STPS15L45C-Y Datasheet Automotive 45 V low drop power Schottky rectifier Features A1 K AEC-Q101 qualified A2 Very small conduction losses K Negligible switching losses Extremely fast switching Low forward voltage drop Avalanche capability specified K A1 PPAP capable A2 ECOPACK 2 compliant component DPAK Applications DC/DC converters Reverse polarity protection Freewheeling diodes Switching diodes Description Dual center tab Schottky rectifier suited for SMPS and high frequency DC to DC converters. Product status link Packaged in DPAK, the STPS15L45C-Y is intended for use in low voltage, high STPS15L45C-Y frequency inverters, freewheeling and polarity protection for automotive application. Product summary Symbol Value I 2 x 7.5 A F(AV) V 45 V RRM T 150 C j(max.) V 0.40 V F(typ.) DS7145 - Rev 3 - November 2018 www.st.com For further information contact your local STMicroelectronics sales office.STPS15L45C-Y Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values, per diode, at 25 C unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 45 V RRM I Forward rms current 10 A F(RMS) Per diode 7.5 I T = 140 C, = 0.5 square wave Average forward current A F(AV) c Per device 15 I Surge non repetitive forward current t = 10 ms sinusoidal 75 A FSM p P Repetitive peak avalanche power t = 10 s, T = 125 C 266 W ARM p j T Storage temperature range -65 to +175 C stg (1) T Operating junction temperature range -40 to +150 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 Max. value Unit Per diode 4 R Junction to case th(j-c) Total 2.4 C/W R Coupling 0.7 th(c) When the diodes 1 and 2 are used simultaneously : T (diode 1) = P(diode1) x R (per diode) + P(diode 2) x R j th(j-c) th(c) Table 3. Static electrical characteristics (per diode) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 500 A j (1) I Reverse leakage current V = V R RRM R T = 125 C - 60 120 mA j T = 25 C - 0.52 j I = 7.5 A F T = 125 C - 0.40 0.46 j T = 25 C - 0.60 j (1) V Forward voltage drop I = 12 A V F F T = 125 C - 0.49 0.57 j T = 25 C - 0.64 j I = 15 A F T = 125 C - 0.53 0.63 j 1. t = 380 s, < 2% p To evaluate the conduction losses, use the following equation: 2 P = 0.29 x I + 0.023 x I F(AV) F (RMS) DS7145 - Rev 3 page 2/10