FERD15S50DJF-TR STMicroelectronics

FERD15S50 Datasheet 50 V, 15 A field effect rectifier Features ST patented rectifier process Stable leakage current over reverse voltage Low forward voltage drop High frequency operation ECOPACK 2 compliant Applications Set-top box Battery charger DC / DC converter Description This single rectifier is based on a patented technology, enabling to achieve the best in class V /I trade-off for a given silicon surface. F R Packaged in PowerFLAT 5x6, the FERD15S50 is optimized for use in rectification and freewheeling operations in switch mode power supplies. Product status FERD15S50 Product summary Symbol Value I 15 A F(AV) V 50 V RRM T 150 C j(max.) V 0.48 V F(typ.) DS9849 - Rev 3 - February 2019 www.st.com For further information contact your local STMicroelectronics sales office.FERD15S50 Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified, anode terminals short circuited) Symbol Parameter Value Unit V Repetitive peak reverse voltage 50 V RRM I Forward rms current 25 A F(RMS) I Average forward current, = 0.5 square wave T = 120 C 15 A F(AV) C I Surge non repetitive forward current t = 10 ms sinusoidal 80 A FSM p T Storage temperature range -65 to +175 C stg (1) T Maximum operating junction temperature +150 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 parameter Symbol Parameter Max. value Unit R Junction to case 2.8 C/W th(j-c) For more information, please refer to the following application note : AN5046 : Printed circuit board assembly recommendations for STMicroelectronics PowerFLAT packages Table 3. Static electrical characteristics (anode terminals short circuited) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 250 650 A j (1) I Reverse leakage current V = V R RRM R T = 125 C - 20 40 mA j T = 25 C - 0.35 j I = 5 A F T = 125 C - 0.30 0.35 j (2) T = 25 C V Forward voltage drop - 0.42 0.48 V F j I = 10 A F T = 125 C - 0.41 0.45 j T = 25 C I = 15 A - 0.48 j F 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 = 0.25 x I + 0.02 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 AN4021: Calculation of reverse losses on a power diode DS9849 - Rev 3 page 2/9