SGS Thomson Microelectronics STPS41L45CT, STPS41L45CR, STPS41L45CG Datasheet
®
STPS41L45CG/CT/CR
LOW DROP POWER SCHOTTKY RECTIFIER
MAIN PRODUCTS CHARACTERISTICS
I
F(AV)
V
RRM
2x20A
45 V
Tj (max) 150 °C
V
(max) 0.47 V
F
FEATURES AND BENEFITS
VERY SMALL CONDUCTION LOSSES
■
NEGLIGIBLE SWITCHING LOSSES
■
EXTREMELY FAST SWITCHING
■
LOW FORWARD VOLTAGE DROP
■
■ LOWTHERMAL RESISTANCE
■ AVALANCHECAPABILITY SPECIFIED
DESCRIPTION
Dual center tab Schottky rectifier suited for 5V
output in off line AC/DC power supplies.
Packaged in D2PAK, I2PAK and TO-220AB this
device is intended for use in low voltage, high
frequency inverters, free-wheeling and polarity
protection applications.
A1
A2
A1
2
PAK
I
STPS41L45CR
K
STPS41L45CG
A2
K
D2PAK
K
TO-220AB
STPS41L45CT
A2
A1
A1
A2
K
ABSOLUTE RATINGS (limiting values, per diode)
Symbol Parameter Value Unit
V
RRM
I
F(RMS)
I
F(AV)
I
FSM
I
RRM
P
ARM
T
stg
Tj
dV/dt
dPtot
*:
Repetitive peak reverse voltage
RMS forward current
Average forward current Tc = 130°C
Surge non repetitive forward current tp = 10 ms sinusoidal
Peak repetitive reverse current tp=2 µs square F=1kHz
Repetitive peak avalanche power tp = 1µs Tj = 25°C
Storage temperature range
Maximum operating junction temperature *
Critical rate of rise reverse voltage
<
dTj Rth j a
July 2003 - Ed : 3A
Per diode
δ = 0.5
Per device
thermal runaway condition for a diode on its own heatsink
−1()
45 V
30 A
20
40
220 A
1A
10000 W
-65 to+175 °C
150 °C
10000 V/µs
A
1/6
STPS41L45CG / STPS41L45CT / STPS41L45CR
THERMAL RESISTANCES
Symbol Parameter Value Unit
R
th(j-c)
R
th(c)
Junction to case
Coupling
When the diodes 1 and 2 are used simultaneously :
∆ Tj(diode 1) = P(diode1) x R
(Per diode) + P(diode 2) x R
th(j-c)
STATIC ELECTRICAL CHARACTERISTICS (per diode)
Symbol Parameter Tests Conditions Min. Typ. Max. Unit
Per diode
Total
th(c)
1.5
0.8
0.1
°C/W
*
I
R
Reverse leakage current Tj = 25°C V
R=VRRM
Tj = 125°C
V
*
F
Forward voltage drop Tj = 25°CI
Tj = 125°C I
Tj=25°CI
Tj = 125°C I
=20A
F
=20A
F
=40A
F
=40A
F
Pulse test : * tp = 380 µs, δ <2%
To evaluate the conduction losses use the following equation :
P=0.28xI
Fig.1:Conductionlossesversus average current.
F(AV)
+ 0.0095 I
F2(RMS)
Fig. 2: Average forward current versus ambient
temperature (δ = 0.5).
1.2 mA
110 220 mA
0.53 V
0.42 0.47
0.68
0.60 0.66
PF(av)(W)
14
12
10
8
6
4
2
0
0 2 4 6 8 10121416182022242628
δ = 0.05
δ = 0.1
δ = 0.2
IF(av)(A)
δ = 0.5
δ
δ = 1
T
=tp/T
2/6
IF(av)(A)
25
20
15
10
5
tp
0
T
tp
=tp/T
δ
0 25 50 75 100 125 150
Rth(j-a)=Rth(j-c)
Rth(j-a)=50°C/W
Tamb(°C)
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