STPS15H100C High voltage power Schottky rectifier Datasheet - production data Description Dual center tab Schottky rectifier suited for A1 switched mode power supply and high frequency K DC to DC converters. A2 Packaged in DPAK, this device is intended for use K in high frequency inverters. K Table 1. Device summary A1 A1 A2 Symbol Value A2 I 2 x 7.5 A DPAK F(AV) V 100 V RRM T 175 C j V (typ) 0.62 V F Features Negligible switching losses Low leakage current Good trade off between leakage current and forward voltage drop Low thermal resistance Avalanche capability specified ECOPACK 2 compliant component for DPAK on demand December 2015 DocID8562 Rev 7 1/8 This is information on a product in full production. www.st.comCharacteristics STPS15H100C 1 Characteristics Table 2. 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) Per diode 7.5 Average forward current, = 0.5, square (1) I T = 135 C A F(AV) c wave Per device 15 I Surge non repetitive forward current t = 10 ms sinusoidal 75 A FSM p t = 10 s, T = 125 C P Repetitive peak avalanche power 475 W p j ARM T Storage temperature range -65 to +175 C stg (2) T Maximum operating junction temperature 175 C j 1. Value based on R max (per diode) th(j-c) dPtot 1 --------------- -------------------------- 2. condition to avoid thermal runaway for a diode on its own heatsink dTj Rthj a Table 3. Thermal resistance Symbol Parameter 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 4. Static electrical characteristics (per diode) Symbol Parameter Test Conditions Min. Typ. Max. Unit T = 25 C 3A j (1) I Reverse leakage current V = V R R RRM T = 125 C 1.3 4 mA j T = 25 C I = 7.5 A 0.8 j F T = 125 C I = 7.5 A 0.62 0.67 j F T = 25 C I = 12 A 0.85 j F (2) V Forward voltage drop V F T = 125 C I = 12 A 0.68 0.73 j F T = 25 C I = 15 A 0.89 j F T = 125 C I = 15 A 0.71 0.76 j F 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 I F(AV) F (RMS) 2/8 DocID8562 Rev 7