Transcript Drive family

s
MASTERDRIVES MC
Motor/Drive Sizing Exercise
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s
MASTERDRIVES MC
Motor/Drive Sizing Exercise
Z
Y
X
3 - Axis Conveyor
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s
MASTERDRIVES MC
Calculation of the Travel gear (X-Axis)
The Following Data is Given:
¨ Mass to be transported
¨ Diameter of drive wheel
¨ Max. speed
¨ Max. acceleration and deceleration
¨ Distance traveled
¨ Cycle time
¨ Mech.. efficiency
¨ Specific travelling resistance
¨ Mech.. accuracy
¨ Overall accuracy required
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m= 400 kg
D= 0.14 m
V max= 1.6 m/s
a max= 6.4 m/s2
s= 2 m
T= 7 s
 mech.= 0.9
w f= 0.1
s mech.= ±0.1 mm
s tot= ±0.2 mm
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MASTERDRIVES MC
Travel Curve (X-Axis)
v
v
Forwards
max
Reverse
Area corresponds to travel distance
t
tb
tk
tv
tp
t tot
T´
v
max
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MASTERDRIVES MC
Calculation of the Time Segments (X-Axis)
Since the travel is symmetrical, we only have to consider the forward movement!
So the new cycle time is:
T
7
T 
  3.5 sec
2
2
Now determine the remaining time values of the curve.
vmax
tb  t v 

amax
tk 
tb
t
 vmax  v
2
2 
vmax
s  vmax 
ttot  tb  tk  tv 
t p  T   ttot 
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1.6
 0.25s
6.4
2  1.6 
0.25
0.25
 1.6 
2
2  1s
1.6
0.25  1  0.25  1.5s
3.5  1.5  2s
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MASTERDRIVES MC
Calculation of the Maximum Speed (X-Axis)
Now we must calculate the Maximum speed under load at the drive wheel.
nLoad max 
vmax  60

 D
1.6  60
 218.27rpm
  0.14
Due to low rpm, a gearbox should be used to better match the motor speed to the load speed. In this case
a gearbox with a transmission ratio of i=10 is chosen, giving a resulting rpm at the motor of:
nMotmax  i  nLoad max  10  218.27  2182.7rpm
Why is a gearbox necessary? Why not just select a motor with a lower rpm?
Answer:
A smaller motor can be used - larger motor mean larger motor inertia
Motor/ Gear box is more economical
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MASTERDRIVES MC
Calculation of the Maximum Torque (X-Axis)
Now we must calculate the Maximum Resistance Torque of the load at the drive wheel.
W  m  g  w f 
D
0.14
 400  9.81  0.1 
 27.47 Nm
2
2
We also need to know the Maximum Acceleration and Deceleration Torque for the Load.
Maximum Acceleration and Deceleration Torque for a Rotational motion is given by:
 a ,br load  J load  load 
Therefore:
 load  amax 
2
2
 6.4 
 91.4s  2
D
0.14
2
J load
D
 m   
2
2
 0.14 
2
400  
  1.96kgm
 2 
 a ,br load  J load  load  1.96  91.4  179.2Nm
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MASTERDRIVES MC
Calculation of the Maximum Torque (X-Axis) Cont.
Now we can get the Maximum Torque on the output side of the gear.
 loadmax  ( a load   W ) 
1
mech
 (179.2  27.47)  1  229.6Nm
0.9
We also need to know the Maximum Acceleration and Deceleration Torque of the gear unit itself.
The Technical Data of the gear unit is as follows:
¨ Gear Ratio
i = 10
¨ Max, Torque
 G = 400 Nm
¨ Torsional Play
 G = 3’
¨ Gear Unit Efficiency  G.= 0.95
- Inertia
J G.= 0.001 Kgm2
Therefore:
 a,br G  J G*   load  i  
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0.001  91.4 10  0.914Nm
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MASTERDRIVES MC
Calculation of the Position Accuracy (X-Axis)
Before we choose a motor, we should determine whether or not the position accuracy required is met.
For the Gear Unit:
s gear 
D  G



360 60
0.14   3

 0.061mm
360 60
I.e. + 0.0305 mm
For the Encoder:
sencoder 
D 

iz
0.14  
 0.04mm
10 1024
With a 2-pole resolver
Total:
stot  smech  s gear  sencoder  0.1  0.0305  0.04  0.1705  0.2mm
Hence, the required accuracy it complied with.
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MASTERDRIVES MC
Calculation of the Position Accuracy of other Feedback Devices
What would be the accuracy if a different feedback device had been chosen?
For a Pulse encoder with 2048 PPR?
sencoder 
D 

iz
0.14  
 0.021mm
10  2048
For a Sin/Cos encoder or Absolute value (ERN/EQN)?
sencoder 
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D 
0.14  

 0.00044mm
5
iz
10 10
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MASTERDRIVES MC
Calculation of the Maximum Torque reflected to the Motor (X-Axis)
Maximum Torque reflected to the motor is determined by adding all the individual torque values.
Therefore:
 Mot max   a Mot   a br G  ( a load   W ) 
  a Mot  0.914  (179.2  27.47) 
Where:
1
10  0.9  0.95
1
i mech G
  a Mot  25.08Nm
 a Mot  J Mot load  i  J Mot  91.4 10  J Mot  914s 2
Hence, before we can figure out the Complete Torque a motor must be chosen. Since we know some of
the torque required, an educated guess must be made in choosing an appropriate motor.
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MASTERDRIVES MC
Calculation of the Maximum Torque reflected to the Motor (X-Axis)
Choose a Synchronous Servo motor (1FT6) based on the information known and the Dynamic Limit
curves.
The first 1FT6 motor with nn=3000 rpm, which satisfies the condition of the Dynamic Limit curve is
1FT6084-8AF7 with the following characteristics:
- Pn = 4.6 kW
- n = 14.7 Nm
- Max = 65 Nm
- Jmot = 0.0065 kgm2
- kTn = 1.34 Nm/A
- mot = 0.92
- 0 = 20 Nm
Explain why this motor was chosen over any other motor.
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MASTERDRIVES MC
Dynamic Limit curve of the 1FT6084-8AF7 Motor (X-Axis)
1FT6084-8AF7
with SIMOVERT MASTERDRIVE MC 3 AC 480V

700
70
600
60
500
400
Intermittent Operating Region
40
300
Torque [Nm]

Torque [lb-in]
50
30
Mot max
200
S1 (100K)
20
100
10
Continous Operating Region
0
0
0
1000
2000
3000
Speed [rpm]
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MASTERDRIVES MC
Dynamic Limit curve of the 1FT6082-8AF7 Motor (X-Axis)
1FT6082-8AF7
with SIMOVERT MASTERDRIVE MC 3 AC 480V

400
45
350
40
35
300

30
Mot max
Intermittent Operating Region
25
200
20
Torque [Nm]
Torque [lb-in]
250
150
15
S1 (100K)
100
10
50
5
Continous Operating Region
0
0
0
1000
2000
3000
Speed [rpm]
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MASTERDRIVES MC
Calculation of the Maximum Torque of the Motor (X-Axis)
The Acceleration Torque of the motor itself is thus:
 a ,br Mot  0.0065  914  5.94Nm
Now the Maximum motor torque can be determined.
 Motmax   Mota  5.94  25.08  31.03Nm
Next a check must be done to ensure that the thermal limits of the motor are not exceeded. This is
accomplished by determining the motor torques at every point of the travel curve.
We already have determined the acceleration torque. 31.03 Nm
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MASTERDRIVES MC
Calculation of the Travel Curve Torques (X-Axis)
We already have determined the acceleration torque.
 Motmax   Mota  5.94  25.08  31.03Nm
The Torque during constant travel now needs to be calculated.
 Motk   W 
1
i mech G
 27.47 
1
 3.21Nm
10  0.9  095
Finally the torque during deceleration is required.
 Mot br   br Mot   a G  ( br Load   W ) 
1
i  (mech G )
sign(  br load  W )
0.9  0.95
 5.94  0.914  (179.2  27.47) 
 19.83Nm
10
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MASTERDRIVES MC
Torque Characteristic Curve (X-Axis)
M
Mot
31.03 Nm
3.21 Nm
t
0.25 s
1s
0.25 s
-19.83 Nm
3.5 s
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MASTERDRIVES MC
Checking the Validity of the selected Motor (X-Axis)
To verify it the motor we selected from the Dynamic Curve Profile is valid, the effective(rms) torque and
speed must be determined.
The Effective Torque is calculated as follows:
 eff 

2
Moti
T
 ti

31.032  0.25  3.212 1  19.832  0.25
 10Nm
3.5
By using the Travel Curve, the Effective Speed is calculated as follows:
nmean 

nB  nE
 ti
2

T
2182.7
2182.7
 0.25  2182.7 1 
 0.25
2
2
 779.5rpm
3.5
By using the S1 Curve for the given motor, a determination can be made whether or not the motor
selected is satisfactory.
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MASTERDRIVES MC
Checking the Validity of the selected Motor (X-Axis)
M/Nm
20
18
16
Since the operating point
is well below the S1 Curve,
the motor selected is suitable.
14
12
Meff 10
8
6
4
2
0
500
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nmean 1000
1500
2000
2500
3000
n/min-1
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MASTERDRIVES MC
Selecting the Inverter (X-Axis)
The Inverter is selected according to the maximum motor current and mean motor current.
Therefore the Max current is given by:
I Motmax 
 Motmax
kTn100
31.03
 23.16 A
1.34

The Mean motor current is calculated as follows:
I Motmean



Moti
 ti
kTn100  T 

31.03  0.25  3.21 1  19.83  0.25
 3.4 A
1.34  3.5
For overload calculations, the motor current during constant travel is calculated:
I Motk 
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 Motk
kTn100

3.21
 2.4 A
1.34
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MASTERDRIVES MC
Selecting the Inverter (X-Axis)
Since the Accelerating and Decelerating times are < 250 ms and the time between is > 750 ms, a check
should be made to see if the 300% overload capability of the Compact Plus unit can be utilized.
From the data calculated select the drive that best fits.
Drive Selected: 6SE7021-0TP50 with an Iun = 10.2 A
Thus, checking:
I Motmax  23.16 A  3  IUn  30 A
I Mot mean  3.4 A  IUn 10.2 A
I Mot k  2.4 A  0.91 IUn  9.3 A
Since the above criteria are met, a correct selection has been made.
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MASTERDRIVES MC
Determination of DC link Currents (X-Axis)
The Max DC link current and the mean DC link current must be determined for later rating of the rectifier
unit. This is done by first determining all of the motor power levels within the travel curve.
Maximum power output during acceleration, calculation:
PMota max 
 Mota  nMotmax
9550
Power output during constant travel, calculation:
PMotk 
 Motk  nMotmax
9550


31.03  2182.7
 7.09kW
9550
3.21 2182.7
 0.734kW
9550
Maximum power output during deceleration, calculation:
PMot br max 
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 Mot br  nMotmax
9550

 19.83  2182.7
 4.53kW
9550
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MASTERDRIVES MC
Power Characteristic Curve (X-Axis)
P
Mot
7.09 kW
0.734 kW
0.25 s
t
1s
0.25 s
Negative area corresponds to regenerative
operation
-4.53 kW
3.5 s
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s
MASTERDRIVES MC
Determination of DC link Currents (X-Axis)
Now that we have the Power Characteristic curve, the maximum power can be determined and the
Maximum DC Link current can be calculated.
Maximum DC Link current during acceleration is given by:
I LinkInvmax 
PMotmax
Mot Inv 1.35 Vline
7090

 12.66 A
0.92  0.98 1.35  460
Now we need to determine the Mean value of DC current, but before we can do this we need the Mean
Power.
The Mean Power output during motor operation is given by:
PMotmean 
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
PMotB  PMotE
2
T
 ti

1
 7.09  0.25  0.734 1
2
 0.463kW
3.5
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MASTERDRIVES MC
Determination of DC link Currents (X-Axis)
Now that we have the Mean Power the mean DC Link currents can be calculated.
The Mean DC Link current is given by:
I LinkInvmean 
PMotmean
Mot Inv 1.35 Vline

463
 0.83A
0.92  0.98 1.35  460
The next step involves determining the Braking Power.
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MASTERDRIVES MC
Determination of Braking Power (X-Axis)
The Braking Power and the Mean Braking Power are calculated for later sizing of the braking resistor.
The Maximum Braking Power is given by:
Pbrmax  PMot br max Mot  Inv   4.53  0.92  0.98  4.08kW
The Mean Braking Power is obtained from the negative characteristic of the motor output. The calculation
is as follows:
Pbr mean 

PMot br B  PMotbrE
2
T
 ti
Mot  Inv
1
 (4.53)  0.25
2

 0.92  0.98  0.146kW
3.5
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MASTERDRIVES MC
Calculation of the Travel Gear (Y-Axis)
Since the Y-Axis is the same as the X-Axis, you will now proceed with all of the same calculations for the
Y-Axis on your own and present the results.
Ha Ha - Just Kidding!
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MASTERDRIVES MC
Calculation of the Lifting Drive (Z-Axis)
Since the Z-Axis is slightly different than the X-Axis and Y-Axis, we will now go over the major differences
and dispense with the entire calculation.
What are the differences?
Travel Distance is represented in Height.
The entire Travel Curve must be taken into consideration because of
the
different torque values for Lifting and Lowering even if it is
Symmetrical.
Rotational Force is translated to Linear Force via a Rack and Pinion.
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MASTERDRIVES MC
Calculation of the Torque of the Lifting Drive (Z-Axis)
Since the Z-Axis is a lifting drive the, the Resistance Torque Calculation is now called the Lifting Torque.
H  m g 
D
0.1
 200  9.81 
 98.1Nm
2
2
Since the Z-Axis is a lifting drive the, ALL Torque cases must be evaluated.
•
•
•
•
•
•
Lifting the load, motor torque during constant travel
Lowering the load, motor torque during constant travel
Lifting the load, motor torque during acceleration
Lifting the load, motor torque during deceleration
Lowering the load, motor torque during acceleration
Lowering the load, motor torque during deceleration
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MASTERDRIVES MC
Torque Characteristic Curve of the Lifting Drive (Z-Axis)
Lifting
M
Mot
Lowering
16.58 Nm
12.7 Nm
11.47 Nm
8.39 Nm
6.37 Nm
4.08 Nm
t
0.6 s
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0.3 s
0.6 s
7s
0.6 s
0.3 s
0.6 s
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MASTERDRIVES MC
Selection of the Inverter for the Lifting Drive (Z-Axis)
The same calculations used for the X & Y - Axis are followed here, too, with one major consideration
change.
What is it?
The acceleration and Deceleration times are > 250ms, so only the 160%
rating can be applied!
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overload
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MASTERDRIVES MC
Selection of the Inverter for the Lifting Drive (Z-Axis)
Once again, since the Z-Axis is a lifting drive the, ALL Power cases must be evaluated for future
determination of DC Link current needed for rectifier selection.
•
•
•
•
•
•
Lifting the load, Power output during constant travel
Lowering the load, Power output during constant travel
Lifting the load, Power output during acceleration
Lifting the load, Power output during deceleration
Lowering the load, Power output during acceleration
Lowering the load, Power output during deceleration
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MASTERDRIVES MC
Power output Characteristic Curve for the Lifting Drive (Z-Axis)
P
Lifting
Mot
Lowering
4.97 kW
3.44 kW
1.91 kW
0.6 s
0.3 s
0.6 s
t
0.6 s
0.3 s
0.6 s
-1.22 kW
-2.52 kW
-3.81 kW
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Negative area corresponds to regenerativ e
operation
7s
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MASTERDRIVES MC
Selection of the Rectifier Unit for the 3-Axis Conveyer
Now that all of the Inverters have been chosen, a rectifier needs to be selected.
Recall that the X and Z -Axis operate simultaneously.
Also, recall that the Y-Axis is much smaller than any of the other axis’ and, thus, it does not need to be
considered in the calculation of the rectifier.
First calculate the maximum DC Link current as follows:
I LinkRe ctmax   I LinkInvmax 
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12.66 A  9.18A  21.84 A
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MASTERDRIVES MC
Selection of the Rectifier Unit for the 3-Axis Conveyer
Next, calculate the Mean value of the DC Link current as follows:
I LinkRe ctmean   I LinkInvmean 
0.83A  0.82 A  1.65 A
Now base on this calculated information, choose a Rectifier.
To maintain a uniform design, a Compact Plus Rectifier should be chosen.
Therefore we should choose:
6SE7024 - 1EP85-0AA0 with IDCn = 41 A
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MASTERDRIVES MC
Selection of the Rectifier Unit for the 3-Axis Conveyer
Confirmation of this choice is given by:
I LinkRe ct max  21.84 A 
1.6  I Linkn  65.6 A
I LinkRe ct mean  1.89 A  I Linkn  41A
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MASTERDRIVES MC
Selection of the Braking Resistor for the 3-Axis Conveyer
What components are needed?
Just the Braking Resistor! Why?
The Chopper is integrated into the Compact Plus Rectifier!
What assumptions can be made?
Since the X & Z-Axis travel simultaneously, both may brake at the same
time.
So, first we need to calculate the maximum Braking Power Level as follows:
Pbrmax   PbrInv   4.08kW  3.32kW  7.4kW
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MASTERDRIVES MC
Selection of the Braking Resistor for the 3-Axis Conveyer
Next, we need to determine the Mean Braking Power as follows:
Pbrmean   PbrInvmean   0.146kW  0.28kW  0.426kW
Now that all the calculation have been made a suitable resistor needs to be chosen.
Therefore, we should choose:
6SE7018 - 0ES87 - 2DC0, with P20 = 5 kW, & R = 80
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MASTERDRIVES MC
Selection of the Braking Resistor for the 3-Axis Conveyer
Checking our selection:
Pbr max  7.4 kW  1.5  P20  7.5kW
Pbr mean  0.426kW 
P20 / 4.5  1.11kW
Since the above conditions have been met, our choice is valid.
Drives Business Unit
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MASTERDRIVES MC
Other requirements for the 3-Axis Conveyer
What other components/options are needed?
Given:
Positioning to be carried out non-centrally.
Need Technology Option, Order # 6SW1700-5AD00-1XX0
All three motions are synchronized.
Need SIMOLINK board, Order # 6SX7010-0FJ00
PROFIBUS communication to PLC.
Need CBP board, Order # 6SX7010-0FF00
A Resolver is used.
Need SBR1/2 board, Order # 6SX7010-0FC00 or
6SX7010-0FC00
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MASTERDRIVES MC
Other requirements for the 3-Axis Conveyer
What other components/options are needed that are not mentioned?
Motor Power Cables - assembled.
Order # 6FX2002-5DA31-???0, ??? = Cable Length
Feed Back Cables - assembled.
Order # 6FX2002-2cf01-???0, ??? = Cable Length
Other Miscellaneous options?
I/O Expansion Boards
RFI Filter
Line Reactor
Capacitor Module
OP1S
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MASTERDRIVES MC
Other requirements for the 3-Axis Conveyer Cont.
Other Miscellaneous options?
-Z +K80, Safety Off - Special Order with extended Lead Time
-Z +C91, 24 VDC Power Supply for Rect. - Special Order with extended
Lead Time
There will be some other Option Codes Available, but they all bring
times with them.
Drives Business Unit
General Motion Control (June 1999)
extended Lead
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MASTERDRIVES MC
Ability of Drive to actually do the 3-Axis Conveyer
Now that everything has been calculated and all the options and accessories have been chosen, can the
drive do the application????
What is the limiting factor?
The Acceleration and Deceleration times and how they relate to the Drive
for the Position Loop.
Process time
Lets Look at the Function Diagram for the Operating Mode MDI [823]
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Ability of Drive to actually do the 3-Axis Conveyer
Position (X)
Speed (F)
U953.32=___ (20)
90 30... 91 39
(90 30)
U550.1 U559.1
0
1
<2>
-999 999 999... 999 999 999 LU
U532 (0)
KK
10
U550.2 U559.2
0
U534 (0)
KK
1
1
1... 100 000 000 [x 10 LU/min]
U533 (0)
KK
U550.3 U559.3
10
0
1
10
Recommended: U953.32=4
<4>
n540.13 MDI block number 0...10;
Following is displayed:
- at standstill: selected MDI block
- during travel: currently traversed
MDI block
- no MDI mode ==> display "0"
0
n540.12
G
X
F
at MD1=3
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Ability of Drive to actually do the 3-Axis Conveyer
Parameter f or setting the sampling time
Value range: 2 ... 20
Factory setting: 20 (block is not calculated)
Parameter v alue Sampling time
Sampling time at 10 kHz
Sampling time at 5 kHz
pulse
f
requency
(T0
=
100
µs)
pulse f requency ( T0 = 200 micro sec)
(T0 = 1/pulse f requency
= 1/P340)
2
T2 =
4 x T0
0.4 ms
0.8 ms
3
T3 =
8 x T0
0.8 ms
01.6 ms
4
T4 =
16 x T0
1.6 ms
3.2 ms
5
T5 =
32 x T0
3.2 ms
6.4 ms
6
T6 =
64 x T0
6.4 ms
12.8 ms
7
T7 =
128 x T0
12.8 ms
25.6 ms
8
T8 =
256 x T0
25.6 ms
51.2 ms
9
T9 =
512 x T0
51.2 ms
102.4 ms
10
T10 =
1024 x T0
102.4 ms
204.8 ms
11 ... 19
Reserv ed f or f uture applications
20
Block is not calculated
Typically the Position Loop should be calculated at least 10 times!
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MASTERDRIVES MC
Ability of Drive to actually do the 3-Axis Conveyer
This gives you a value of 32 ms for accurate positioning.
Lets check this value with the Acceleration and Deceleration time required by the application.
Position Time  32 ms 
250 ms
Since the Positioning Time of the Position Loop is less than the required Acceleration and Deceleration
time, “WE CAN DO THAT”.
Drives Business Unit
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Drives and
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s
MASTERDRIVES MC
Motor/Drive Sizing Exercise
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