Page 235 - Demo

V012024 • © 2024 Portescap. Specifications subject to change without notice.Chopper Driver – No Gearbox – With Ramping AccelerationThe load inertia of 20 × 10-7 kg-m2 has to be moved by an angle of 0.50 rad (θ) in 20 msec using an acceleration ramp. We will use a triangular speed profile with an acceleration time of 10 msec for a shaft movement of .25 rad. The speed required is calculated as follows. Acceleration = α = 2 × (θ/∆t2) = 2 × .25 rad /(0.01 sec)2 = 5,000 rad/s2 Speed = ω = (5,000 rad/s2) × 0.01 sec = 50 rad/s nmotor = 50 rad/s × 60/2π rpm-s/rad = 477.5 rpmThe torque necessary to accelerate the load is: τload = Jload × α [Nm] τload = 20 × 10-7 kg-m2× 5,000 rad/s2 = .010 NmWith a triangular speed profile this requires a peak speed up to 477.5 rpm, with a load torque of 10 mNm. Therefore the peak power required for the load alone is 0.50 W. P = τ × ω = .010 Nm × 50 rad/s = .50 WattsLet us first consider the P430 with rotor inertia of 3 × 10-7 kgm2. We have:Jtotal = Jload + Jmotor = (20 × 10-7 kg-m2) + (3 × 10-7 kg-m2) = 23 × 10-7 kg-m2τtotal = Jtotal × α = 23 × 10-7 kg-m2× 5,000 rad/s2 = .0115 Nm = 11.5 mNmThe curve shows that P430-258-005 with current limit of 1.6 A and 24 V can easily do the job.In fact, at around 500 rpm the motor can provide up to 43 mNm. With a safety factor of 75% we could easily provide 32 mNm with the P430 and the motion time could be reduced as followed:αmax = τmax / Jtotal = 32 x 10-3 Nm / 23 x 10-7 kg-m2 = 13,913 rad/s2.∆t = √(2 x θ / αmax) = √(2 x 0.25 rad / 13,913 rad/s2) = 6.0 msec0102030400.01.02.03.04.05.06.07.00 1000 2000 4000 6000 8000 100000 600 1200 2400 3600 4800 6000oz-in mNmP430 258 013 ParallelTorque vs Speed, Half-step ModerpmppsP430 258 013 @ 24V P430 258 013 @ 24V, 36VP430 258 013 @ 36VPull-Out Torque Pull-In Torqueω10 ms 10 msSpeed Profile235