For a 1.8°, 2-phase bipolar stepper motor, the stepping rate is 100 steps/sec. The rotational speed of the motor in RPM is

Question

Question

This question was previously asked in

ISRO (VSSC) Technical Assistant Electrical 2017 Official Paper

Answer 1

Concept:

Speed of motor \( = \frac{{β f}}{{360}}\)

Step angle in a stepper motor,

\(β = \frac{{\left( {{P_s} - {P_r}} \right)}}{{{P_s}{P_r}}} \times 360\)

Where, Ps is the number of stator poles

Pr is the number of rotor poles

f is stepping rate

Calculation:

Given-

β = 1.8°, f = 100 steps/sec

Speed of motor \( = \frac{{1.8 \times 100}}{{360}} = 0.5\;rps\)

Hence the rotational speed of the motor in RPM = 0.5 x 60 = 30