2024年3月16日发(作者:吉利帝豪gl图片)
元器件交易网
?
STPS30L60CW/CT/CG/CR
POWER SCHOTTKY RECTIFIER
MAINPRODUCTCHARACTERISTICS
I
F(AV)
V
RRM
Tj(max)
V
F
(max)
2x15A
60V
150°C
0.56V
A1
K
A2
FEATURESANDBENEFITS
s
A2
K
A1
K
A1
A2
s
s
s
NEGLIGIBLESWITCHINGLOSSES
LOWFORWARDVOLTAGEDROP
LOWTHERMALRESISTANCE
AVALANCHECAPABILITYSPECIFIED
TO-247
STPS30L60CW
TO-220AB
STPS30L60CT
DESCRIPTION
DualcentertapSchottkyrectifierssuitedfor
SwitchedModePowerSuppliesandhigh
edin
TO-220,D
2
PAK,I
2
PAKandTO-247thisdeviceis
intendedforuseinhighfrequencyinverters.
A2
K
A1
K
A1
A2
D
2
PAK
STPS30L60CG
I
2
PAK
STPS30L60CR
ABSOLUTERATINGS(limitingvalues,perdiode)
Symbol
V
RRM
I
F(RMS)
I
F(AV)
I
FSM
I
RRM
P
ARM
T
stg
Tj
dV/dt
*:
Parameter
Repetitivepeakreversevoltage
RMSforwardcurrent
Averageforwardcurrent
Surgenonrepetitiveforwardcurrent
Repetitivepeakreversecurrent
Repetitivepeakavalanchepower
Storagetemperaturerange
Maximumoperatingjunctiontemperature*
Criticalrateofriseofreversevoltage
Tc=130°C
δ=0.5
Perdiode
Perdevice
Value
60
30
15
30
230
2
7800
-65to+175
150
10000
Unit
V
A
A
A
A
W
°C
°C
V/?s
tp=10msSinusoidal
tp=2?ssquareF=1kHz
tp=1?sTj=25°C
dPtot
1
thermalrunawayconditionforadiodeonitsownheatsink
<
dTjRth
(
j
?
a
)
1/6
July2003-Ed:3B
元器件交易网
STPS30L60CW/CT/CG/CR
THERMALRESISTANCES
Symbol
R
th(j-c)
R
th(c)
Junctiontocase
Parameter
Perdiode
Total
Coupling
Value
1.5
0.8
0.1
Unit
°C/W
°C/W
Whenthediodes1and2areusedsimultaneously:
?Tj(diode1)=P(diode1)xR
th(j-c)
(Perdiode)+P(diode2)xR
th(c)
STATICELECTRICALCHARACTERISTICS(perdiode)
Symbol
I
R
*
Parameter
Reverseleakagecur-
rent
Forwardvoltagedrop
TestsConditions
Tj=25°C
Tj=125°C
Tj=25°C
Tj=125°C
Tj=25°C
Tj=125°C
Pulsetest:*tp=380?s,δ<2%
.
480
Unit
?A
mA
V
V
R
=V
RRM
77
I
F
=15A
I
F
=15A
I
F
=30A
I
F
=30A
0.65
0.5
130
0.6
0.56
0.75
0.7
V
F
*
Toevaluatethemaximumconductionlossesusethefollowingequation:
P=0.42xI
F(AV)
+0.009I
F
2
(RMS)
Fig.1:Averageforwardpowerdissipationversus
averageforwardcurrent(perdiode).
12
10
8
6
4
2
IF(av) (A)
0
δ
=tp/T
tp
T
Fig.2:Averageforwardcurrentversusambient
temperature(δ=0.5,perdiode).
IF(av)(A)
Rth(j-a)=Rth(j-c)
PF(av)(W)
δ= 0.1
δ= 0.05
δ= 0.2
δ= 0.5
δ= 1
161820
18
16
14
12
10
8
6
4
2
0
Rth(j-a)=15°C/W
T
δ
=tp/T
tp
Tamb(°C)
5025
2/6
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