STPS5H100UFN Datasheet 5 A - 100 V power Schottky rectifier Features Low profile design package height of 1.1 mm Wettable flanks for automatic visual inspection Low forward voltage drop Avalanche capability ECOPACK2 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 SMB Flat Notch, the STPS5H100UFN provides a high level of performance in a compact and flat package which can withstand very high operating junction temperature. Product status link STPS5H100UFN Product summary Symbol Value I F(AV) 5 A V 100 V RRM T (max.) 175 C j V (typ.) 0.545 V F DS13227 - Rev 1 - January 2020 www.st.com For further information contact your local STMicroelectronics sales office.STPS5H100UFN Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 100 V RRM I T = 115 C Average forward current, = 0.5 square wave 5 A F(AV) L I t = 10 ms sinusoidal Surge non repetitive forward current 190 A FSM p P t = 10 s, T = 125 C Repetitive peak avalanche power 518 W ARM p j T Storage temperature range -65 to +175 C stg (1) T +175 C j Maximum operating junction temperature 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 lead 6.6 C/W th(j-l) For more information, please refer to the following application note: AN5088: Rectifiers thermal management, handling and mounting recommendations Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 8 A j (1) V = V I Reverse leakage current R R RRM 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 DS13227 - Rev 1 page 2/8