Question 6.5.3: Calculating Motor-Amplifier Requirements The trapezoidal pro...
Calculating Motor-Amplifier Requirements
The trapezoidal profile requirements for a specific application are given in the following table, along with the load and motor data. Determine the motor and amplifier requirements.
| Profile data | ||
| Cycle times | t_{1} = 0.2 s, t_{2} = 0.4 s, t_{f} = 0.6 s | |
| Load data | ||
| Inertia I_{L} = 4 × 10^{−3} kg · m^{2} | Displacement θ_{Lf} = 10 \pi rad | |
| Torque T_{L} = 0.1 N · m | Reduction ratio N = 2 | |
| Motor data | ||
| Resistance R = 2 Ω | Torque constant K_{T} = 0.3 N · m/A | |
| Inductance L = 3 × 10^{−3} H | Damping c = 0 | |
| Inertia I_{m} = 10^{−3} kg · m^{2} | ||
| Time constants | 1.56 × 10^{−3} s and 0.043 s |
The total inertia I is the sum of the motor inertia and the reflected load inertia. Thus,
I = I_{m} + \frac{I_{L}}{N^{2}} = 10^{−3} + \frac{4 × 10^{−3}}{2^{2}} = 2 × 10^{−3} kg · m^{2}
Because the reduction ratio is N = 2, the required motor displacement is N θ_{L f} = 2(10π) = 20π rad, and the load torque as felt at the motor shaft is T_{d} = T_{L} /N = 0.1/2 = 0.05 N · m.
The motor’s energy consumption per cycle is found from (6.5.13).
E = \frac{R}{K^{2}_{T}} \left(\frac{2 I^{2}θ^{2}_{f}}{t_{1} t^{2}_{2}} + T^{2}_{d} t_{f} \right) (6.5.13)
E = \frac{2}{(0.3)^{2}} \left[ \frac{2(4 × 10^{−6})(20π)^{2}}{0.2(0.4)^{2}} + (0.05)^{2} 0.6 \right] = 22 J/cycle
The power consumption is 22/t_{f} = 37 J/s, or 37 W.
Equation (6.5.12) shows that the maximum speed for the trapezoidal profile is
ω_{max} = \frac{θ_{f}}{t_{2}} = 50π rad/s
which is 50π(60)/(2π) = 1500 rpm. So the motor’s maximum permissible speed must be greater than 1500 rpm.
The rms torque is found from (6.5.17) to be
T_{rms} = \sqrt{\frac{2 I^{2} θ^{2}_{f}}{t_{f} t_{1}t^{2}_{2}} + T^{2}_{d}} (6.5.17)
T_{rms} = \sqrt{\frac{2(4 × 10^{−6})(20π)^{2}}{0.6(0.2)(0.4)^{2}} + (0.05)^{2}} = 1.28 N · m
Use (6.5.16) to compute the maximum required torque.
T_{max} = I α_{max} + T_{d} = I \frac{ω_{max}}{t_{1}} + T_{d} = I \frac{θ_{f}}{t_{1} t_{2}} + T_{d} (6.5.16)
T_{max} = \frac{2 × 10^{−3} (20π)}{0.2(0.4)} + 0.05 = 1.57 + 0.05 = 1.62 N · m
Note that the load torque contributes little to T_{max}. Most of the required torque is needed to accelerate the inertia.
The system time constants are obtained from the roots of (6.4.11), which are s = −643 and s = −23.3. The system must be fast enough to respond to the profile command.
L_{a} I s^{2} + (R_{a} I + c L_{a})s + cR_{a} + K_{b} K_{T} = 0 (6.4.11)
Its largest time constant, 1/23.3 = 0.043 s, is less than one-fourth of the ramp time t_{1} = 0.2, so the system is fast enough.
The amplifier requirements are calculated as follows. Note that because the motor data is given in SI units, K_{b} = K_{T} = 0.3 . From (6.5.18), (6.5.19), and (6.5.20),
i_{max} = \frac{T_{max}}{K_{T}} (6.5.18)
i_{rms} = \frac{T_{rms}}{K_{T}} (6.5.19)
v_{max} = R i_{max} + K_{b} ω_{max} (6.5.20)
i_{max} = \frac{1.62}{0.3} = 5.4 A
i_{rms} = \frac{1.28}{0.3} = 4.27 A
v_{max} = 2(5.4) + 0.3(50π) = 10.8 + 47.1 = 57.9 V
Note that most of the required voltage is needed to oppose the back emf.