At each commutation, energy is required to drive current into the phase that was previously unattached and is now attached to a power supply rail. The energy stored into the inductance is:
Energy has units of Joules and a Joule equals a Watt*second. A Volt*Ampere equals a Watt.
The switched inductance is half the motor inductance so the energy stored in the inductor is:
where L is the motor inductance.
Energy is required at every commutation, which occurs at a frequency of:
Fcom = N[kRev/min]*1000/k*min/(60sec)*Poles/2[elec rev/mech rev]* 6/elec rev
Fcom = N[kRev/min]* Poles*50
The power used to provide energy to the top switched inductance is:
P = E* Fcom = 1/4LI2*Fcom = 1/4LI2* N[kRev/min]* Poles*50
The average voltage used drive the switch inductance is.
V = P/I = 1/4LI* N[kRev/min]* Poles*50
This average voltage must be subtracted from the supply voltage in the equation for speed.
Substituting we get an equation with N on both sides:
N = (V-12.5*L*I*Poles*N-I*R)/Ke.
Solving for N:
N = 2*(V-I*R)/(2*Ke+25*L*I*Poles).
The average current supplied by the energy stored in the inductance released from the power supply rail is:
Iave = P/V = 1/4LI2* Fcom/V
The average measured current is then:
Im = (I-L*I2*Fcom)/(4*V)
Next: Final comments on the motor model