Inductive effects on Motor Performance: Quantitative Analysis

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:

 E=1/2LI2.

 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:

 E=1/4LI2.

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

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