STPS5S100SFY Datasheet Automotive 100 V, 5 A low I power Schottky rectifier r Features AEC-Q101 qualified PPAP capable V guaranteed from -40 C to 175 C RRM Low leakage current Avalanche capability specified Very low conduction losses High junction temperature capability Low profile design - 1.1 mm package typical height Wettable flanks for automatic visual inspection ECOPACK2 compliant component Applications DC / DC converter Auxiliary Power supply Freewheeling function Reverse battery protection Description Product status link The 5 A, 100 V power Schottky rectifier has been designed for automotive STPS5S100SFY applications. Packaged in PSMC (TO-277A), the STPS5S100SFY provides a high level of Product summary performance in a compact and flat package which can withstand high operating Symbol Value junction temperature. I F(AV) 5 A V 100 V RRM T (max.) 175 C j V (typ.) 0.590 V F DS13593 - Rev 1 - December 2020 www.st.com For further information contact your local STMicroelectronics sales office.STPS5S100SFY Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified with 2 anode terminals short-circuited) Symbol Parameter Value Unit V Repetitive peak reverse voltage (T = -40 C to +175 C) 100 V RRM j I Average forward current, = 0.5 T = 160 C 5 A F(AV) c I t = 10 ms sinusoidal Surge non repetitive forward current 110 A FSM p P t = 10 s, T = 125 C Repetitive peak avalanche power 290 W ARM p j T Storage temperature range -65 to +175 C stg (1) T Operating junction temperature range -40 to +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 1.46 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 - 2.5 A j (1) I Reverse leakage current V = V R RRM R T = 125 C - 0.85 2.0 mA j T = 25 C - 0.720 j I = 2.5 A F T = 125 C - 0.525 0.595 j (2) V Forward voltage drop V F T = 25 C - 0.820 j I = 5 A F T = 125 C - 0.590 0.670 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 = 0.520 x I + 0.030 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 DS13593 - Rev 1 page 2/9