STPS2545CT STMicroelectronics

STPS2545C Power Schottky rectifier Datasheet - production data A1 K Description A2 Dual center tab Schottky rectifier suited for switch K mode power supplies and high frequency DC to DC converters. A2 This device is especially intended for use in low K A1 voltage, high frequency inverters, free-wheeling TO-220AB and polarity protection applications. K K Table 1: Device summary Symbol Value A2 A2 IF(AV) 2 x 12.5 A A1 A1 DPAK V 45 V RRM Tj(max.) 175 C Features V (typ.) 0.5 V F Very small conduction losses Negligible switching losses Extremely fast switching Low thermal resistance Avalanche capability specified High frequency operation ECOPACK 2 compliant component (on demand for DPAK) May 2017 DocID8736 Rev 5 1/11 www.st.com This is information on a product in full production. Characteristics STPS2545C 1 Characteristics Table 2: Absolute ratings (limiting values, per diode, at 25 C, unless otherwise specified) Symbol Parameter Value Unit Repetitive peak reverse voltage VRRM 45 V I Forward rms current 30 A F(RMS) T = 160 C Per diode 12.5 C Average forward current = 0.5, IF(AV) A square wave TC = 155 C Per device 25 Surge non repetitive forward I tp = 10 ms sinusoidal 200 A FSM current P Repetitive peak avalanche power tp = 10 s, Tj = 125 C 345 W ARM 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 parameters Symbol Parameter Max. value Unit Per diode 1.6 R Junction to case C/W th(j-c) Total 1.1 Rth(c) Coupling 0.6 C/W 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 Tj = 25 C - 125 A (1) I Reverse leakage current V = V R R RRM T = 125 C - 9 25 mA j Tj = 125 C IF = 12.5 A - 0.50 0.57 (1) V Forward voltage drop T = 25 C - 0.84 V F j IF = 25 A Tj = 125 C - 0.65 0.72 Notes: (1) Pulse test: t = 380 s, < 2% p To evaluate the conduction losses, use the following equation: 2 P = 0.42 x I + 0.012 x I F(AV) F (RMS) 2/11 DocID8736 Rev 5