STPS40L45CGY-TR STMicroelectronics

STPS40L45C-Y Automotive power Schottky rectifier Datasheet production data Features A1 Low forward voltage drop meaning very small K conduction losses A2 Low switching losses allowing high frequency K operation Avalanche capability specified AEC-Q101 qualified A2 A1 Description DPAK Dual center tap Schottky barrier rectifier designed STPS40L45CGY for high frequency switched mode power supplies and DC to DC converters. Packaged in DPAK, this device is intended for Table 1. Device summary use in low voltage, high frequency inverters, free- Symbol Value wheeling and polarity protection for automotive applications. I 2 x 20 A F(AV) V 45 V RRM T (max) 150 C j V (max) 0.49 V F June 2012 Doc ID 023224 Rev 1 1/7 This is information on a product in full production. www.st.com 7Characteristics STPS40L45C-Y 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) = 0.5 per device 40 I Surge non repetitive forward current t = 10 ms sinusoidal 230 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 Operating junction temperature -40 to + 150 C j dV/dt Critical rate of rise of reverse voltage 10000 V/s 1 dPtot < 1. condition to avoid thermal runaway for a diode on its own heatsink Rth(j-a) dTj 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(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 0.6 mA j Reverse leakage (1) I V = V R R RRM current 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 (1) V 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) 2/7 Doc ID 023224 Rev 1