®
LOW DROP POWER SCHOTTKY RECTIFIER
MAIN PRODUCT CHARACTERISTIC S
STPS10L25D/G
I
F(AV)
V
RRM
10 A
25 V
Tj (max) 150 °C
(max) 0.35 V
V
F
FEATURES AND BENEFITS
VERY LOW FORWARD VOLTAGE DR OP FOR
LESS POWER DISSIPATION
OPTIMIZED CONDUCTION / REVERSE
LOSSES TRADE-OFF WHICH MEANS THE
HIGHEST EFFICIENCY IN THE APPLICATIONS
DESCRIPTION
Single Schottky rectifier suited to 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)
TO-220AC
STPS10L25D
K
A
A
K
NC
D2PAK
STPS10L25G
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 30 A
Average forward current Tc = 140°C δ = 0.5 10 A
Surge non repetitive forward current tp = 10 ms Sinus oidal 200 A
Repetitive peak reverse current tp=2 µs square F=1kHz 1 A
Non repetitive peak reverse c urrent tp = 100 µs square 3 A
Storage temperature range - 65 to + 150
Tj Maximum operating junction temperature * 150 °C
dV/dt Critical rate of rise of reverse voltage 10000 V/µs
dPtot
* :
June 1999 - Ed : 3B
dTj
<
1
Rth(j−a
thermal runaway condition for a diode on its own heatsink
)
°
C
1/5
STPS10L25D/G
THERMAL RE SISTA NC E
Symbol Parameter Value Unit
R
th (j-c)
Junction to case 1.5
°
STATIC ELECTRICAL CHARACTE RISTICS
Symbol Tests Conditions Tests Conditions Min. Typ. Max. Unit
* Reverse leakage current Tj = 25°CV
I
R
= V
R
RRM
800
Tj = 125°C 135 260 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%
= 10 A 0.46 V
F
= 10 A 0.30 0.35
F
= 20 A 0.55
F
= 20 A 0.41 0.48
F
To evaluate the maximum conduction losses use the following equation :
P = 0.22 x I
F(AV)
+ 0.013 I
F2(RMS)
C/W
µ
A
Fig.1 :
Average forward power dissipation versus
average forward current.
PF(av)(W)
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
01234567891011
δ = 0.05
2/5
δ = 0.1
IF(av) (A)
δ = 0.2
δ = 0.5
δ
δ = 1
T
=tp/T
Fig.2 :
Average forward current versus ambient
temperature ( δ = 0.5).
IF(av)(A)
12
10
8
6
4
2
tp
0
0 25 50 75 100 125 150
δ
=tp/T
T
tp
Rth(j-a)=Rth(j-c)
Rth(j-a)=50°C/W
Tamb(°C)