STPS60150C Datasheet 150 V power Schottky rectifier Features A1 K A2 High junction temperature capability K Low leakage current High voltage capabilities Low thermal resistance High frequency operation A2 Avalanche specification K ECOPACK 2 compliant A1 TO-220AB Applications Switching diode SMPS DC/DC converter Telecom power Description This dual center tab Schottky rectifier is suited for high frequency switched mode power supplies. Packaged in TO-220AB, the STPS60150C combines high current rating and low Product status volume to enhance both reliability and power density of the application. STPS60150C Product summary I 2 x 30 A F(AV) V 150 V RRM T 175 C j(max.) V 0.72 V F(typ.) DS4090 - Rev 2 - June 2018 www.st.com For further information contact your local STMicroelectronics sales office.STPS60150C Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values, per diode at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 150 V RRM I Forward rms current 60 A F(RMS) T = 145 C Per diode 30 C I Average forward current, = 0.5, square wave A F(AV) T = 135 C Per device 60 C I Surge non repetitive forward current t = 10 ms sinusoidal 270 A FSM p P Repetitive peak avalanche power t = 10 s, T = 125 C 1245 W ARM p j T Storage temperature range -65 to +175 C stg (1) T +175 C j Maximum operating junction temperature 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 Value Symbol Parameter Unit Max. Per diode 1.0 R Junction to case C/W th(j-c) Total 0.7 R Coupling 0.4 C/W th(c) When the diodes 1 and 2 are used simultaneously: T = P x R (per diode) + P x R j (diode1) (diode1) th(j-c) (diode2) th(c) Table 3. Static electrical characteristics (per diode) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 3 15 A j (1) I Reverse leakage current V = V R RRM R T = 125 C - 3 10 mA j T = 25 C - 0.94 j I = 30 A F T = 125 C - 0.72 0.76 j (2) V Forward voltage drop V F T = 25 C - 0.97 1.05 j I = 60 A F T = 125 C - 0.86 0.92 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 = 0.6 x I + 0.0053 x I F(AV) F (RMS) For more information, please refer to the following application notes related to the power losses : AN604: Calculation of conduction losses in a power rectifier DS4090 - Rev 2 page 2/10