Question 6.5.1: No-Load Speed and Stall Torque The parameter values for a ce...
No-Load Speed and Stall Torque
The parameter values for a certain motor are
K_{T} = K_{b} = 0.05 N · m/Ac = 10^{−4} N · m ·s/rad R_{a} = 0.5 Ω
The manufacturer’s data states that the motor’s maximum speed is 3000 rpm, and the maximum armature current it can withstand without demagnetizing is 30 A.
Compute the no-load speed, the no-load current, and the stall torque. Determine whether the motor can be used with an applied voltage of v_{a} = 10 V.
For v_{a} = 10 V, (6.5.5) and (6.5.6) give
i_{a} = \frac{c V_{a} + K_{b} T_{L}}{c R_{a} + K_{b}K_{T}} (6.5.5)
ω = \frac{K_{T} V_{a} − R_{a} T_{L}}{c R_{a} + K_{b} K_{T}} (6.5.6)
i_{a} = 0.392 + 19.61 T_{L} A ω = 196.1 − 196.1 T_{L} rad/s
The no-load speed is found from the second equation with T_{L} = 0. It is 196.1 rad/s, or 1872 rpm, which is less than the maximum speed of 3000 rpm. The corresponding no-load current is i_{a} = 0.392 A, which is less than the maximum allowable current of 30 A. The no-load current is required to provide a motor torque K_{T} i_{a} to cancel the damping torque cω.
The stall torque is found by setting ω = 0. It is T_{L} = 1 N · m. The corresponding stall current is i_{a} = 20 A, which is less than the maximum allowable current.