C&H Technology PS21265-AP User Manual
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6
PS21265-P / PS21265-AP
Intellimod™ Module
Dual-In-Line Intelligent Power Module
20 Amperes/600 Volts
Powerex, Inc., 173 Pavilion Lane, Youngwood, Pennsylvania 15697 (724) 925-7272 www.pwrx.com
08/10 Rev. 01
0
45
35 40
15
5 10
20 25 30
COLLECTOR-CURRENT, I
C
, (AMPERES)
COLLECTOR-EMITTER
SATURATION VOLTAGE,
V
CE(sat)
, (VOLTS)
COLLECTOR-EMITTER
SATURATION VOLTAGE CHARACTERISTICS
(TYPICAL)
V
D
= V
DB
=15V
V
IN
= 5V
T
j
= 25°C
T
j
= 125°C
2.0
2.2
2.4
0.2
0.4
0.6
0
1.4
1.2
2.6
1.6
1.8
0.8
1.0
FREE-WHEEL DIODE
FORWARD CHARACTERISTICS
(TYPICAL)
COLLECTOR CURRENT, I
C
, (AMPERES)
SWITCHING LOSS,
P
SW(on)
, (mJ/PULSE)
10
2
SWITCHING LOSS (ON) VS.
COLLECTOR CURRENT
(TYPICAL - N-SIDE)
10
1
10
-1
10
-2
10
0
COLLECTOR CURRENT, I
C
, (AMPERES)
SWITCHING LOSS,
P
SW(off)
, (mJ/PULSE)
10
2
10
1
10
1
SWITCHING LOSS (OFF) VS.
COLLECTOR CURRENT
(TYPICAL - N-SIDE)
10
1
10
0
10
0
10
-1
10
-2
10
0
COLLECTOR CURRENT, I
C
, (AMPERES)
SWITCHING LOSS,
P
SW(on)
, (mJ/PULSE)
SWITCHING LOSS (ON) VS.
COLLECTOR CURRENT
(TYPICAL - P-SIDE)
COLLECTOR CURRENT, I
C
, (AMPERES)
REVERSE RECO
VER
Y CURRENT
, I
rr
, (AMPERES)
REVERSE RECOVERY TIME,
t
rr
, (ns)
10
0
10
2
10
1
COLLECTOR CURRENT, I
C
, (AMPERES)
REVERSE RECOVERY CHARACTERISTICS
(TYPICAL - P-SIDE)
REVERSE RECOVERY CHARACTERISTICS
(TYPICAL - P-SIDE)
10
3
10
2
10
1
REVERSE RECOVERY TIME,
t
rr
, (ns)
COLLECTOR CURRENT, I
C
, (AMPERES)
REVERSE RECOVERY CHARACTERISTICS
(TYPICAL - N-SIDE)
COLLECTOR CURRENT, I
C
, (AMPERES)
REVERSE RECO
VER
Y CURRENT
, I
rr
, (AMPERES)
10
0
10
2
10
1
REVERSE RECOVERY CHARACTERISTICS
(TYPICAL - N-SIDE)
V
CC
= 300V
V
D
= 15V
V
IN
= 0 ⇔ 5V
T
j
= 25°C
T
j
= 125°C
10
2
10
1
10
0
0
45
35 40
15
5 10
20 25 30
COLLECTOR-CURRENT, I
C
, (AMPERES)
EMITTER-COLLECTOR VOLTAGE,
V
EC
, (VOLTS)
V
IN
= 0V
T
j
= 25°C
T
j
= 125°C
2.0
2.2
2.4
0.2
0.4
0.6
0
1.4
1.2
2.6
1.6
1.8
0.8
1.0
10
0
10
2
10
1
10
3
10
2
10
1
10
0
10
2
10
1
V
CC
= 300V
V
D
= V
DB
= 15V
V
IN
= 0 ⇔ 5V
T
j
= 25°C
T
j
= 125°C
10
2
10
1
10
0
V
CC
= 300V
V
D
= V
DB
= 15V
V
IN
= 0 ⇔ 5V
T
j
= 25°C
T
j
= 125°C
V
CC
= 300V
V
D
= 15V
V
IN
= 0 ⇔ 5V
T
j
= 25°C
T
j
= 125°C
V
CC
= 300V
V
D
= 15V
V
IN
= 0 ⇔ 5V
P
SW(on)
= 25°C
P
SW(on)
125°C
V
CC
= 300V
V
D
= V
DB
= 15V
V
IN
= 0 ⇔ 5V
P
SW(on)
= 25°C
P
SW(on)
= 125°C
V
CC
= 300V
V
D
= 15V
V
IN
= 0 ⇔ 5V
P
SW(off)
= 25°C
P
SW(off)
= 125°C
10
2
10
1
10
1
10
0
10
-1
10
-2
10
0