303
Inverter Positioning Section 5-3
Example of
Calculating
Conversion Factor
Conditions
• Power Supply Frequency for One Motor Revolution per Second: 2 Hz
(PLC Setup)
• Number of Encoder Pulses for One Motor Revolution: 1,000 (PLC Setup)
• Rotary encoder resolution: 1,000 pulses/revolution (encoder specifica-
tion)
• High-speed counter multiplier: x4 (PLC Setup)
• Gear ratio between motor and encoder shafts: 1/4 (machine specifica-
tion)
• Error Counter Cycle: 12 ms (PLC Setup)
Calculation
The calculation performed inside the PLC is as shown below.
Serial Communications The command value calculated above is used in the Modbus-RTU command
frame, adjusting for the frequency unit. (See note.)
Refer to
6-3-3 Modbus-RTU Easy Master Function
and to the inverter manual
for details on Modbus-RTU communications.
Note If the frequency command unit set in the inverter is 0.1 Hz, divide the com-
mand frequency in A23 or A33 by 10.
Analog Output The following example is for the CP1W-DA041.
The analog output resolution is 6,000, so the command value calculated
above is multiplied by 6,000 divided by the inverter’s maximum output fre-
quency.
CP1L
PLC Setup
Output value
Automatic
calculation
Inverter frequency
command value
Auxiliary Area
A23 or A33
Unit: Hz
Auxiliary Area
A20/A21 or
A30/A31
Unit: 0.01 Hz
A23/A33
Inverter frequency
command value
(0.1-Hz increments)
2
1000 × 0.012
Output value: Hz
A20/A21
A30/A31
Output value: Hz
A20/A21
A30/A31
17
(To 0.01-Hz
increments)
100
CP1L
Automatic
calculation
Auxiliary Area
A23/A33
Inverter
frequency
command value
Serial communications
Unit: 0.01 Hz
To Next Page
Loading...
