motor speed is an equilibrium state.

Say we're talking about a conventional permanent magnet motor. Lighten load, and the motor spins faster. Ergo, current has less time to build in each winding. Therefore the motor draws less current, since the light load needs reduced power.

Or...
Increase load, speed slows. This makes longer time for current to build in a winding, until it is switched out. Ergo, the motor draws more current because of the heavier load.

For a similar reason, a motor draws greater current when starting and when stalled.

Anyway for a given load, there is a point between the two extremes, where the mechanical torque matches electrical power.

I don't know if it is exactly the same with brushless motors, as to how forces interact.

Thanks, so in a BLDC running at nominal speed, and fully loaded, just after the instant that a coil is switched in, the current in that coil has immediately risen up to the full current level in that coil?

That is, I am wondering what is the di/dt of the coil current when each coil is switched in? Certainly with stepper motors, there is a di/dt whereby the higher the DC driving voltage, the quicker the current builds up in the bipolar stepper motor coil.

Is this the process?...the current is flowing in a coil, that coil's IGBT's switch off, but the current keeps flowing but now in the anti-parallel diodes, then when the next coil gets switched in, the current simply starts flowing in that coil?

in a BLDC running at nominal speed, and fully loaded, just after the instant that a coil is switched in, the current in that coil has immediately risen up to the full current level in that coil?

Click to expand...

That will hardly happen with constant voltage drive. With voltage squarewave, you get an exponential (first order low-pass) current settling according to motor L/R time constant. And there will of course a difference between motor emf and terminal voltage according to the load torque.

thanks,
page 17 of the following actually depicts the way that the current rises……

http://documentation.renesas.com/doc/DocumentServer/U18028EU1V1UME0.pdf

…though I am trying to work out the rise time of the current in a coil. I presume it obeys the following equation?

i(t) = I(nom) * [1 – e^(-Rt/L) ]

Where:
I(nom) = { [V(DC link)] – (Coil Back EMF) } / (Coil resistance)
R = Coil resistance (its actually the inductance of 2 coils in series)
L = inductance of coil (its actually the inductance of 2 coils in series)

Do you agree?

Yes, that's the current equation if both terminal voltage and emf have a rectangular waveform. The actual emf waveform might be different.

To properly model a motor drive you cant just assume the motor is a set of passive time-invariant components. You must determine its time-varying emf using its equations of motion and loading.