STPS5H100SF STMicroelectronics

STPS5H100SF Datasheet 100 V power Schottky rectifier Features Low profile design package height of 1.1 mm typ. Wettable flanks for automatic visual inspection Low forward voltage drop Avalanche capability ECOPACK 2 compliant Applications Switching diode Notebook adapter LED lighting DC/DC converter Description This high voltage Schottky barrier rectifier has been optimized for use in high frequency miniature DC/DC converters, reverse battery protection, battery chargers and adaptors. Packaged in PSMC (TO-277A), the STPS5H100SF provides a high level of performance in a compact and flat package which can withstand very high operating junction temperature. Product status link STPS5H100SF Product summary Symbol Value I 5 A F(AV) V 100 V RRM T (max.) 175 C j V (typ.) 0.545 V F DS12694 - Rev 1 - July 2018 www.st.com For further information contact your local STMicroelectronics sales office.STPS5H100SF Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified, anode terminals short- circuited) Symbol Parameter Value Unit V Repetitive peak reverse voltage 100 V RRM I Average forward current, = 0.5 square wave T = 160 C 5 A F(AV) c I Surge non repetitive forward current t = 10 ms sinusoidal 230 A FSM p P Repetitive peak avalanche power t = 10 s, T = 125 C 518 W ARM p j T Storage temperature range -65 to +175 C stg (1) T Maximum operating junction temperature +175 C j 1. (dP /dT ) < (1/R ) condition to avoid thermal runaway for a diode on its own heatsink. tot j th(j-a) Table 2. Thermal resistance parameters Symbol Parameter Typ. Unit R Junction to case 2.1 C/W th(j-c) For more information, please refer to the following application note: AN5088: Rectifiers thermal management, handling and mounting recommendations Table 3. Static electrical characteristics (anode terminals short-circuited) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 8 A j (1) I Reverse leakage current V = V R RRM R T = 125 C - 1.5 5 mA j T = 25 C - 0.640 j I = 2.5 A F T = 125 C - 0.480 0.540 j (2) V Forward voltage drop V F T = 25 C - 0.745 j I = 5 A F T = 125 C - 0.545 0.610 j 1. Pulse test: t = 5 ms, < 2% p 2. Pulse test: t = 380 s, < 2% p To evaluate the conduction losses, use the following equation: 2 P = 0.470 x I + 0.028 x I F(AV) F (RMS) For more information, please refer to the following application notes related to the power losses: AN604: Calculation of conduction losses in a power rectifier AN4021: Calculation of reverse losses in a power diode DS12694 - Rev 1 page 2/9