FERD40M45C Field effect rectifier Datasheet - production data Description A1 This dual rectifier is based on a proprietary K technology that achieves the best in class V /I F R A2 for a given silicon surface. K 2 Packaged in TO-220AB, and D PAK, this device K is intended to be used in switch mode power supplies, or automotive applications A2 Table 1. Device summary A2 K A1 A1 I 2 x 20 A F(AV) 2 TO-220AB D PAK V 45 V RRM FERD40M45CG FERD40M45CT V (typ) 0.34 V F Features ST advanced rectifier process Stable leakage current over reverse voltage Low forward voltage drop High frequency operation November 2013 DocID024889 Rev 1 1/10 This is information on a product in full production. www.st.comCharacteristics FERD40M45C 1 Characteristics Table 2. Absolute ratings (limiting values, per diode at 25 C, unless otherwise stated) Symbol Parameter Value Unit V Repetitive peak reverse voltage 45 V RRM I Forward rms current 40 A F(RMS) T =150 C Per diode 20 c I Average forward current, = 0.5 A F(AV) T =140 C Per device 40 c I Surge non repetitive forward current t = 10 ms sinusoidal 275 A FSM p T Storage temperature range -65 to + 175 C stg 2 175 TO-220AB, D PAK Maximum operating junction T C (1) 2 j temperature D PAK (DC forward current without 200 reverse bias, t = 1 hour) dPtot 1 --------------- -------------------------- 1. condition to avoid thermal runaway for a diode on its own heatsink dTj Rthj a Table 3. Thermal resistances Symbol Parameter Value Unit Per diode 1.6 R Junction to case C/W th (j-c) Total 1.1 R Coupling 0.5 C/W th(c) When the diodes 1 and 2 are used simultaneously: T (diode 1) = P(diode1) x R (Per diode) + P(diode2) x R . j th(j-c) th(c) Table 4. Static electrical characteristics (per diode) Symbol Parameter Test Conditions Min. Typ. Max. Unit T = 25 C 650 A j (1) I Reverse leakage current V = V R R RRM T = 125 C 25 50 mA j T = 25 C 0.38 0.415 j I = 10 A F T = 125 C 0.34 0.37 j (2) V Forward voltage drop V F T = 25 C 0.46 0.50 j I = 20 A F T = 125 C 0.46 0.50 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 = 0.31 x I + 0.0095 I F(AV) F (RMS) 2/10 DocID024889 Rev 1