STPS40120CT STMicroelectronics

STPS40120C Datasheet 120 V power Schottky rectifier A1 Features K A2 High junction temperature capability Avalanche rated Low leakage current K K Good trade-off between leakage current and forward voltage drop ECOPACK 2 compliant A2 A2 K K A1 A1 Applications TO-220AB TO-220AB narrow leads Switching diode SMPS DC/DC converter LED lighting Notebook adapter Description This dual center tap Schottky rectifier is ideally suited for high frequency switch mode power supply. Packaged in TO-220AB and TO-220AB narrow leads, the STPS40120C is optimized for use in notebook and LCD adapters, desktop SMPS, providing in these applications a margin between the remaining voltages applied on the diode and the voltage capability of the diode. Product status link STPS40120C Product summary Symbol Device I 2 x 20 A F(AV) V 120 V RRM T (max.) 175 C j V (typ.) 0.69 V F DS4231 - Rev 4 - June 2018 www.st.com For further information contact your local STMicroelectronics sales office.STPS40120C Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C unless otherwise specified, per diode) Symbol Parameter Value Unit V Repetitive peak reverse voltage 120 V RRM I Forward rms current 30 A F(RMS) Per diode 20 I Average forward current , T = 145 C, = 0.5 A F(AV) c Per device 40 I Surge non repetitive forward current t = 10 ms sinusoidal 200 A FSM p P Repetitive peak avalanche power t = 10 s, T = 125 C 755 W ARM p j T Storage temperature range -65 to +175 C stg (1) T Maximum operating junction temperature 175 C j 1. (dP /dT ) < (1/R ) condition to avoid thermal runaway for a diode on its own heatsink. tot j th(j-a) Table 2. Thermal resistance parameters Symbol Parameter Max. value Unit Per diode 1.6 R Junction to case th(j-c) Total 0.85 C/W R Coupling 0.1 th(c) When the diodes 1 and 2 are used simultaneously: T (diode 1) = P(diode 1) x R (per diode) + P(diode 2) x R j th(j-c) th(c) Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 25 A j (1) I Reverse leakage current V = V R RRM R T = 125 C - 4 12 mA j T = 25 C - 0.73 j I = 7.5 A F T = 125 C - 0.57 0.61 j T = 25 C - 0.90 j (2) V Forward voltage drop I = 20 A V F F T = 125 C - 0.69 0.73 j T = 25 C - 1.00 j I = 40 A F T = 125 C - 0.83 0.88 j 1. Pulse test: t = 5 ms, < 2% p 2. Pulse test: t = 380 s, < 2% p To evaluate the conduction losses, use the following equation: 2 P = 0.58 x I + 0.0075 x I F(AV) F (RMS) For more information, please refer to the following application notes related to the power losses : DS4231 - Rev 4 page 2/11