10 Fx062PA075SA P993F0x D2 19

10-FZ062PA075SA-P993F08
10-F0062PA075SA-P993F09
datasheet
Output Inverter Application
flow PHASE0
600 V / 75 A
General conditions
3phase SPWM
V GEon = 15 V
V GEoff = -15 V
R gon = 2 Ω
R goff = 2 Ω
Figure 1
IGBT
Figure 2
Typical average static loss as a function of output current
P loss = f(I out)
120
Ploss (W)
160
Ploss (W)
FWD
Typical average static loss as a function of output current
P loss = f(I out)
140
100
Mi*cosfi = 1
120
Mi*cosf i= -1
80
100
60
80
60
40
40
20
20
Mi*cosfi = -1
0
At
Tj =
20
150
40
60
80
100
0
120
140
Iout (A)
At
Tj =
°C
Mi*cosφ from -1 to 1 in steps of 0,2
20
150
40
60
80
100
120
Iout (A)
140
°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)
Ploss (W)
Ploss (W)
Mi*cosfi = 1
0
0
40,0
35,0
fsw = 16kHz
P loss = f(I out)
20,0
18,0
16,0
fsw = 16kHz
30,0
14,0
25,0
12,0
10,0
20,0
8,0
15,0
6,0
10,0
4,0
5,0
2,0
fsw = 2kHz
fsw = 2kHz
0,0
0,0
0
At
Tj =
20
150
40
60
80
100
0
120Iout (A) 140
At
Tj =
°C
DC link = 320
V
f sw from 2 kHz to 16 kHz in steps of factor 2
copyright Vincotech
20
150
40
60
80
100
120
Iout (A)
140
°C
DC link = 320
V
f sw from 2 kHz to 16 kHz in steps of factor 2
1
04 Jun. 2016 / Revision 2
10-FZ062PA075SA-P993F08
10-F0062PA075SA-P993F09
datasheet
Output Inverter Application
flow PHASE0
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)
120
Iout (A)
600 V / 75 A
Ts = 60°C
100
120
Ts = 60°C
100
80
80
Ts = 100°C
60
60
40
40
20
20
0
-1,0
At
Tj =
Ts = 100°C
0
-0,8
-0,6
-0,4
150
-0,2
0,0
0,2
0,4
0,6
0,8
1,0
Mi*cos φ
1
At
Tj =
°C
DC link = 320
V
f sw =
4
kHz
T s from
60 °C to 100 °C in steps of 5 °C
10
150
DC link = 320
Mi*cos φ =0,8
T s from
Figure 7
Phase
100
°C
V
60 °C to 100 °C in steps of 5 °C
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)
fsw (kHz)
100
90
-0,60
80
-0,40
70
-0,20
60
Ts = 60°C
100,0-110,0
80,0-90,0
70,0-80,0
0,00
60,0-70,0
50,0-60,0
40,0-50,0
Mi*cosfi
90,0-100,0
50
0,20
40
0,40
30
0,60
20
0,80
10
1,00
0
Ts = 100°C
30,0-40,0
1
2
4
8
16
32
64
1
fsw
(kHz)
At
Tj =
10
fsw (kHz)
150
°C
At
Tj =
DC link = 320
Ts =
80
V
°C
DC link = 320
V
T s from
60 °C to 100 °C in steps of 5 °C
Mi =
copyright Vincotech
2
150
100
°C
0
04 Jun. 2016 / Revision 2
10-FZ062PA075SA-P993F08
10-F0062PA075SA-P993F09
datasheet
Output Inverter Application
flow PHASE0
Figure 9
Inverter
Figure 10
Inverter
Typical efficiency as a function of output power
efficiency=f(P out)
efficiency (%)
Typical available peak output power as a function of
heatsink temperature
P out=f(T s)
Pout (kW)
600 V / 75 A
30,0
25,0
100,0
99,0
2kHz
98,0
2kHz
97,0
20,0
16kHz
96,0
15,0
95,0
16kHz
94,0
10,0
93,0
92,0
5,0
91,0
0,0
90,0
60
65
At
Tj =
DC link =
Mi =
cos φ =
f sw from
70
75
150
°C
320
1
V
80
85
90
95
100
Ts ( o C)
0,0
At
Tj =
DC link =
Mi =
cos φ =
f sw from
0,80
2 kHz to 16 kHz in steps of factor 2
Figure 11
5,0
10,0
150
°C
320
1
V
15,0
20,0
25,0
30,0
35,0
Pout (kW)
0,80
2 kHz to 16 kHz in steps of factor 2
Inverter
Overload (%)
Typical available overload factor as a function of
motor power and switching frequency P peak / P nom = f( P nom , f sw)
500
450
400
350
300
250
200
Motor nominal power (HP/kW)
Switching frequency (kHz)
150
100
5,00 / 3,68
7,50 / 5,52
10,00 / 7,36
15,00 / 11,03
20,00 / 14,71
25,00 / 18,39
1
538
359
269
179
135
0
2
531
354
265
177
133
0
4
517
344
258
172
129
0
8
489
326
245
163
122
0
16
439
293
220
146
0
0
At
Tj =
150
DC link = 320
°C
V
Mi =
1
cos φ =
f sw from
Ts =
0,8
1 kHz to 16kHz in steps of factor 2
80
°C
Motor eff =0,85
copyright Vincotech
3
04 Jun. 2016 / Revision 2
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