STPS15H100CBY-TR STMicroelectronics

STPS15H100C-Y Datasheet Automotive high voltage power Schottky rectifier Features A1 AEC-Q101 qualified K Negligible switching losses A2 Low leakage current Good trade-off between leakage current and forward voltage drop K Low thermal resistance Avalanche capability specified PPAP capable A1 A2 Description Dual center tab Schottky rectifier suited for switched mode power supply and high DPAK frequency DC to DC converters. Packaged in DPAK, the STPS15H100C-Y is intended for use in high frequency LED head lamp circuits for automotive applications. Product status STPS15H100C-Y Product summary Symbol Value I 2 x 7.5 A F(AV) V 100 V RRM T 175 C j(max.) V 0.67 V F(max.) DS6880 - Rev 2 - April 2018 www.st.com For further information contact your local STMicroelectronics sales office.STPS15H100C-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 100 V RRM I Forward rms current 10 A F(RMS) T = 150 C, = 0.5 square wave Per diode 7.5 c I Average forward current A F(AV) T = 145 C, = 0.5 square wave Per device 15 c 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 475 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 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 - 3 A j (1) I V = V Reverse leakage current R R RRM T = 125 C - 1.3 4 mA j T = 25 C - 0.8 j I = 7.5 A F T = 125 C - 0.62 0.67 j T = 25 C - 0.85 j (2) V Forward voltage drop I = 12 A V F F T = 125 C - 0.68 0.73 j T = 25 C - 0.89 j I = 15 A F T = 125 C - 0.71 0.76 j 1. t = 5 ms, < 2% p 2. t = 380 s, < 2% p To evaluate the conduction losses, use the following equation: 2 P = 0.58 x I + 0.012 x I F(AV) F (RMS) DS6880 - Rev 2 page 2/9