STPS8H100G-TR STMicroelectronics

STPS8H100 High voltage power Schottky rectifier Datasheet - production data A K Features Negligible switching losses K K High junction temperature capability Low leakage current Good trade off between leakage current and forward voltage drop A A A Insulated package: TO-220AC K NC NC TO-220FPAC, insulating voltage = 2 2000 V sine RMS D PAK Avalanche capability specified ECOPACK 2 compliant component for DPAK on demand A Description K TO-220FPAC Schottky barrier rectifier designed for high frequency compact switched mode power supplies such as adaptors and on-board DC-DC converters. Table 1: Device summary Symbol Value IF(AV) 8 A V 100 V RRM Tj (max) 175 C V (typ) 0.56 V F December 2015 DocID5387 Rev 12 1/14 www.st.com This is information on a product in full production. Characteristics STPS8H100 1 Characteristics Table 2: Absolute ratings (limiting values, at 25 C, unless otherwise specified) Symbol Parameter Value Unit VRRM Repetitive peak reverse voltage 100 V I Forward rms current 30 A F(RMS) 2 Average forward TO-220AC, D PAK TC = 165 C IF(AV) current = 0.5, 8 A TO-220FPAC T = 150 C C square wave Surge non I repetitive forward tp = 10 ms sinusoidal 250 A FSM current Repetitive peak P tp = 10 s, T= 125 C 750 W ARM j avalanche power T Storage temperature range -65 to + 175 C stg (1) Tj Maximum operating junction temperature + 175 C Notes: (1) (dP /dT ) < (1/R ) condition to avoid thermal runaway for a diode on its own heatsink. tot j th(j-a) Table 3: Thermal parameter Symbol Parameter Value Unit 2 TO-220AC, D PAK 1.6 Junction to case Rth(j-c) C/W TO-220FPAC 4 Table 4: Static electrical characteristics Symbol Parameter Test conditions Min. Typ. Max. Unit Tj = 25 C - 4.5 A (1) IR Reverse leakage current VR = VRRM T = 125 C - 2.0 6.0 mA j Tj = 25 C - 0.71 IF = 8 A T = 125 C - 0.56 0.58 j Tj = 25 C - 0.77 (2) VF Forward voltage drop IF = 10 A V T = 125 C - 0.59 0.64 j Tj = 25 C - 0.81 IF = 16 A T = 125 C - 0.65 0.68 j Notes: (1) Pulse test: tp = 5 ms, < 2% (2) Pulse test: t = 380 s, < 2% p To evaluate the conduction losses use the following equation: 2 P = 0.48 x I + 0.0125 I F(AV) F (RMS) 2/14 DocID5387 Rev 12