WEG CFW100 - Application 5 - Simple Application 2 (Nominal Speed, Inverted Analog Input)

15
APPLICATION EXAMPLES
15.1.5 Application 5 - Simple application 2 (nominal speed, inverted analog input)
This example describes an application where the analogic input signal corresponds as frequency reference. Thus,
the total excursion of analog signal represents the drive of the motor from its minimum frequency to its maximum
frequency, as presented on Figure 15.6 on page 15-6. Here the analog input is inverted when compared to
Application 1. On Table 15.5 on page 15-6 are shown the parameters that are used to the correct setting.
Requirements:
Motor 1 HP, 220 V, 2.9 A, 1725 rpm, 60 Hz
Min. Frequency = 0 Hz
Max. Frequency = 60 Hz
Parametrization:
Table 15.5: Parameters for application 5
Analog Input AI1 AI2 Potenciometer Value
Min. Frequency P133 0.0 Hz
Max. Frequency P134 60.0 Hz
Max. Output Voltage P142 100.0 %
Interm. Output Voltage P143 50.0 %
Field Weak. Freq. P145 60.0 Hz
Intermediate Freq. P146 30.0 Hz
Selection source P220 0
Reference source P221 = 1 P221 = 2 P221 = 3 According application*
Rotation sel. P223 = 0 P223 = 0 P223 = 1 According application
Signal function P231 P236 P241 0
Gain P232 P237 P242 1.000
Input signal P233 P238 - 2
Offset P234 = 0 % P239 = 0 % P244 = -100.0 % According application**
(*) Consult Chapter 7 COMMAND AND REFERENCES on page 7-1.
(**) For AIx consult Chapter 9.1 ANALOG INPUTS on page 9-1 , for potenciometer this parameter is not available.
Example:
P134 (60 Hz)
P133 (0 Hz)
0 ....................................... 10 V
0 ....................................... 20 mA
4 ....................................... 20 mA
10 V .................................... 0
20 mA ................................. 0
20 mA ............................... 4 mA
AIx Signal
Output
frequency
For AI1 set to 10-0 V (P233=2)
and an analog input of 7.5 V:
P018(%) = 100.0 % -
(
7.5 V
10 V
x (100.0 %) + 0.0 %
)
x 1.000 = 25.0 %
Output Freq. = P018 x P134 = 25.0% x 60.0 Hz = 15.0 Hz
For potentiometer acessory with an analog input of 7.5 V one
need P244 = -100.0 % and P223 = 1:
P018(%) =
(
7.5 V
10 V
x (-100.0 %) + 0.0 %
)
x 1.000 = -25.0 %
Output Freq. = P018 x P134 = -25.0% x 60.0 Hz = -15.0 Hz
Figure 15.6: Result for application 5
15-6 | Micro Mini Drives