STPS40L45CG STMicroelectronics

STPS40L45C Low drop power Schottky rectifier Datasheet - production data Description A1 K Dual center tap Schottky barrier rectifier designed for high frequency switched mode power supplies A2 and DC to DC converters. 2 Packaged in TO-220AB, TO-247 and D PAK K these devices are intended for use in low voltage, high frequency inverters, free-wheeling and polarity protection applications. A2 A1 Table 1. Device summary 2 D PAK I 2 x 20 A STPS40L45CG F(AV) V 45 V RRM T (max) 150 C j V (max) 0.49 V F A2 A2 K K A1 A1 TO-220AB TO-247 STPS40L45CT STPS40L45CW Features Low forward voltage drop meaning very small conduction losses Low switching losses allowing high frequency operation Avalanche capability specified October 2013 DocID6857 Rev 5 1/11 This is information on a product in full production. www.st.comCharacteristics STPS40L45C 1 Characteristics Table 2. Absolute Ratings (limiting values, per diode) Symbol Parameter Value Unit V Repetitive peak reverse voltage 45 V RRM I Forward rms current 30 A F(RMS) T =130 C Per diode 20 c I Average forward current A F(AV) 40 = 0.5 Per device I Surge non repetitive forward current t = 10 ms Sinusoidal 220 A FSM p I Repetitive peak reverse current t = 2 s square F = 1 kHz 2 A RRM p I Non repetitive peak reverse current t = 100 s square 3 A RSM p P Repetitive peak avalanche power t = 1 s T = 25 C 8100 W ARM p j T Storage temperature range -65 to + 150 C stg (1) T Maximum operating junction temperature 150 C j dV/dt Critical rate of rise of reverse voltage 10000 V/s dPtot 1 --------------- -------------------------- 1. 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 1.5 R Junction to case C/W th (j-c) Total 0.8 R Coupling 0.1 C/W th(c) When the diodes 1 and 2 are used simultaneously: T (diode 1) = P(diode1) x R (Per diode) + P(diode2) 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 0.6 mA j (1) I Reverse leakage current V = V R R RRM T = 125 C 140 280 mA j T = 25 C I = 20 A 0.53 j F T = 125 C I = 20 A 0.42 0.49 j F ( ) V 1. Forward voltage drop V F T = 25 C I = 40 A 0.69 j F T = 125 C I = 40 A 0.6 0.7 j F 1. Pulse test: t = 380 s, < 2% p To evaluate the conduction losses use the following equation: 2 P = 0.28 x I + 0.0105 I F(AV) F (RMS) 2/11 DocID6857 Rev 5