10 Fx127PA025SC L159E0x D1 19

10-F0127PA025SC-L159E09
10-FZ127PA025SC-L159E08
datasheet
1200 V / 25 A
Output Inverter Application
flow 7PACK 0
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)
60
Ploss (W)
80
Ploss (W)
FWD
Typical average static loss as a function of output
current
P loss = f(I out)
Mi*cosfi = 1
70
50
Mi*cosf i= -1
60
40
50
30
40
30
20
20
10
10
Mi*cosfi = -1
Mi*cosfi = 1
0
0
0
5
10
15
20
25
30
35
40
45
0
50
5
10
15
20
25
30
35
40
Iout (A)
At
Tj =
150
At
Tj =
°C
Mi*cosφ from -1 to 1 in steps of 0,2
50
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)
45
Iout (A)
70,0
fsw = 16kHz
P loss = f(I out)
25,0
20,0
fsw = 16kHz
60,0
50,0
15,0
40,0
10,0
30,0
20,0
5,0
10,0
fsw = 2kHz
fsw = 2kHz
0,0
0,0
0
At
Tj =
5
150
10
15
20
25
30
35
40
0
45
50
Iout (A)
At
Tj =
°C
DC link = 600
V
f sw from
2 kHz to 16 kHz in steps of factor 2
copyright Vincotech
5
150
10
15
20
25
30
35
40
45
50
Iout (A)
°C
DC link = 600
V
f sw from
2 kHz to 16 kHz in steps of factor 2
1
24 Jul. 2015 / Revision 1
10-F0127PA025SC-L159E09
10-FZ127PA025SC-L159E08
datasheet
1200 V / 25 A
Output Inverter Application
flow 7PACK 0
Figure 5
Phase
Figure 6
Typical available 50Hz output current
as a function Mi*cosφ
I out = f(Mi*cos φ )
Iout (A)
Iout (A)
40
Th = 60°C
35
30
Phase
Typical available 50Hz output current
as a function of switching frequency I out = f(f sw)
40
Th = 60°C
35
30
Th = 100°C
25
25
20
20
15
15
10
10
5
5
Th = 100°C
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)
40
Th = 60°C
35
-0,60
30
-0,40
28,0-32,0
-0,20
24,0-28,0
0,00
20,0-24,0
25
Mi*cosfi
32,0-36,0
20
0,20
16,0-20,0
12,0-16,0
0,40
8,0-12,0
0,60
15
10
Th = 100°C
5
0,80
1,00
1
2
4
8
16
32
0
64
1
10
100
fsw
At
Tj =
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
24 Jul. 2015 / Revision 1
10-F0127PA025SC-L159E09
10-FZ127PA025SC-L159E08
datasheet
1200 V / 25 A
Output Inverter Application
flow 7PACK 0
Figure 9
Inverter
Figure 10
Inverter
Typical efficiency as a function of output power
efficiency=f(Pout)
efficiency (%)
Pout (kW)
Typical available peak output power as a function of
heatsink temperature
P out=f(T h)
18,0
2kHz
16,0
100,0
99,0
2kHz
98,0
14,0
97,0
12,0
96,0
16kHz
10,0
95,0
16kHz
8,0
94,0
6,0
93,0
4,0
92,0
2,0
91,0
0,0
60
65
70
75
80
85
90
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
90,0
95
100
Th (oC)
0,0
10,0
15,0
20,0
25,0
Pout (kW)
°C
Figure 11
5,0
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)10,00 / 7,36
5,00 / 3,68
7,50 / 5,52
15,00 / 11,03
20,00 / 14,71
25,00 / 18,39
1
374
250
187
125
0
0
2
374
250
187
125
0
0
4
374
250
187
125
0
0
8
374
250
187
125
0
0
16
294
196
147
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
24 Jul. 2015 / Revision 1
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