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

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