# Cranking out motor size

Oct. 5, 2000
Consider a motor application that requires a continuous torque T of 0.

Consider a motor application that requires a continuous torque T of 0.003 N-m at 5,000 rpm, with both factors referred to the motor shaft. The available power supply voltage is 12 Vdc with maximum allowable steady-state current I of 0.5 A.

First, calculate the required torque constant Kt using the torque and current. Thus, Kt = T/I = 0.003/0.5 = 0.006 N-m/A. Because Kt is expressed in the SI system of units, (N-m/A), the voltage constant Ke is numerically the same with units of V/rad/sec. Also, = 5,000 rpm X 6.28/60 = 523 rad/sec.

Next, calculate the required armature resistance, R = (V Ke)/I = (12 523 0.006)/0.5 = 17.7 Ω Finally, calculate the motor constant, Km = Kt/R 0.5 = 0.006/(17.7) 0.5 = 0.0014 N-m/W 0.5 .

Referring to the table, calculate Km for each of the motors listed:

• Motor 1; Km = 0.001/(5) .5 = 0.00045 N-m/W 0.5
• Motor 2; Km = 0.005/(9.5) .5 = 0.0016 N-m/W 0.5
• Motor 3; Km = 0.012/(10) .5 = 0.0038 N-m/W 0.5

For the required Km of 0.0014 N-m/W 0.5 , motor 1 does not meet the specification, motor 2 is adequate, and motor 3 has the best margin.

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