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Rockwell Automation Publication 750-AT006D-EN-P - January 2022
Chapter 6 Active Front End Tuning
nameplate data. These values can be directly entered in the configuration parameters.
summarizes the source selection
parameters.
In applications where there is an additional inductor L
ext
in [H] or that use very long cables, the following steps can be used.
1.
Calculate the value of the inductance in percentage based on the feeder transformer/generator kVA and output voltage V
rat
as
follows:
2. Therefore, the total impedance of the transformer and the inductor in percentage can be calculated as follows:
Notice that system configuration allows potential configuration for two different power sources, source A and source B. The selection
of each source as active can be done through 13:31 [AC Line Source]. A typical example for this configuration is when the converter
can be supplied by two different sources, such as the utility and a backup generator.
3. The value calculated in
can be directly entered into the 13:34 [AC Line Imped %A] or 13:35 [AC Line Imped %B]. The ratings of
the transformer/feeder can be set using 13:32 [AC Line KVA A] for source A or 13:33 [AC Line KVA B] for source B.
Bandwidth and damping
The Bandwidth B
W
of the current regulator determines how fast the control regulator responds during transient conditions. While high
bandwidth enhances system response, specifically the current control loop, relatively high bandwidth values can trigger a resonance
condition or cause instability. The default values of the bandwidth and damping are adequate for most of the applications. In some
applications where the source impedance is relatively large, the current regulator may need to be detuned; in other words reduce B
W
to
alleviate resonance effects. This will not significantly affect the dynamic performance of the power converter. See
for more information.
The default bandwidth of the current regulator for FR5 and FR6 is set to 150 Hz; it is set to 300 Hz for other frames. The converter can
maintain an excellent dynamic performance at relatively low bandwidths (100…150 Hz), provided that the disturbance rejection is properly
configured. If needed, the proportional gains K
P
and the integral gains K
I
for the current regulators in the d-axis and the q-axis of the current
regulator loops can be directly set for fine tuning. Setting the B
W
and damping of the current control loop is more straightforward and it is
recommended for most of applications.
Table 26 - Source Selection and Configuration Parameter Settings
Parameter No.
Parameter Name Description
13:31
[AC Line Source]
AC Line Source Select
Enter a value to select which set of Line Impedance and KVA parameters the control uses. Parameter values
are as follows.
• AC Line A (0) – selects the A parameters. These are 13:32 [AC Line kVA A] and 13:34 [AC Line Imped% A].
• AC Line B (1) – selects the B parameters. These are 13:33 [AC Line kVA B] and 13:35 [AC Line Imped% B].
When parameter 0:136 [DI AC Line Source] is configured, the product ignores 13:31 [DI AC LineSource].
Important:
coordinate changes to this parameter with logic that switches the AC input between the two
sources, and do not change the value of this parameter while the line-side converter is modulating.
13:32
[AC Line kVA A]
Apparent Power Rating
AC Line Source A
Enter a value for the kVA of the transformer or generator associated with AC Line A.
13:33
[AC Line kVA B]
Apparent Power Rating
AC Line Source B
Enter a value for the kVA of the transformer or generator associated with AC Line B.
13:34
[AC Line Imped% A]
Impedance AC Line
Source A
Enter a value for the percent impedance of the transformer or generator associated with AC Line A.
• This should include the transformer or generator plus any inductor between there and the line-side
converter.
13:35
[AC Line Imped% B]
Impedance AC Line
Source B
Enter a value for the percent impedance of the transformer or generator associated with AC Line B.
• This should include the transformer or generator and any inductor between there and the line-side
converter.
Z
ind pct
–
L
ext
kVA
2
f
nom
1000
V
rat
2
---------------------------------------------------------
=
Z
pct
Z
pct
Z
ind pct
–
+
=
