Drives Business Unit General Motion Control (June 1999)
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1
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Motor/Drive Sizing Exercise
Drives Business Unit General Motion Control (June 1999)
s
2
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Motor/Drive Sizing Exercise
Y
X
Z
3 - Axis Conveyor
Drives Business Unit General Motion Control (June 1999)
s
3
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Calculation of the Travel gear (X-Axis)
The Following Data is Given:¨ Mass to be transported m= 400 kg¨ Diameter of drive wheel D= 0.14 m¨ Max. speed V max= 1.6 m/s¨ Max. acceleration and deceleration a max= 6.4 m/s2
¨ Distance traveled s= 2 m¨ Cycle time T= 7 s¨ Mech.. efficiency mech.= 0.9¨ Specific travelling resistance w f= 0.1
¨ Mech.. accuracy s mech.= ±0.1 mm
¨ Overall accuracy required s tot= ±0.2 mm
Drives Business Unit General Motion Control (June 1999)
s
4
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Travel Curve (X-Axis)
t t
t
T´
b v
tot
v
vmax Area corresponds to travel distance
Forwards Reverse
t
vmax
t kt p
Drives Business Unit General Motion Control (June 1999)
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5
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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:
sec5.32
7
2
TT
Now determine the remaining time values of the curve.
max
max
a
vtt vb s25.0
4.6
6.1
max
maxmax 22v
tv
tvs
t
vb
k s16.1
225.0
6.1225.0
6.12
s5.125.0125.0
totp tTt
vkbtot tttt
s25.15.3
Drives Business Unit General Motion Control (June 1999)
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6
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Calculation of the Maximum Speed (X-Axis)
Now we must calculate the Maximum speed under load at the drive wheel.
D
vnLoad
60maxmax rpm27.218
14.0
606.1
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:
rpmnin LoadMot 7.218227.21810maxmax
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 inertiaMotor/ Gear box is more economical
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Calculation of the Maximum Torque (X-Axis)Now we must calculate the Maximum Resistance Torque of the load at the drive wheel.
We also need to know the Maximum Acceleration and Deceleration Torque for the Load.
2
Dwgm fW Nm47.27
2
14.01.081.9400
D
aload
2max
2
2
DmJ load
loadloadloadbra J ,
24.9114.0
24.6 s
22
96.12
14.0400 kgm
Nm2.1794.9196.1
Maximum Acceleration and Deceleration Torque for a Rotational motion is given by:
loadloadloadbra J ,
Therefore:
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Calculation of the Maximum Torque (X-Axis) Cont.
Now we can get the Maximum Torque on the output side of the gear.
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
mech
Wloadaload 1)(max Nm6.229
9.0
1)47.272.179(
Therefore:
iJ loadGGbra *, Nm914.0104.91001.0
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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 Encoder:
Total:
60360G
gear
Ds
mm061.0
60
3
360
14.0
I.e. + 0.0305 mm
For the Gear Unit:
zi
Dsencoder
mm04.0
102410
14.0
With a 2-pole resolver
encodergearmechtot ssss mm2.01705.004.00305.01.0
Hence, the required accuracy it complied with.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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?
For a Sin/Cos encoder or Absolute value (ERN/EQN)?
zi
Dsencoder
mm021.0
204810
14.0
zi
Dsencoder
mm00044.0
1010
14.05
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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:
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.
GmechWloadaGbraMotaMot i
1
)(max
95.09.010
1)47.272.179(914.0
Mota NmMota 08.25
2914104.91 sJJiJ MotMotloadMotMota Where:
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Dynamic Limit curve of the 1FT6084-8AF7 Motor (X-Axis)
Mot max
1FT6084-8AF7with SIMOVERT MASTERDRIVE MC 3 AC 480V
0
100
200
300
400
500
600
700
0 1000 2000 3000
Speed [rpm]
Torq
ue
[lb
-in
]
0
10
20
30
40
50
60
70
Torq
ue
[Nm
]
S1 (100K)
Intermittent Operating Region
Continous Operating Region
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Dynamic Limit curve of the 1FT6082-8AF7 Motor (X-Axis)
1FT6082-8AF7with SIMOVERT MASTERDRIVE MC 3 AC 480V
0
50
100
150
200
250
300
350
400
0 1000 2000 3000
Speed [rpm]
Torq
ue
[lb
-in
]
0
5
10
15
20
25
30
35
40
45
Torq
ue
[Nm
]
S1 (100K)
Intermittent Operating Region
Continous Operating Region
Mot max
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Calculation of the Maximum Torque of the Motor (X-Axis)
The Acceleration Torque of the motor itself is thus:
NmMotbra 94.59140065.0,
NmaMotMot 03.3108.2594.5max
Now the Maximum motor torque can be determined.
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
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Calculation of the Travel Curve Torques (X-Axis)
We already have determined the acceleration torque.
NmaMotMot 03.3108.2594.5max
The Torque during constant travel now needs to be calculated.
Finally the torque during deceleration is required.
Gmech
WkMot i 1
Nm21.30959.010
147.27
)()(
1)(
WloadbrsignGmech
WLoadbrGaMotbrbrMoti
Nm83.1910
95.09.0)47.272.179(914.094.5
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Torque Characteristic Curve (X-Axis)
MMot
31.03 Nm
3.21 Nm
-19.83 Nm
0.25 s 1 s 0.25 s
3.5 s
t
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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.
By using the Travel Curve, the Effective Speed is calculated as follows:
The Effective Torque is calculated as follows:
T
tiiMoteff
2 Nm10
5.3
25.083.19121.325.003.31 222
T
tnn
ni
EB
mean2 rpm5.779
5.3
25.027.2182
17.218225.027.2182
By using the S1 Curve for the given motor, a determination can be made whether or not the motor selected is satisfactory.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Checking the Validity of the selected Motor (X-Axis)
M/Nm
Meff
n/min-1nmean
12
10
8
6
4
2
01000 2000 30001500 2500500
14
16
18
20
Since the operating pointis well below the S1 Curve,
the motor selected is suitable.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selecting the Inverter (X-Axis)The Inverter is selected according to the maximum motor current and mean motor current.
The Mean motor current is calculated as follows:
Therefore the Max current is given by:
100
maxmax
Tn
MotMot k
I
A16.2334.1
03.31
Tk
tI
Tn
iiMot
meanMot100
A4.3
5.334.1
25.083.19121.325.003.31
For overload calculations, the motor current during constant travel is calculated:
100Tn
kMotkMot k
I
A4.234.1
21.3
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selecting the Inverter (X-Axis)
Thus, checking:
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.
AIAI UnMot 30316.23max
AIAI UnmeanMot 2.104.3
AIAI UnkMot 3.991.04.2
From the data calculated select the drive that best fits.
Drive Selected: 6SE7021-0TP50 with an Iun = 10.2 A
Since the above criteria are met, a correct selection has been made.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Determination of DC link Currents (X-Axis)
Power output during constant travel, calculation:
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:
Maximum power output during deceleration, calculation:
9550
maxmax
MotaMotaMot
nP
kW09.7
9550
7.218203.31
9550
maxMotkMotkMot
nP
kW734.0
9550
7.218221.3
9550
maxmax
MotbrMotbrMot
nP
kW53.4
9550
7.218283.19
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Power Characteristic Curve (X-Axis)
0.25 s
1 s
0.25 s
PMot
t
7.09 kW
0.734 kW
-4.53 kW
Negative area corresponds to regenerativeoperation
3.5 s
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Determination of DC link Currents (X-Axis)
The Mean Power output during motor operation is given by:
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:
lineInvMot
MotInvLink V
PI
35.1max
max A66.12
46035.198.092.0
7090
T
tPP
Pi
EMotBMot
meanMot2 kW463.0
5.3
1734.025.009.721
Now we need to determine the Mean value of DC current, but before we can do this we need the Mean Power.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Determination of DC link Currents (X-Axis)
The next step involves determining the Braking Power.
Now that we have the Mean Power the mean DC Link currents can be calculated.
The Mean DC Link current is given by:
lineInvMot
meanMotmeanInvLink V
PI
35.1A83.0
46035.198.092.0
463
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Determination of Braking Power (X-Axis)
The Mean Braking Power is obtained from the negative characteristic of the motor output. The calculation is as follows:
The Braking Power and the Mean Braking Power are calculated for later sizing of the braking resistor.
The Maximum Braking Power is given by:
InvMotbrMotbr PP maxmax kW08.498.092.053.4
InvMot
iEbrMotBbrMot
meanbr T
tPP
P
2
kW146.098.092.05.3
25.0)53.4(21
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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!
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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.
NmD
gmH 1.982
1.081.9200
2
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Torque Characteristic Curve of the Lifting Drive (Z-Axis)
MMot 16.58 Nm
11.47 Nm
6.37 Nm4.08 Nm
8.39 Nm
12.7 Nm
0.6 s 0.3 s 0.6 s 0.6 s 0.3 s 0.6 s7 s
Lifting Lowering
t
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selection of the Inverter for the Lifting Drive (Z-Axis)
Drives Business Unit General Motion Control (June 1999)
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32
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selection of the Inverter for the Lifting Drive (Z-Axis)
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Power output Characteristic Curve for the Lifting Drive (Z-Axis)
PMot
0.6 s 0.3 s 0.6 s
0.6 s 0.3 s 0.6 s
7 s
Lifting Lowering
t
4.97 kW
3.44 kW
1.91 kW
-1.22 kW
-2.52 kW
-3.81 kW
Negative area corresponds to regenerativeoperation
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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:
maxmaxRe InvLinkctLink II AAA 84.2118.966.12
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selection of the Rectifier Unit for the 3-Axis Conveyer
Next, calculate the Mean value of the DC Link current as follows:
meanInvLinkmeanctLink II Re AAA 65.182.083.0
Now base on this calculated information, choose a Rectifier.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selection of the Rectifier Unit for the 3-Axis Conveyer
Confirmation of this choice is given by:
AI ctLink 84.21maxRe AI nLink 6.656.1
AI meanctLink 89.1Re AI nLink 41
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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:
Invbrbr PP max kWkWkW 4.732.308.4
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selection of the Braking Resistor for the 3-Axis Conveyer
Next, we need to determine the Mean Braking Power as follows:
meanInvbrmeanbr PP kWkWkW 426.028.0146.0
Now that all the calculation have been made a suitable resistor needs to be chosen.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Selection of the Braking Resistor for the 3-Axis Conveyer
kWPbr 4.7max kWP 5.75.1 20
kWP meanbr 426.0 kWP 11.15.4/20
Checking our selection:
Since the above conditions have been met, our choice is valid.
Drives Business Unit General Motion Control (June 1999)
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40
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Other requirements for the 3-Axis Conveyer
Drives Business Unit General Motion Control (June 1999)
s
41
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Other requirements for the 3-Axis Conveyer
Drives Business Unit General Motion Control (June 1999)
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42
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Other requirements for the 3-Axis Conveyer Cont.
Drives Business Unit General Motion Control (June 1999)
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43
MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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????
Lets Look at the Function Diagram for the Operating Mode MDI [823]
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Ability of Drive to actually do the 3-Axis Conveyer
U953.32=___ (20)
1 0
0 1 10
U550.2 U559.2
0 1 10
U532 (0)
KKU550.3 U559.3
0 1 10
U533 (0)
KK
U534 (0)
KK
Position (X) Speed (F)
X
n540.12
F
MDI block number 0...10;Following is displayed:- at standstill: selected MDI block- during travel: currently traversed MDI block - no MDI mode ==> display "0"
U559.1 U550.1
<2>Recommended: U953.32=4
<4>
n540.13
-999 999 999... 999 999 999 LU 1... 100 000 000 [x 10 LU/min]90 30... 91 39
(90 30)
G
at MD1=3
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Ability of Drive to actually do the 3-Axis Conveyer
T2 = 4 x T0T3 = 8 x T0T4 = 16 x T0T5 = 32 x T0T6 = 64 x T0T7 = 128 x T0T8 = 256 x T0T9 = 512 x T0T10 = 1024 x T0
23456789
1011 ... 19
20
Parameter for setting the sampling timeValue range: 2 ... 20Factory setting: 20 (block is not calculated)Parameter value Sampling time
(T0 = 1/pulse frequency = 1/P340)
Sampling time at 10 kHzpulse frequency (T0 = 100 µs)
Reserved for future applicationsBlock is not calculated
0.4 ms0.8 ms1.6 ms3.2 ms6.4 ms
12.8 ms25.6 ms51.2 ms
102.4 ms
Sampling time at 5 kHzpulse frequency ( T0 = 200 micro sec)
0.8 ms01.6 ms
3.2 ms6.4 ms
12.8 ms25.6 ms51.2 ms
102.4 ms204.8 ms
Typically the Position Loop should be calculated at least 10 times!
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
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.
msTimePosition 32 ms250
Since the Positioning Time of the Position Loop is less than the required Acceleration and Deceleration time, “WE CAN DO THAT”.
Drives Business Unit General Motion Control (June 1999)
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MASTERDRIVES MC
Drives and Standard Productsfrom Siemens
Motor/Drive Sizing Exercise
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