Engineering Guide CDA3000
3-59
3 Selection of inverter module
DE
EN
1
2
3
4
5
6
7
A
Calculation examples
Appli-
cation
Acceleration
Current at
V = constant
Deceleration
Stopping
time
Effective
inverter
capacity
utilization
Permissible
Current
Time
Current
Time
Current
Time
Yes
No
1
1.8
.
I
N
15 s
0
0
1.8
.
I
N
15 s
70 s
I
eff
< I
N
X
2
1.8
.
I
N
15 s
0.3
.
I
N
75 s
1.8
.
I
N
15 s
0 s
I
eff
< I
N
X
3
1.5
.
I
N
30 s
0
0
1.5
.
I
N
30 s
80 s
I
eff
< I
N
X
4
1.5
.
I
N
1 s
0.7
.
I
N
3 s
1.5
.
I
N
1 s
1 s
I
eff
< I
N
X
5
1.8
.
I
N
0.2 s
0,2
.
I
N
0.5 s
1.8
.
I
N
0.2 s
0.45 s
I
eff
< I
N
X
6
1.8
.
I
N
0.2 s
0.3
.
I
N
0.3 s
1.8
.
I
N
0.2 s
0.2 s
I
eff
> I
N
X
7
1.8
.
I
N
0.1 s
0.3
.
I
N
0.3 s
1.8
.
I
N
0.1 s
0.2 s
I
eff
< I
N
X
8
1.7
.
I
N
0.1 s
0
0
1.7
.
I
N
0.1 s
0.4 s
I
eff
< I
N
X
Table 3.28
Calculation example for the effective inverter current