STPS30M100ST STMicroelectronics

STPS30M100S Datasheet 100 V power Schottky rectifier Features Low forward voltage drop A Good trade-off between leakage current and forward voltage drop K A High frequency operation Avalanche capability specified K ECOPACK 2 compliant A A Applications K K A A Switching diode TO-220FPAB TO-220AB SMPS DC/DC converter LED lighting Desktop power supply Description This rectifier is suited for high frequency switch mode power supply. Housed in TO-220AB and TO-220FPAB packages the STPS30M100S is optimized for use in notebook and game station adapters, providing in these applications a good efficiency at both low and high load. Product status link STPS30M100S Product summary Symbol Value I 30 A F(AV) V 100 V RRM T (max.) 150 C j V (typ.) 0.605 V F DS6171 - Rev 4 - June 2018 www.st.com For further information contact your local STMicroelectronics sales office.STPS30M100S 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 Forward rms current 60 A F(RMS) I Average forward current 30 A F(AV) I Surge non repetitive forward current t = 10 ms sinusoidal 300 A FSM p P Repetitive peak avalanche power t = 10 s , T = 125 C 1900 W ARM p j T Storage temperature range -65 to +175 C stg (1) T Maximum operating junction temperature 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 TO-220AB 1 R Junction to case C/W th(j-c) TO-220FPAB 4 Table 3. Static electrical characteristics (anode terminals short circuited) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 175 A j V = V R RRM T = 125 C - 20 50 mA j (1) I Reverse leakage current R T = 25 C - 60 A j V = 70 V R T = 125 C - 10 20 mA j T = 25 C - 0.475 j I = 5 A F T = 125 C - 0.385 j T = 25 C - 0.555 j I = 10 A F T = 125 C - 0.475 j (2) V Forward voltage drop V F T = 25 C - 0.620 0.660 j I = 15 A F T = 125 C - 0.525 0.565 j T = 25 C - 0.740 0.800 j I = 30 A F T = 125 C - 0.605 0.655 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 = 0.475 x I + 0.006 x I F(AV) F (RMS) For more information, please refer to the following application notes related to the power losses : DS6171 - Rev 4 page 2/11