STPS3H100UF STMicroelectronics

STPS3H100 Datasheet 3 A - 100 V power Schottky rectifier Features Negligible switching losses High junction temperature capability Low leakage current Good trade-off between leakage current and forward voltage drop Avalanche capability specified ECOPACK2 compliant Applications Switching diode SMPS DC/DC converter LED lighting Description This 100 V power Schottky rectifier is ideal for switch mode power supplies, for 12 V rails and high frequency converters. Packaged in SMB and SMB Flat, the STPS3H100 is optimized for use in consumer and computer applications where low drop forward voltage is required to reduce power dissipation. Product status link STPS3H100 Product summary I 3 A F(AV) V 100 V RRM T (max.) 175 C j V (typ.) 0.63 V F DS6597 - Rev 3 - January 2020 www.st.com For further information contact your local STMicroelectronics sales office.STPS3H100 Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 100 V RRM T = 115 C SMB l I Average forward current, = 0.5 square wave 3 A F(AV) T = 140 C SMB Flat l I t = 10 ms sinusoidal Surge non repetitive forward current 75 A FSM p t = 10 s, p P Repetitive peak avalanche power 172 W ARM T = 125 C j T Storage temperature range -65 to +175 C stg (1) T 175 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 parameter Symbol Parameter Max. value Unit SMB 25 R Junction to lead C/W th(j-l) SMB Flat 15 For more information, please refer to the following application note: AN5088: Rectifiers thermal management, handling and mounting recommendations Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 1.00 A j (1) I Reverse leakage current V = V R RRM R T = 125 C - 0.40 1.00 mA j T = 25 C - 0.84 j I = 3 A F T = 125 C - 0.63 0.68 j (2) V Forward voltage drop V F T = 25 C - 0.92 j I = 6 A F T = 125 C - 0.71 0.76 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 = 0.6 x I + 0.027 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 on a power diode DS6597 - Rev 3 page 2/11