STPS15L30CDJFTR STMicroelectronics

STPS15L30CDJF Low drop power Schottky rectifier Features Very small conduction losses A1 Negligible switching losses Extremely fast switching Low forward voltage drop K Low thermal resistance A2 High avalanche capability specified Description K K Dual center tap Schottky rectifier suited for switch mode power supply and high frequency DC to DC converters. Packaged in PowerFLAT, this device is intended for use in low voltage, high frequency A1 inverters, free-wheeling and polarity protection A2 applications. A2 A1 PowerFLAT 5x6 STPS15L30CDJF Table 1. Device summary Symbol Value I 2 x 7.5 A F(AV) V 30 V RRM T (max) 150 C j V (typ) 0.34 V F TM: PowerFLAT is a trademark of STMicroelectronics May 2011 Doc ID 15664 Rev 4 1/7 www.st.com 7 Characteristics STPS15L30CDJF 1 Characteristics Table 2. Absolute ratings (limiting values, per diode) Symbol Parameter Value Unit V Repetitive peak reverse voltage 30 V RRM I Forward rms current 10 A F(RMS) Per diode 7.5 I Average forward current = 0.5 T = 140 C A F(AV) c Per device 15 I Surge non repetitive forward current t = 10 ms sinusoidal 75 A FSM p I Peak repetitive reverse current t = 2 s square F= 1 kHz 1 A RRM p P Repetitive peak avalanche power t = 1 s T = 25 C 2800 W ARM p j T Storage temperature range -65 to + 175 C stg (1) T Maximum operating junction temperature 150 C j 1 dPtot < 1. condition to avoid thermal runaway for a diode on its own heatsink Rth(j-a) dTj Table 3. Thermal resistance Symbol Parameter Value Unit Per diode 2.5 R Junction to case th(j-c) Total 1.6 C/W R Coupling 0.7 th(c) When 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 -- 1 mA j Reverse leakage (1) I V = V R R RRM current T = 125 C - 70 140 mA j T = 25 C I = 7.5 A - - 0.48 j F T = 125 C I = 7.5 A - 0.34 0.39 j F (1) V Forward voltage drop V F T = 25 C I = 15 A - - 0.57 j F T = 125 C I = 15 A - 0.44 0.51 j F 1. Pulse test: t = 380 s, < 2% p To evaluate the conduction losses use the following equation: 2 P = 0.27 x I + 0.016 I F(AV) F (RMS) 2/7 Doc ID 15664 Rev 4