Ironcore and Ironless Linear Servo Motors
Motor Sizing Example
Let's assume we want to move horizontally a mass of 6 kg point to point for a distance of 100 mm (X) in 205 msec including settling time (Tm) to +/- 1 micron. Total travel is 400 mm, and a dwell time of 200 msec is needed after each move.
Move profile
We will further assume an estimated settling time of 30 msec (Tst).Now, let's also assume a 25 msec smoothing time (Tj) (time for the current to go ramp up linearly from zero to the full peak current).
So the move cycle time (Tc) is 205 + 200 = 405 msec
Using move formulas note in previous section:
T (msec) = Tm - (Tst + Tj)
T (msec) = 205 - (30 + 25) = 150 msec
T (msec) = 205 - (30 + 25) = 150 msec
We will assume an efficient trapezoidal profile (1/3, 1/3, 1/3)
Acceleration needed here (see previous move formula):
A = (4.5)*100*10-3 /(0.15)2
A = 20 m/sec2 (about 2 "g")
V = (1.5)*0.1/0.15
V = 1.0 m/sec
Jerk = 20/25*10-3
Jerk = 800 m/sec3
A = 20 m/sec2 (about 2 "g")
V = (1.5)*0.1/0.15
V = 1.0 m/sec
Jerk = 20/25*10-3
Jerk = 800 m/sec3
The acceleration and deceleration time becomes (150/3) + 25 = 75 msec
Since the smooth time is 25 msec, the time at constant acceleration is 75 - (2*25) = 25 msec
The time at constant speed is now (150/3) - 25 = 25 msec
Linear Motor Selection
We can estimate the acceleration force of the load only (see previously mentioned formula) at 2g*9.81*6 kg = 117 N.Based on this we can select coil LC-50-100-D (peak force = 318 N, continuous force = 139 N) assuming a coil mounting plate of 1 kg.
Total moving mass: 6 kg (load) + 1 kg (plate) + 1.63 kg (coil mass) = 8.63 kg
Coil magnetic attraction Force Fa = 690 N, Coil resistance = 3.76 ohm, Coil Force constant 30.3 N/Ap, Thermal Resistance
1.3°C/W, Back Emf 35.8 Vp/m/sec, Inductance p-p 36 mH, Electrical cycle length 50 mm
We assume a good set of linear bearings with µ=0.005 and 20 N of friction.
| Friction Force | Ff (N) = 8.63*9.81*[sin(0) + 0.005*cos(0)] + 690*0.005 + 20 = 24 N |
|---|---|
| Inertial Force | Fi (N) = 8.63*20 = 173 N |
| Lets neglect the damping force | Fd = 0 |
| Total Acceleration Force | Fta (N) = 173 + 24 = 197 N |
| Total Constant Velocity Force | Ftcv (N) = 24 N |
| Total Deceleration Force | Ftd (N) = 173 - 24 = 149 N |
| Total Dwell Force | Ftdw (N) = 0 N |
| RMS Force | Frms (N) = √[{1972 *(25*2/3+25)+242 *25+1492 *(25*2/3+25)+ (1972 +1492 )*0.25*30}/405] |
| Frms (N) = 86.3 N | |
| RMS Current | Ica = 86.3/30.3 = 2.85 Amp 0-p |
| Ica = 2.85/√2 = 2.01 Amp rms | |
| Peak Current | Ipa = 197/30.3 = 6.5 Amp (0-p) |
| Ipa = 6.5/√2 = 4.6 Amp rms | |
| Motor Coil Temperature | Tc (°C) = 25 + 1/[1/(1.5*3.76*2.01 2 *1.3)-1/259.5]= 58.4°C |
| Motor Resistance Hot | Rhot = 3.76*[234.5+58.4]/(234.5+25) =4.25 ohm |
| Motor Power Losses | Pl (W) = 1.5*4.25 *2.012 = 26 W |
| Voltage due to Back Emf | Vbemf = 1.2*1 m/sec * 35.8 Vp/m/sec = 43 V |
| Voltage due to R*I | Vri = 0.866*4.25 (ohm) * 6.5 (Amp 0-p) = 24 V |
| Voltage due to Inductance | VL = 5.44*1.2*1 (m/sec)*36*6.5/50 = 31 V |
| Bus Voltage needed | Vbus = 1.15*√[(43+24)2 + 31 2 ] = 85 Vdc |
Notes:
- Vbus is a worst case since we have assumed no phase advance and no jerk time (max speed at max acceleration).
- An Allen Bradley Digital Servo Drives Ultra 3000 Model 2098-DSD-010 with 5 A (0-peak) continuous and 15 A (0-peak) of peak current will do the job here with either AC input 115 Vac or 230 Vac.