STPS30100ST Datasheet 100 V power Schottky rectifier Features A K Low V F A Good trade-off between leakage current and forward voltage drop K High frequency operation Avalanche capability specified ECOPACK 2 compliant A K A Applications Switching diode TO-220AB SMPS DC/DC converter LED lighting Desktop power supply Description This single Schottky rectifier is ideal for high frequency switch mode power supply. Housed in a TO-220AB package, the STPS30100ST is optimized for use in notebook and game station adaptors, providing an improved efficiency at both low and high load. Product status link STPS30100ST Product summary Symbol Value I 30 A F(AV) V 100 V RRM T (max.) 150 C j V (typ.) 0.605 V F DS5095 - Rev 2 - May 2018 www.st.com For further information contact your local STMicroelectronics sales office.STPS30100ST Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values with terminals 1 and 3 short circuited at T = 25 C, unless amb otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 100 V RRM I RMS forward current 60 A F(RMS) I T = 125 C, = 0.5 Average forward current 30 A F(AV) c 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 150 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 Max. value Unit R Junction to case 1 C/W th(j-c) Table 3. Static electrical characteristics (terminals 1 and 3 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 : AN604: Calculation of conduction losses in a power rectifier DS5095 - Rev 2 page 2/9