FERD40U50CFP STMicroelectronics

FERD40U50C Field effect rectifier Datasheet - production data Description This dual rectifier is based on a proprietary . technology that achieved the best in class V /I F R for a given silicon surface. Packaged in TO-220FPAB, this device is intended to be used in rectification and freewheeling operations in switch-mode power supplies. Table 1. Device summary Symbol Value . % 2 )3 7 I 2 x 20 A F(AV) V 50 V RRM T (max) +175 C j Features V (typ) 0.43 V F ST advanced rectifier process Stable leakage current over reverse voltage Reduced leakage current Low forward voltage drop High frequency operation Insulated package: TO-220FPAB Insulating voltage: 2000 V sine RMS June 2015 DocID027943 Rev 1 1/8 This is information on a product in full production. www.st.comCharacteristics FERD40U50C 1 Characteristics Table 2. Absolute ratings (limiting values, per diode, at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 50 V RRM I Forward rms current 40 A F(RMS) T = 120 C Per diode 20 c I Average forward current, = 0.5 A F(AV) T = 90 C Per device 40 c I Surge non repetitive forward current t = 10 ms sinusoidal 250 A FSM p T Storage temperature range -65 to + 175 C stg (1) T Maximum operating junction temperature 175 C j dPtot 1 --------------- -------------------------- 1. < condition to avoid thermal runaway for a diode on its own heatsink. dTj Rth()j a Table 3. Thermal resistance Symbol Parameter Value (max) Unit Per diode 4.1 R Junction to case th(j-c) Total 3.3 C/W R Coupling 2.4 th(c) When 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 4. Static electrical characteristics (per diode) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C 0.8 mA j V = V R RRM T = 125 C 30 60 mA j (1) I Reverse leakage current R T = 25 C 460 A j V = 35 V R = 125 C 20 40 mA T j T = 125 C I = 5 A 0.25 j F T = 125 C I = 10 A 0.33 j F T = 25 C 0.41 0.46 j (2) V Forward voltage drop I = 15 A V F F T = 125 C 0.39 0.43 j T = 25 C 0.44 0.49 j I = 20 A F = 125 C 0.43 0.48 T 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 = 0.329 x I + 0.007 I F(AV) F (RMS) 2/8 DocID027943 Rev 1