SIEMENS
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Siemens D 21.4 · 2017
Tools and configuration
Configuration notes
Power units
13
■
Configuration
(continued)
Example 1
:
A Power Module with rated line current
I
e
is directly connected
to a low-voltage transformer and the PCC is the transformer con-
nection terminal. The equation for the ratio between rated line
current
I
e
of the Power Module and rated current
I
rated
of the
transformer is
I
e
= 0.25 ×
I
rated
. The per unit voltage drop
u
k
of
the 400 V transformer is 4 %. If the transformer is loaded with its
rated current
I
rated
, the voltage drop across impedance
Z
s
is
9.2 V (corresponding to 4 % of the phase voltage
U
o
= 230 V).
u
k
= (
Z
s
×
I
rated
) / 230 V = 0.04
The following formula applies to the rated line current
I
e
of the
Power Module:
I
e
=
k
×
I
rated
The per unit voltage drop across the transformer when loaded
with
I
e
is thus:
u
k
=
Z
s
×
I
e
/
U
o
=
Z
s
×
k
×
I
rated
/
U
o
With the specified ratio between
I
e
and
I
rated
, the per unit voltage
drop is calculated as
u
k
= 1 % or 2.3 V. In relation to the
Power Module, this transformer therefore functions like a line
impedance in accordance with
u
k
= 1 %.
The magnitude of system disturbance in converter systems is
assessed on the basis of short-circuit power ratio
R
sc
:
R
sc
=
S
c
ν
/
P
According to this definition in accordance with EN 60146-1,
P
is
the fundamental-wave apparent power drawn by the converter.
S
c
ν
is the short-circuit power drawn from the mains in the event
of a short-circuit on the terminals U1, V1, W1. Since the ohmic
components of impedances are negligible in practice,
Z
n
≈
j
X
n
applies
S
c
ν
≈
3 ×
U
o
2
/
X
n
and thus
R
sc
≈
3 ×
U
o
2
/ (
X
n
×
P
)
The short-circuit power ratio
R
sc
is therefore dependent on the
current output power
P
of the converter and is determined by
network impedance
X
n
.
If we assume the power to be
P
≈
3 ×
U
o
×
I
e
=
√
3 ×
U
rated
×
I
e
the short-circuit power ratio
R
sc
is in inverse proportion to the per
unit voltage drop
u
k
across the effective line impedance.
R
sc
≈
3 ×
U
o
2
/ (
X
n
×
P
) = 3 ×
U
o
2
/ (
X
n
× 3 ×
U
o
×
I
e
) =
U
o
/ (
X
n
×
I
e
) = 1 /
u
k
The short-circuit power ratio for example 1 is
R
sc
≈
100 if no line
reactor is installed (
Z
e
= 0).
Note
:
The term "short-circuit power ratio" as used in technical standards
is not a harmonized definition. The short-circuit power ratio
R
sce
defined according to IEC 61000-3-12 is calculated from the short-
circuit power
S
SC
at the PCC referred to the power
S
equ
= 3 ×
U
o
×
I
e
consumed by the load.
Basic Line Modules and Power Modules are designed with a
rectifier bridge on the line side. An inherent feature of the principle
of rectification with load-side capacitance for DC link voltage
smoothing are harmonics in the line current which result in a non-
sinusoidal power input. The diagram shows the basic current
waveform of a Power Module or Basic Line Module as a function
of short-circuit power ratio
R
sc
.
Active Line Modules generate virtually no current harmonics
(Active Mode) at all and are employed when system disturbance
needs to be minimized, e.g. stipulation of IEEE 519 that THD
(Total Harmonic Distortion) must be < 10 %.
The SIZER for Siemens Drives engineering tool calculates the
line harmonic distortion on the basis of the supply data entered
and lists them against the limit values of relevant standards.
Line current of a Basic Line Module or Power Module as a function of the
short-circuit power ratio
R
sc
The rms of the line current
I
e
for which the line-side components
must be rated comprises fundamental wave
I
e1
and the current
harmonics, which increase in relation to the rise in short-circuit
power ratio
R
sc
. If the DC link power
P
d
has been calculated
(see
Line Modules)
, the required line-side active power is a known
quantity with Line Module efficiency, or the rectifier efficiency in
the case of a Power Module. However, this active power is con-
nected only with the current fundamental wave
I
e1
. The rms of
the line current
I
e
is always greater than
I
e1
as a result of the cur-
rent harmonics. The following applies for a short-circuit power
ratio
R
sc
= 100:
I
e
≈
1.3 ×
I
e1
The apparent power of a transformer selected to supply the
drive must be greater than the drive power by a factor of about
1.3.
The harmonic currents produce only alternating power, but no
active power. The following applies to the apparent power
S
on
the line side:
S
2
=
P
2
+
Q
1
2
+
D
2
•
with active component
P
= 3 ×
U
o
×
I
1
× cos
ϕ
1
, which is solely produced by the
current fundamental
•
reactive component
Q
1
= 3 ×
U
o
×
I
1
× sin
ϕ
1
•
and the distortion component
The ratio between active power and apparent power is referred
to as power factor
λ
or total power factor:
0
U
0
U
R
= approx. 10
sc
R
= approx. 100
sc
G_D211_EN_00127a
e
e
3 ×
U
0
=
D
=
=
P
P
S
P
2
2
2
1
+
D
+
Q
© Siemens AG 2017