STPS16170CR STMicroelectronics

STPS16170C High voltage power Schottky rectifier Datasheet production data Description . This dual center tab Schottky rectifier is suited for high frequency switched mode power supplies Table 1. Device summary . Symbol Value I 2 x 8 A F(AV) V 170 V RRM . 3 T 175 C j(max) . V 0.70 V F (Typ) 3 Features High junction temperature capability Good trade off between leakage current and forward voltage drop Low leakage current Avalanche capability specified ECOPACK 2 compliant component for DPAK and DPAK on demand December 2015 DocID12540 Rev 4 1/12 This is information on a product in full production. www.st.com ,3Characteristics STPS16170C 1 Characteristics Table 2. Absolute ratings (limiting values per diode at T = 25 C unless otherwise stated) amb Symbol Parameter Value Unit V Repetitive peak reverse voltage 170 V RRM I Forward rms current 20 A F(RMS) Per diode 8 I Average forward current, = 0.5, square wave T = 150 C A F(AV) c Total 16 I Surge non repetitive forward current t = 10 ms sinusoidal 75 A FSM p (1) P Repetitive peak avalanche power t = 10 s, T = 125 C 335 W ARM p j T Storage temperature range -65 to + 175 C stg (2) T Maximum operating junction temperature 175 C j 1. For pulse time duration derating, please refer to Figure 3. More details regarding the avalanche energy measurements and diode validation in the avalanche are provided in the application notes AN1768 and AN2025. 1 dPtot 2. < condition to avoid thermal runaway for a diode on its own heatsink dTj Rth(j-a) Table 3. Thermal parameters Symbol Parameter Value Unit Per diode 3 R Junction to case th(j-c) Total 1.8 C/W R Coupling 0.6 th(c) When the diodes 1 and 2 are used simultaneously: Tj(diode 1) = P(diode1) x R (Per diode) + P(diode 2) x R th(j-c) th(c) Table 4. Static electrical characteristics (per diode) Symbol Parameter Test conditions Min. Typ Max. Unit T = 25 C -- 15 A j (1) I Reverse leakage current V = V R R RRM T = 125 C - - 15 mA j T = 25 C - - 0.92 j I = 8 A F T = 125 C - 0.70 0.75 j (2) V Forward voltage drop V F T = 25 C -- 1.0 j I = 16 A F T = 125 C - 0.80 0.86 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 = 0.64 x I + 0.014 x I F(AV) F (RMS) 2/12 DocID12540 Rev 4