10 F1127PA035SC L168E09 D1 19

10-F1127PA035SC-L168E09
Output Inverter Application
flow 7PACK 1
1200 V / 35 A
General conditions
3phase SPWM
V GEon = 15 V
V GEoff = -15 V
R gon = 16 Ω
R goff = 16 Ω
Figure 1
IGBT
Figure 2
Typical average static loss as a function of output
current
P loss = f(I out)
70
Ploss (W)
100
Ploss (W)
FWD
Typical average static loss as a function of output
current
P loss = f(I out)
Mi*cosfi = 1
90
Mi*cosf i= -1
60
80
50
70
60
40
50
30
40
30
20
20
10
10
Mi*cosfi = 1
Mi*cosfi = -1
0
0
0
10
20
30
40
50
60
0
70
10
20
30
40
50
60
Iout (A)
At
Tj =
150
At
Tj =
°C
Mi*cosφ from -1 to 1 in steps of 0,2
150
°C
Mi*cosφ from -1 to 1 in steps of 0,2
Figure 3
IGBT
Typical average switching loss
as a function of output current
Figure 4
FWD
Typical average switching loss
as a function of output current
P loss = f(I out)
80,0
Ploss (W)
Ploss (W)
70
Iout (A)
fsw = 16kHz
70,0
P loss = f(I out)
35,0
fsw = 16kHz
30,0
60,0
25,0
50,0
20,0
40,0
15,0
30,0
10,0
20,0
5,0
10,0
fsw = 2kHz
fsw = 2kHz
0,0
0,0
0
10
20
30
40
50
60
0
70
10
20
30
40
50
At
Tj =
150
At
Tj =
°C
DC link = 600
V
f sw from
2 kHz to 16 kHz in steps of factor 2
copyright Vincotech
60
70
Iout (A)
Iout (A)
150
°C
DC link = 600
V
f sw from
2 kHz to 16 kHz in steps of factor 2
1
27 Jul. 2015 / Revision 1
10-F1127PA035SC-L168E09
Output Inverter Application
flow 7PACK 1
Figure 5
Phase
Figure 6
Typical available 50Hz output current
as a function Mi*cosφ
I out = f(Mi*cos φ )
Phase
Typical available 50Hz output current
as a function of switching frequency I out = f(f sw)
Iout (A)
60
Iout (A)
1200 V / 35 A
Th = 60°C
60
Th = 60°C
50
50
40
40
Th = 100°C
Th = 100°C
30
30
20
20
10
10
0
0
-1,0
-0,8
At
Tj =
-0,6
-0,4
150
-0,2
0,0
0,2
0,4
0,6
1
0,8
1,0
Mi*cos φ
10
100
fsw (kHz)
At
Tj =
°C
DC link = 600
V
f sw =
4
kHz
T h from
60 °C to 100 °C in steps of 5 °C
150
°C
DC link = 600
V
Mi*cos φ =0,8
T h from
60 °C to 100 °C in steps of 5 °C
Figure 7
Phase
Figure 8
Typical available 50Hz output current as a function of
Mi*cos φ and switching frequency
I out = f(f sw, Mi*cos φ )
Phase
Typical available 0Hz output current as a function
of switching frequency
I outpeak = f(f sw)
Iout (Apeak)
-1,00
-0,80
Iout (A)
60
50
-0,60
40,0-47,0
-0,20
33,0-40,0
0,00
26,0-33,0
0,20
19,0-26,0
0,40
12,0-19,0
0,60
Th = 60°C
40
Mi*cosfi
47,0-50,0
-0,40
30
20
Th = 100°C
10
0,80
1,00
1
2
4
8
16
32
0
64
1
fsw
At
Tj =
10
100
fsw (kHz)
150
°C
At
Tj =
DC link = 600
Th =
80
V
°C
DC link = 600
V
T h from
60 °C to 100 °C in steps of 5 °C
Mi =
copyright Vincotech
2
150
°C
0
27 Jul. 2015 / Revision 1
10-F1127PA035SC-L168E09
Output Inverter Application
flow 7PACK 1
Figure 9
Inverter
Figure 10
Inverter
Typical efficiency as a function of output power
efficiency=f(Pout)
efficiency (%)
Typical available peak output power as a function of
heatsink temperature
P out=f(T h)
Pout (kW)
1200 V / 35 A
25,0
2kHz
100,0
20,0
99,0
2kHz
98,0
97,0
15,0
96,0
16kHz
95,0
16kHz
10,0
94,0
93,0
5,0
92,0
91,0
0,0
60
65
70
75
80
85
90
95
90,0
100
0,0
5,0
10,0
15,0
20,0
25,0
Th (oC)
At
Tj =
150
DC link =
Mi =
cos φ=
f sw from
600
V
1
0,80
2 kHz to 16 kHz in steps of factor 2
35,0
Pout (kW)
°C
Figure 11
30,0
At
Tj =
150
DC link =
Mi =
cos φ=
f sw from
600
V
1
0,80
2 kHz to 16 kHz in steps of factor 2
°C
Inverter
Overload (%)
Typical available overload factor as a function of
motor power and switching frequency P peak / P nom=f(P nom,fsw)
400
350
300
250
200
150
Switching frequency (kHz)
100
Motor nominal power (HP/kW)
7,50 / 5,52
10,00 / 7,36
15,00 / 11,03
20,00 / 14,71
25,00 / 18,39
30,00 / 22,07
1
349
262
175
131
0
0
2
349
262
175
131
0
0
4
349
262
175
131
0
0
8
313
235
157
117
0
0
16
246
184
123
0
0
0
At
Tj =
150
DC link =
Mi =
cos φ=
f sw from
Th =
600
V
1
0,8
1 kHz to 16kHz in steps of factor 2
80
°C
°C
Motor eff =0,85
copyright Vincotech
3
27 Jul. 2015 / Revision 1