SF08E60_15 Datasheet, PDF (1/2 Pages) SeCoS Halbleitertechnologie GmbH – 8 Amp Super Fast Rectifier

Elektronische Bauelemente
SF08E60
Voltage 600V
8 Amp Super Fast Rectifier
RoHS Compliant Product
A suffix of “-C” specifies halogen free
FEATURES
Fast switching for high efficiency
Low forward voltage drop
High current capability
Low reverse leakage current
High surge current capability
MECHANICAL DATA
Case: Molded plastic
Plastic Materials used Carries Underwriters
Laboratory Flammability Classification 94V-O
Lead Temperature for Soldering Purposes：
265℃ Max. for 10 Seconds
TO-220A
B
N
D
E
2
MA
13
O
H
JC
K
G
L
L
F
Dimensions in millimeters
REF.
Millimeter
Min. Max.
REF.
Millimeter
Min. Max.
A
14.68 15.50
H
3.57 4.20
B
9.7 10.4
J
-
1.30
C
13.06 14.62
K
0.72 0.96
D
4.22 4.98
L
4.84 5.32
1
3
E
1.14 1.38
M
2.48 2.98
F
2.20 2.98
N
φ 3.7 φ 3.9
G
0.27 0.55
O
1.12 1.37
MAXIMUM RATINGS AND ELECTRICAL CHARACTERISTICS
(Rating 25°C ambient temperature unless otherwise s pecified. Single phase half wave, 60Hz, resistive or inductive load.
For capacitive load, de-rate current by 20%.)
Parameter
Symbol
Rating
Maximum Recurrent Peak Reverse Voltage
Working Peak Reverse Voltage
Maximum DC Blocking Voltage
Maximum Average Forward Rectified Current
Peak Repetitive Forward Current
Peak Forward Surge Current, 8.3 ms single half sine-wave
Typical Thermal Resistance
Operating and Storage Temperature Range
VRRM
VRSM
VDC
IF
IFM
IFSM
R_θJC
TJ,TSTG
600
600
600
8
35
29
2
150, -55~150
Unit
V
V
V
A
A
A
°C /W
°C
ELECTRICAL CHARACTERISTICS
Parameter
Symbol
Typ.
Max. Unit
Test Condition
Maximum Instantaneous Forward
Voltage
VF
1.5
1.7
V
IF = 8 A, TJ = 25°C
1.3
-
IF = 8 A, TJ = 125°C
Maximum DC Reverse Current
at Rated DC Blocking Voltage ³
Max. Reverse Recovery Time ¹
Typical Junction Capacitance ²
IR
0.002
0.2
mA ^TJ=25°C
-
20
TJ=100°C
TRR
22
35
nS
CJ
53
-
pF
NOTES:
1. Reverse recovery test conditions IF= 0.5A, IR= 1.0A, IRR= 0.25A.
2. Measured at 1MHz and applied reverse voltage of 4.0V D.C.
3. Pulse Test：Pulse Width = 300 µs, Duty Cycle ≦ 2.0%.
4. To evaluate the maximum conduction losses use the following equation：PF(av) = 0.95 X IF(av) + 0.04 X I²F(RMS)
http://www.SeCoSGmbH.com/
17-Jan-2014 Rev.A
Any changes of specification will not be informed individually.
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