Application 2 - Simple Application (Overspeed) - WEG CFW100 Programming Manual

15.1.2 Application 2 - Simple application (overspeed)

This example describes an application where the analogic input signal corresponds as frequency reference. Thus,
the total excursion of analogic signal represents the drive of the motor from its minimum frequency to its maximum
frequency, as presented on
(or fielf weakeaning speed). On
setting.
Requirements:
Motor 1 HP, 220 V, 2.9 A, 1725 rpm, 60 Hz
Min. Frequency = 0 Hz
Max. Frequency = 80 Hz
Parametrization:
Analog Input
Min. Frequency
Max. Frequency
Max. Output Voltage
Interm. Output Voltage
Field Weak. Freq.
Intermediate Freq.
Selection source
Reference source
Rotation sel.
Signal function
Gain
Input signal
Offset
(*) Consult Chapter 7 COMMAND AND REFERENCES on page
(**) For AIx consult Chapter 9.1 ANALOG INPUTS on page 9-1
Example:
Output
frequency
P134
(80 Hz)
P133
(0 Hz)
0 ....................................... 10 V
0 ....................................... 20 mA
4 ....................................... 20 mA
10 V .................................... 0
20 mA ................................. 0
20 mA ............................... 4 mA
Figure 15.3 on page 15-3. In this example, maximum speed is over the nominal speed
Table 15.2 on page 15-3
Table 15.2: Parameters for application 2
AI1 AI2
P221 = 1 P221 = 2
P231 P236
P232 P237
P233 P238
P234 P239
7-1.
, for potenciometer this parameter is not available.
AIx Signal
Figure 15.3: Result for application 2
are shown the parameters that are used to the correct
Potenciometer
P133
P134
P142
P143
P145
P146
P220
P221 = 3 According application*
P223
P241
P242
- According application**
P244
For AI1 set to 0-10 V (P233=0)
and an analog input of 5 V:
(
5 V
P018(%) = x (100.0 %) + 0.0 %
10 V
Output Freq. = P018 x P134 = 50.0 % x 80.0 Hz = 40.0 Hz
APPLICATION EXAMPLES
Value
0.0 Hz
80.0 Hz
100.0 %
50.0 %
60.0 Hz
30.0 Hz
0
0
0
1.000
0 %
)
x 1.000 = 50.0 %
Micro Mini Drives | 15-3
15