10 Fx127PA008SC L156E0x D1 19

10-F0127PA008SC-L156E09
10-FZ127PA008SC-L156E08
datasheet
1200 V / 8 A
Output Inverter Application
flow 7PACK 0
General conditions
3phase SPWM
V GEon = 15 V
V GEoff = -15 V
R gon = 32 Ω
R goff = 32 Ω
Figure 1
IGBT
Figure 2
Typical average static loss as a function of output
current
P loss = f(I out)
FWD
Typical average static loss as a function of output
current
P loss = f(I out)
Ploss (W)
14
Ploss (W)
25
Mi*cosfi = 1
12
Mi*cosf i= -1
20
10
15
8
6
10
4
5
2
Mi*cosfi = 1
Mi*cosfi = -1
0
0
0
2
4
6
8
10
12
14
0
16
2
4
6
8
10
12
Iout (A)
At
Tj =
149
16
Iout (A)
At
Tj =
°C
Mi*cosφ from -1 to 1 in steps of 0,2
149
°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)
20,0
Ploss (W)
Ploss (W)
14
fsw = 16kHz
18,0
16,0
P loss = f(I out)
9,0
fsw = 16kHz
8,0
7,0
14,0
6,0
12,0
5,0
10,0
4,0
8,0
3,0
6,0
2,0
4,0
1,0
2,0
fsw = 2kHz
fsw = 2kHz
0,0
0,0
0
At
Tj =
2
149
4
6
8
10
12
0
14
16
Iout (A)
At
Tj =
°C
DC link = 600
V
f sw from
2 kHz to 16 kHz in steps of factor 2
copyright Vincotech
2
149
4
6
8
10
12
14
16
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-F0127PA008SC-L156E09
10-FZ127PA008SC-L156E08
datasheet
1200 V / 8 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)
12
Iout (A)
Th = 60°C
Th = 100°C
10
Phase
Typical available 50Hz output current
as a function of switching frequency I out = f(f sw)
14
Th = 60°C
12
10
Th = 100°C
8
8
6
6
4
4
2
2
0
-1,0
-0,8
-0,6
-0,4
-0,2
0,0
0,2
0,4
0,6
0,8
0
1,0
1
10
100
Mi*cos φ
At
Tj =
149
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
149
°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)
14
Th = 60°C
12
-0,60
10
-0,40
11,0-12,0
-0,20
8
0,00
9,0-10,0
0,20
Mi*cosfi
10,0-11,0
6
Th = 100°C
8,0-9,0
0,40
7,0-8,0
4
0,60
2
0,80
fsw
1
2
4
8
16
1,00
32
0
64
1
10
100
fsw (kHz)
At
Tj =
149
°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
149
°C
0
24 Jul. 2015 / Revision 1
10-F0127PA008SC-L156E09
10-FZ127PA008SC-L156E08
datasheet
1200 V / 8 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)
6,0
2kHz
5,0
100,0
99,0
2kHz
98,0
16kHz
97,0
4,0
96,0
3,0
16kHz
95,0
94,0
2,0
93,0
92,0
1,0
91,0
0,0
60
65
70
75
80
85
90
At
Tj =
149
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
°C
Figure 11
1,0
2,0
3,0
4,0
5,0
6,0
At
Tj =
149
DC link =
Mi =
cos φ=
f sw from
600
V
1
0,80
2 kHz to 16 kHz in steps of factor 2
7,0
8,0
9,0
Pout (kW)
°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)3,00 / 2,21
1,50 / 1,10
2,00 / 1,47
5,00 / 3,68
7,50 / 5,52
10,00 / 7,36
1
399
300
200
120
0
0
2
399
300
200
120
0
0
4
399
300
200
120
0
0
8
399
300
200
120
0
0
16
399
300
200
120
0
0
At
Tj =
149
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
Similar pages