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E6581315
•If the “Frequency point selection” function is disabled (=)
The operation frequency (frequency command value) of the inverters are calculated using the fol- lowing equations, with the received data in the following equation used as the data received from the master inverter when inverters are operated under the control of a master inverter
fc( Hz ) = | Slave recieve data(%)⋅ Slave side FH |
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| Example: |
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| Maximum frequency | Operation frequency command value | ||||||||
| Master (Fc) |
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| 100.00Hz (10000) |
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| 50.00Hz (5000) | |||||||
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| 90.00Hz (9000) |
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| 45.00Hz (4500) | |||||||
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| 80.00Hz (8000) |
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| 40.00Hz (4000) | |||||||
Master send data：fc(%) = | Master side fc ⋅ 10000 | = | 5000×10000 | = 5000 = 50% | ||||||||||
Master side FH | 10000 | |||||||||||||
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Slave1 : fc( Hz ) = | 5000 ⋅ 9000 | = 4500 = 45Hz |
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10000 |
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Slave 2 : fc( Hz ) = |
| 5000 ⋅ 8000 | = 4000 = 40Hz |
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10000 |
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•If the “Frequency point selection” function is enabled (≠)
When inverters are operated under the control of a mater inverter, the operation frequency (fre- quency command value) of the slave inverters are calculated using the following equations.
When inverters are operated under the control of a computer, read “command from the master inverter” in the following equations as “command from the computer.”
fc( Hz ) = |
| Po int 2 frequency − Po int1 frequency | ⋅（Master command(%) − Po int1）＋Po int1 frequency | |||||||||||||||||
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| Po int 2 − Po int1 |
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| (Hz) |
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| Example: | Units: Frequency unit 1 = 0.01Hz, Point setting unit 1 = 0.01% | ||||||||||||||||||
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| Maximum | Point 1 |
| Point 1 fre- |
| Point 2 set- |
| Point 2 | Frequency |
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| frequency | setting |
| quency |
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| frequency | (Fc) |
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| Master (Fc) | 100.00Hz | － |
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| － | 50.00Hz |
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| (5000) |
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| Slave 1 | 100.00Hz | 0.00% |
| 0.00Hz |
| 100.00% |
| 90.00Hz | 45.00Hz |
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| (10000) |
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| (10000) |
| (9000) | (4500) |
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| Slave 2 | 100.00Hz(1 | 0.00% |
| 0.0Hz |
| 100.00%(10 |
| 80.00Hz | 40.00Hz |
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| 0000) |
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| 000) |
| (8000) | (4000) |
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Data sent by the master inverter |
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Master send data : fc(%) = | Master side fc ⋅10000 | = | 5000×10000 | = 5000 = 50% |
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| Master side FH |
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Both slaves 1 and 2: Result of a conversion made on the slave side |
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fc( Hz ) = | Slave receive data(%) ⋅ Slave side FH | = | 5000 ⋅10000 | = 5000 = 50Hz |
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| 10000 |
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| 10000 |
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Both slaves 1 and 2: Result of a conversion to % made prior to a conversion to point frequency | ||||||||||||||||||||
fc(%) = | fc( Hz ) ⋅10000 | = | 5000 ⋅10000 | = 5000 = 50% |
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| 10000 |
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| Slave side FH |
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Results of conversions to point frequency (for the equation used, see above.)
Slave1 : fc( Hz ) = |
| 9000 − 0 |
| ⋅ ( 5000 − 0 )＋0 = 4500 = 45Hz | |
10000 − 0 | |||||
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Slave 2 : fc( Hz ) = |
| 8000 − 0 | ⋅ ( 5000 − 0 )＋0 = 4000 = 40Hz | ||
10000 − 0 | |||||
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