HIGH VOLTAGE POWER SCHOTTKY RECTIFIER
MAINPRODUCT CHARACTERISTICS
STPS80H100TV
I
F(AV)
V
RRM
2 x 40 A
100 V
Tj (max) 150 °C
V
(max) 0.65 V
F
FEATURESAND BENEFITS
NEGLIGIBLESWITCHINGLOSSES
HIGHJUNCTIONTEMPERATURECAPABILITY
LOW LEAKAGECURRENT
GOODTRADEOFFBETWEENLEAKAGECUR-
RENTANDFORWARDVOLTAGEDROP
AVALANCHERATED
LOW INDUCTANCEPACKAGE
INSULATEDPACKAGE:
Insulatedvoltage= 2500V
(RMS)
Capacitance= 45 pF
DESCRIPTION
High voltag e d ua l Schot tky barrier rectifier
designe d for hig h freque ncy telecom and
computer Switched Mode Power Suppl ies
andotherpowerconverters.
K2K1A2
A1
ISOTOP
TM
Packaged in ISOTOP, this device is intendedfor
use in medium voltage operation, and particularly, in high frequency circuitries where low
switching losses and low noise are required.
ABSOLUTE RATINGS
(limitingvalues,per diode)
Symbol Parameter Value Unit
V
RRM
I
F(RMS)
I
F(AV)
I
FSM
I
RRM
I
RSM
T
Repetitivepeak reversevoltage 100 V
RMSforward current 125 A
Averageforwardcurrent Tc= 120°C
δ = 0.5
Perdiode
Perdevice
Surgenon repetitiveforward current tp= 10 mssinusoidal 700 A
Repetitivepeak reversecurrent tp= 2µs squareF = 1kHz 2 A
Non repetitivepeakreverse current tp= 100 µs square 5 A
Storagetemperaturerange -55 to+150°C
stg
40
80
Tj Maximumoperating junctiontemperature * 150 °C
dV/dt Criticalrate of riseof reversevoltage 10000 V/µs
dPtot
*:
dTj
July 1999- Ed:3A
<
1
Rth(j−a
thermal runawayconditionfor a diodeon its own heatsink
)
A
1/4
STPS80H100TV
THERMAL RESISTANCES
Symbol Parameter Value Unit
R
th (j-c)
R
th (c)
Whenthe diodes1 and 2 areused simultaneously:
∆ Tj(diode1) = P(diode1)x R
STATICELECTRICALCHARACTERISTICS(perdiode)
Symbol Parameter TestsConditions Min. Typ. Max. Unit
I
R
V
F
Junctionto case Per leg 1 °C/W
Total 0.55
0.1
(Perdiode) + P(diode2) x R
th(j-c)
* Reverseleakage current Tj = 25°CV
th(c)
R=VRRM
Coupling
Tj = 125°C 7 25 mA
** Forwardvoltagedrop Tj = 25°CI
Tj = 125°CI
Tj = 25°CI
Tj = 125°CI
= 40 A 0.78 V
F
= 40 A 0.61 0.65
F
= 80 A 0.89
F
= 80 A 0.7 0.74
F
20 µA
Pulse test : * tp = 5 ms,δ <2%
** tp = 380µs, δ <2%
To evaluate the maximum conduction losses use the followingequation :
P = 0.56 x I
Fig. 1: Averageforward powerdissipation versus
averageforwardcurrent(per diode).
PF(av)(W)
35
30
25
20
15
10
5
0
0 5 10 15 20 25 30 35 40 45 50
F(AV)
+ 0.0022 xI
δ = 0.1
δ= 0.05
F2(RMS)
δ = 0.2
IF(av) (A)
δ = 0.5
δ
=tp/T
δ =1
Fig. 2: Average forward current versus ambient
temperature(δ=0.5,per diode).
IF(av)(A)
50
45
40
35
30
25
20
T
tp
15
10
5
=tp/T
δ
0
0 25 50 75 100 125 150
Rth(j-a)=5°C/W
T
tp
Rth(j-a)=Rth(j-c)
Tamb(°C)
2/4