STPS15L60CB-TR STMicroelectronics

STPS15L60C Power Schottky rectifier Datasheet - production data Description Dual center tab Schottky rectifier suited for . switched mode power supply and high frequency DC to DC converters. Packaged in DPAK, this device is intended for use . in low voltage, high frequency inverters, free- . wheeling and polarity protection applications. Table 1. Device summary Symbol Value I 2 x 7.5 A . 3 F(AV) V 60 V RRM T 150 C j(max) Features V (typ) 0.52 V F Very small conduction losses Negligible switching losses Low forward voltage drop Avalanche specification ECOPACK 2 compliant component for DPAK on demand December 2015 DocID8621 Rev 4 1/8 This is information on a product in full production. www.st.comCharacteristics STPS15L60C 1 Characteristics Table 2. Absolute ratings (limiting values per diode at 25 C unless otherwise stated) Symbol Parameter Value Unit V Repetitive peak reverse voltage 60 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 265 W p j ARM T Storage temperature range -65 to +175 C stg (2) T Maximum operating junction temperature 150 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 resistances 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 200 A j (1) I Reverse leakage current V = V R R RRM T = 125 C 45 60 mA j T = 25 C 0.62 j I = 7.5 A F T = 125 C 0.52 0.57 j T = 25 C 0.76 j (1) V Forward voltage drop I = 12 A V F F T = 125 C 0.62 0.68 j T = 25 C 0.82 j I = 15 A F T = 125 C 0.66 0.72 j 1. Pulse test: t = 380 s, < 2% p To evaluate the conduction losses use the following equation: 2 P = 0.32 x I + 0.027 I F(AV) F (RMS) 2/8 DocID8621 Rev 4