10 xZ062PA100SA P994F08x 19

10-FZ062PA100SA-P994F08
10-PZ062PA100SA-P994F08Y
datasheet
Output Inverter Application
flow PHASE0
600 V / 100 A
General conditions
3phase SPWM
VGEon = 15 V
VGEoff = -15 V
Rgon = 4 Ω
Rgoff = 4 Ω
Figure 1
IGBT
Figure 2
FWD
Typical average static loss as a function of output current
Typical average static loss as a function of output current
Ploss = f(Iout)
Ploss = f(Iout)
140
Ploss (W)
Ploss (W)
180
160
140
120
Mi*cosf i= -1
Mi*cosfi = 1
100
120
80
100
80
60
60
40
40
20
20
Mi*cosfi = -1
Mi*cosfi = 1
0
0
0
At
Tj =
20
40
150
60
80
100
120
0
140
160
Iout (A)
At
Tj =
°C
Mi*cosφ from -1 to 1 in steps of 0,2
40
150
60
80
100
120
140
Iout (A)
160
°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
Ploss = f(Iout)
Ploss (W)
Ploss (W)
20
60,0
Ploss = f(Iout)
25,0
fsw = 16kHz
fsw = 16kHz
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
At
Tj =
20
150
40
60
80
100
120
0
140
Iout (A) 160
At
Tj =
°C
DC link = 320
V
fsw from 2 kHz to 16 kHz in steps of factor 2
copyright Vincotech
20
150
40
60
80
100
120
140
160
Iout (A)
°C
DC link = 320
V
fsw from 2 kHz to 16 kHz in steps of factor 2
1
02 Dec. 2015 / Revision 2
10-FZ062PA100SA-P994F08
10-PZ062PA100SA-P994F08Y
datasheet
Output Inverter Application
flow PHASE0
Figure 5
Phase
Figure 6
Typical available 50Hz output current
as a function Mi*cosφ
Iout = f(Mi*cos φ)
Phase
Typical available 50Hz output current
as a function of switching frequency Iout = f(fsw)
Iout (A)
160
Iout (A)
600 V / 100 A
Th = 60°C
160
140
140
120
120
Th = 60°C
100
100
Th = 100°C
80
80
60
60
40
40
20
20
0
-1,0
At
Tj =
0
-0,8
-0,6
-0,4
150
DC link = 320
fsw =
4
Th from
Th = 100°C
-0,2
0,0
0,2
0,4
0,6
0,8
1,0
Mi*cos φ
1
10
°C
At
Tj =
V
kHz
DC link = 320
Mi*cos φ =0,8
60 °C to 100 °C in steps of 5 °C
150
Th from
Figure 7
Phase
V
60 °C to 100 °C in steps of 5 °C
Phase
Typical available 0Hz output current as a function
of switching frequency
Ioutpeak = f(fsw)
Iout (Apeak)
-1,00
-0,80
120
100
-0,60
Th = 60°C
-0,40
80
120,0-135,0
105,0-120,0
100
°C
Figure 8
Typical available 50Hz output current as a function of
Mi*cos φ and switching frequency
Iout = f(fsw, Mi*cos φ)
Iout (A)
fsw (kHz)
90,0-105,0
75,0-90,0
0,00
60,0-75,0
45,0-60,0
Mi*cosfi
-0,20
60
Th = 100°C
0,20
30,0-45,0
40
0,40
0,60
20
0,80
0
1,00
1
2
4
8
16
32
64
1
fsw
(kHz)
At
Tj =
10
fsw (kHz)
150
°C
At
Tj =
DC link = 320
Th =
80
V
°C
DC link = 320
V
Th from
60 °C to 100 °C in steps of 5 °C
Mi =
copyright Vincotech
2
150
100
°C
0
02 Dec. 2015 / Revision 2
10-FZ062PA100SA-P994F08
10-PZ062PA100SA-P994F08Y
datasheet
Output Inverter Application
flow PHASE0
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
Pout=f(Th)
Pout (kW)
600 V / 100 A
40,0
35,0
100,0
99,0
2kHz
98,0
2kHz
30,0
97,0
16kHz
25,0
96,0
20,0
95,0
16kHz
94,0
15,0
93,0
10,0
92,0
5,0
91,0
0,0
90,0
60
65
70
75
80
85
90
At
Tj =
150
DC link =
Mi =
cos φ=
fsw from
320
V
1
0,80
2 kHz to 16 kHz in steps of factor 2
95
100
Th ( o C)
0,0
°C
Figure 11
5,0
10,0
15,0
20,0
25,0
30,0
At
Tj =
150
DC link =
Mi =
cos φ=
fsw from
320
V
1
0,80
2 kHz to 16 kHz in steps of factor 2
35,0
40,0
45,0
Pout (kW)
°C
Inverter
Overload (%)
Typical available overload factor as a function of
motor power and switching frequency Ppeak / Pnom=f(Pnom,fsw)
450
400
350
300
250
200
Switching frequency (kHz)
150
Motor nominal power (HP/kW)
100
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
484
363
242
181
145
121
2
476
357
238
178
143
119
4
461
345
230
173
138
115
8
431
323
216
162
129
0
16
378
284
189
142
114
0
At
Tj =
150
DC link = 320
Mi =
1
cos φ=
fsw from
Th =
°C
V
0,8
1 kHz to 16kHz in steps of factor 2
80
°C
Motor eff =0,85
copyright Vincotech
3
02 Dec. 2015 / Revision 2
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