FERD2045SB-TR STMicroelectronics

FERD2045S 45 V field-effect rectifier diode Datasheet - production data Description A This single rectifier is based on a proprietary K A technology that achieves the best in class V /I F R trade-off for a given silicon surface. K K Therefore it can advantageously replace 45 V low voltage Schottky diodes. Packaged in DPAK, this device is intended to be A A used in rectification and freewheeling operations in power supplies. A A Table 1: Device summary DPAK Symbol Value IF(AV) 20 A Features V 45 V RRM ST advanced rectifier process VF(typ.) 0.29 V Stable leakage current over reverse voltage T(max.) 150 C j Low forward voltage drop High frequency operation ECOPACK 2 compliant component for DPAK on demand January 2018 DocID031305 Rev 1 1/8 www.st.com This is information on a product in full production. Characteristics FERD2045S 1 Characteristics Table 2: Absolute ratings (limiting values at 25 C, unless otherwise specified, anode terminals short-circuited) Symbol Parameter Value Unit VRRM Repetitive peak reverse voltage 45 V I Forward rms current 40 A F(RMS) Average forward current = 0.5, square I T = 125 C 20 A F(AV) C wave I Surge non repetitive forward current t = 10 ms sinusoidal 180 A FSM p Tstg Storage temperature range -65 to +175 C (1) T Maximum operating junction temperature range -40 to +150 C j Notes: (1) (dPtot/dTj) < (1/Rth(j-a)) condition to avoid thermal runaway for a diode on its own heatsink. Table 3: Thermal resistance parameters Symbol Parameter Value Unit R Junction to case 1.4 C/W th(j-c) Table 4: Static electrical characteristics (anode terminals short circuited) Symbol Parameter Test conditions Min. Typ. Max. Unit Tj = 25 C - 100 300 A VR = 35 V T = 125 C - 12 24 mA j (1) IR Reverse leakage current Tj = 25 C - 200 600 A VR = VRRM T = 125 C - 18 40 mA j Tj = 25 C - 0.35 IF = 5 A T = 125 C - 0.29 j Tj = 25 C - 0.41 0.45 (2) VF Forward voltage drop IF = 10 A V T = 125 C - 0.38 0.42 j Tj = 25 C - 0.51 0.55 IF = 20 A T = 125 C - 0.52 0.57 j Notes: (1) Pulse test: tp = 5 ms, < 2% (2) Pulse test: t = 380 s, < 2% p To evaluate the maximum conduction losses use the following equation: 2 P = 0.27 x I + 0.015 x I F(AV) F (RMS) 2/8 DocID031305 Rev 1