SGS Thomson Microelectronics STPS5L25B-1, STPS5L25B Datasheet
®
LOW DROP POWER SCHOTTKY RECTIFIER
MAIN PRODUCT CHARACTERISTIC S
I
F(AV)
V
RRM
Tj (max) 150° C
(max) 0.35 V
V
F
FEATURES AND BENEFITS
VERY LOW F O RW ARD VOLTAG E DROP FOR
LESS POWER DISSIPATION AND REDUCED
HEATSINK
OPTIMIZED CONDUCTION/REVERSE LOSSES
TRADE-OFF WHICH MEANS THE HIGHEST
EFFICIENCY IN THE APPLICATIONS
HIGH POWER SURFACE MOUNT MINIATURE
PACKAGE
AVALANCHE R ATE D
5 A
25 V
4
STPS5L25B
STPS5L25B/B-1
2
3
2
1
NC
DPAK
3
4(TAB)
4
IPAK
STPS5L25B-1
3
2
1
NC
DESCR IPT ION
Single Schottky rectifier suited t o Switched Mode
Power Supplies and high frequency DC to DC converters.
This device is especially intended for use as a Rectifier at the secondary of 3.3V SMPS units.
ABSOLUTE RATINGS
(limiting values)
Symbol Parameter Value Unit
V
RRM
I
F(RMS)
I
F(AV)
I
FSM
I
RRM
I
RSM
T
stg
Repetitive peak reverse voltage 25 V
RMS forward current 7 A
Average forward current Tc = 145°C
δ
= 0.5 5 A
Surge non repetitive forward current tp = 10 ms Sinus oidal 75 A
Repetitive peak reverse current tp=2 µs square F=1k Hz 1 A
Non repetitive peak reverse c urrent tp = 100 µs square 2 A
Storage temperature range - 65 to + 150
Tj Maximum operating junction temperature * 150
dV/dt Critical rate of rise of reverse voltage 10000 V/µs
* :
dPtot
dTj
<
1
Rth(j−a
thermal runaway condition for a diode on its own heatsink
)
°
C
°
C
August 1999 - Ed: 3A
1/5
STPS5L25B/B-1
THERMAL RESISTANCES
Symbol Parameter Value Unit
R
th(j-c)
Junction to case 2.5
STATIC ELECTRICAL CHARACTER ISTICS
Symbol Tests Conditions Tests Conditions Min. Typ. Max. Unit
* Reverse leakage current Tj = 25°CV
I
R
= V
R
RRM
350
Tj = 125°C 55 115 mA
* Forward voltage drop Tj = 25°CI
V
F
Tj = 125°CI
Tj = 25°CI
Tj = 125°CI
Pulse test : * tp = 380 µs, δ < 2%
= 5 A 0.47 V
F
= 5 A 0.31 0.35
F
= 10 A 0.59
F
= 10 A 0.41 0.50
F
To evaluate the maximum conduction losses use the following equation :
P = 0.2 x I
F(AV)
+ 0.030 I
F2(RMS)
°
C/W
µ
A
Fig. 1:
Average forward power dissipation versus
average forward current.
PF(av)(W)
2.5
2.0
1.5
1.0
0.5
0.0
0123456
δ = 0.05
δ = 0.1
δ = 0.2
IF(av) (A)
δ = 0.5
δ
=tp/T
δ = 1
T
tp
Fig. 2:
Average forward current versus ambient
temperature (δ=0.5).
IF(av)(A)
6
5
4
δ
=tp/T
Rth(j-a)=70°C/W
T
tp
3
2
1
0
0 25 50 75 100 125 150
Rth(j-a)=Rth(j-c)
Tamb(°C)
2/5
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