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For a brushed DC motor, there are the well-known differential equations describing the electrical and mechanical behaviour:

\$V_a-i_aR_a-L_a\frac{d}{dt}i_a-k_e\omega_a=0\$

\$k_ti_a-J\frac{d}{dt}\omega_a-B\omega_a-T_L=0\$

For my purpose, I'm interested in steady-state performance. I can solve for \$V_a\$ and \$\omega_a\$ for a given input voltage by setting the derivative terms to zero.

Brushless motors are more complicated in that it's the frequency of the input voltages that determine the speed/current/torque. I wonder if there's an equivalent method for deriving the steady-state performance of a BLDC. Most manufacturers, including hobby motors, provide data on motor constants, no load current and internal resistance, so I figure this must be possible.

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  • \$\begingroup\$ \$i_a\$ shud be \$i_a(t)\$ so the derivative does not get deleted. Usually that impedance is 10% of rated load Req = V(t)/I(t) s.s. at full mechanical load \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 at 18:54
  • \$\begingroup\$ At SS, the change in angular speed should be zero, so the second equation forces the armature current to be steady. \$\endgroup\$ – Jimmy Jun 29 at 19:59
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Brushless motors are more complicated in that it's the frequency of the input voltages that determine the speed/current/torque.

This is true for stepper motors and PM synchronous motors powered by AC, but PM BLDC motors usually have their commutation controlled by Hall sensors or back-emf zero crossings. So it is the speed/current/torque of the motor itself which determines the commutation frequency - same as in a brushed DC motor.

The formulas for brushed and brushless DC motors are the same.

Most manufacturers, including hobby motors, provide data on motor constants, no load current and internal resistance, so I figure this must be possible.

Yes, however 'hobby' motor constants are often not very accurate, and using them can result in quite large differences between calculated and actual performance.

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yes it is possible.

Brush commutation of the voltage source with some ESR from battery or voltage regulator causes dips in source voltage when Ra (aka DCR) approaches ESR of the battery which controls lock rotor or Isc or surge at full voltage start.

BLDC drivers still have a ESR or Rs combined with an array of Caps using ESR with \$\tau_1,\tau_2,\tau_3\$ to Vdd and may include choke impedance in CCM mode with feedback transfer function to define source impedance. Neglecting this is why users have so many problems using SMPS driving BLDC motors with start currents 10x rated load. ...But I digress, because you are only concerned with steady-state.

Looking at various BLDC motor torque vs Hp or V vs I at RPM tells you as you know, kRPM/V is no load and Vin - BEMF (@ RPM) gives you net winding voltage to give SS power. Often MPT or maximum Mechanical Power Transfer is at 50% no load RPM. The maximum Torque starts at 0 RPM and is 0 torque at max RPM or is 2x torque and current , if you inadvertently reverse direction at full speed. (With no mech. load)

Meanwhile the peak torque and peak efficiency have higher than 50% no load RPM.

Excitation current is the current with no mechanical load is often 10% of rated max current so in your model or equations you should expect no load power at full RPM =10% of rated power and start power is 1000% of rated power. This 10x factor can be up to 12x in very efficient powerful motors, or 8x is more lossy motors. Motor L/R ratios are important for RPM and commutation frequency as well,as this defines the rate of change in AC current before each next commutation and thus these are well defined ratios in a family of motors and depend on the number of poles, RPM and RdsOn of the driver. For L/(RdsOn+DCR)

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  • \$\begingroup\$ So it sounds like the above motor equations apply to the steady state case for a BLDC if we assume Va (Vin) is an average of some sort? \$\endgroup\$ – Jimmy Jun 29 at 20:11
  • \$\begingroup\$ Va is your DC source ( with some ESR, ESL, C) impedance, assume 0 for now. But is(t) is the result from motor dynamic impedance \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 at 20:15

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