FERD40L60C 60 V field-effect rectifier diode Datasheet - production data Description A1 K The device is based on a proprietary technology A2 that achieves the best in class V /I trade-off for a F R given silicon surface. K K This 60 V rectifier has been optimized for use in confined applications where both efficiency and thermal performance are key. A2 This device is suitable for use in adapters and A1 A2 K chargers. A1 TO-220AB DPAK Table 1: Device summary Symbol Value IF(AV) 2 x 20 A Features V 60 V RRM ST advanced rectifier process VF (typ.) 0.375 V Stable leakage current over reverse voltage T (max.) 150 C j Reduced leakage current Low forward voltage drop High frequency operation September 2017 DocID030964 Rev 1 1/12 www.st.com This is information on a product in full production. Characteristics FERD40L60C 1 Characteristics Table 2: Absolute ratings (limiting values at 25 C, per diode, unless otherwise specified) Symbol Parameter Value Unit VRRM Repetitive peak reverse voltage 60 V I Forward rms current 60 A F(RMS) Per diode 20 Average forward current = 0.5, IF(AV) TC = 130 C A square wave Per device 40 Surge non repetitive forward IFSM tp = 10 ms sinusoidal 250 A current Tstg Storage temperature range -65 to +175 C (1) T Maximum operating junction temperature +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 Max. value Unit Per diode 1.3 R Junction to case th(j-c) Per device 0.8 C/W Coupling Rth(c) 0.3 Table 4: Static electrical characteristics, per diode Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 1.1 j V = V R RRM (1) IR Reverse leakage current Tj = 125 C - 40 80 mA T = 125 C V = 45 V - 23 46 j R Tj = 25 C - 0.32 0.36 IF = 4 A Tj = 125 C - 0.26 0.3 Tj = 25 C - 0.4 0.445 (2) VF Forward voltage drop IF = 10 A V T = 125 C - 0.375 0.425 j Tj = 25 C - 0.49 0.545 IF = 20 A T = 125 C - 0.515 0.58 j Notes: (1) Pulse test: tp = 5 ms, < 2% (2) Pulse test: t = 380 s, < 2% p To evaluate the conduction losses use the following equation: 2 P = 0.27 x I + 0.0155 x I F(AV) F (RMS) 2/12 DocID030964 Rev 1