STPS41L45CG-TR STMicroelectronics

STPS41L45CG/CT/CR LOW DROP POWER SCHOTTKY RECTIFIER MAIN PRODUCTS CHARACTERISTICS A1 I 2x20A K F(AV) A2 V 45 V RRM Tj (max) 150 C V (max) 0.47 V F FEATURES AND BENEFITS A2 n VERY SMALL CONDUCTION LOSSES A2 K K A1 n NEGLIGIBLE SWITCHING LOSSES A1 n EXTREMELY FAST SWITCHING 2 TO-220AB I PAK n LOW FORWARD VOLTAGE DROP STPS41L45CT STPS41L45CR n LOW THERMAL RESISTANCE n AVALANCHE CAPABILITY SPECIFIED K DESCRIPTION Dual center tab Schottky rectifier suited for 5V A2 output in off line AC/DC power supplies. A1 2 Packaged in D PAK, I PAK and TO-220AB this 2 2 device is intended for use in low voltage, high D PAK frequency inverters, free-wheeling and polarity STPS41L45CG protection applications. ABSOLUTE RATINGS (limiting values, per diode) Symbol Parameter Value Unit V 45 V RRM Repetitive peak reverse voltage I 30 A F(RMS) RMS forward current I 20 F(AV) Average forward current Tc = 130C Per diode A = 0.5 40 Per device I 220 A FSM Surge non repetitive forward current tp = 10 ms sinusoidal I 1A RRM Peak repetitive reverse current tp=2 s square F=1kHz P 10000 W ARM Repetitive peak avalanche power tp = 1s Tj = 25C T -65 to+175 C stg Storage temperature range Tj 150 C Maximum operating junction temperature * dV/dt 10000 V/s Critical rate of rise reverse voltage dPtot 1 *: < thermal runaway condition for a diode on its own heatsink dTj Rth()j - a July 2003 - Ed : 3A 1/6STPS41L45CG / STPS41L45CT / STPS41L45CR THERMAL RESISTANCES Symbol Parameter Value Unit R Per diode 1.5 C/W th(j-c) Junction to case Total 0.8 R 0.1 th(c) 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) th(c) STATIC ELECTRICAL CHARACTERISTICS (per diode) Symbol Parameter Tests Conditions Min. Typ. Max. Unit I * 1.2 mA R Reverse leakage current Tj = 25C V =V R RRM 110 220 mA Tj = 125C V * 0.53 V F Forward voltage drop Tj = 25CI =20A F 0.42 0.47 Tj = 125C I =20A F 0.68 Tj=25CI =40A F 0.60 0.66 Tj = 125C I =40A F Pulse test : * tp = 380 s, <2% To evaluate the conduction losses use the following equation : 2 P=0.28xI + 0.0095 I F(AV) F (RMS) Fig. 1: Conduction losses versus average current. Fig. 2: Average forward current versus ambient temperature ( = 0.5). PF(av)(W) IF(av)(A) 14 25 = 0.5 = 0.2 = 0.1 12 Rth(j-a)=Rth(j-c) = 0.05 = 1 20 10 15 8 6 10 4 Rth(j-a)=50C/W T 5 T 2 Tamb(C) IF(av)(A) =tp/T tp =tp/T tp 0 0 0 2 4 6 8 1012 1416 1820 2224 2628 0 25 50 75 100 125 150 2/6