STPS10L40C Low drop power Schottky rectifier Datasheet - production data Description A1 K Dual center tap Schottky rectifier suited for switch A2 mode power supply and high frequency DC to DC converters. K 2 Packaged either in TO-220AB and D PAK, this device is especially intended for use in low voltage, high frequency inverters, free wheeling and polarity protection applications. A2 Table 1: Device summary K A1 Symbol Value TO-220AB IF(AV) 2 x 5 A VRRM 40 V K K Tj (max.) 150 C V (typ.) 0.36 V F A2 A2 A1 A1 2 D PAK Features Low forward voltage drop meaning very small conduction losses Low dynamic losses as a result of the schottky barrier Avalanche capability specified ECOPACK 2 compliant component for DPAK on demand April 2016 DocID9433 Rev 7 1/12 www.st.com This is information on a product in full production. Characteristics STPS10L40C 1 Characteristics Table 2: Absolute ratings (limiting values, per diode, at 25 C, unless otherwise specified) Symbol Parameter Value Unit VRRM Repetitive peak reverse voltage 40 V I Forward rms current 20 A F(RMS) Per diode 5 Average forward current = 0.5, IF(AV) TC = 140 C A square wave Per device 10 I Surge non repetitive forward current t = 10 ms sinusoidal 150 A FSM p PARM Repetitive peak avalanche power tp = 10 s, Tj = 125 C 190 W T Storage temperature range -65 to +150 C stg (1) Tj Maximum operating junction temperature +150 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 Value Unit Per diode 3.0 Junction to case Rth(j-c) C/W Total 1.7 R Coupling - 0.35 C/W th(c) When the diodes 1 and 2 are used simultaneously: Tj (diode1) = P(diode1) x Rth(j-c) (per diode) + P(diode2) x Rth(c) Table 4: Static electrical characteristics (per diode) Symbol Parameter Test conditions Min. Typ. Max. Unit Tj = 25 C - 0.2 mA (1) I Reverse leakage current V = V R R RRM T = 100 C - 8 25 mA j Tj = 25 C IF = 5 A - 0.53 T = 100 C I = 5 A - 0.36 0.46 j F (1) VF Forward voltage drop V Tj = 25 C IF = 10 A - 0.67 T = 125 C I = 10 A - 0.49 0.59 j F Notes: (1) Pulse test: tp = 380 s, < 2% To evaluate the conduction losses use the following equation: 2 P = 0.33 x IF(AV) + 0.026 IF (RMS) 2/12 DocID9433 Rev 7