FERD20L60C 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 10 A Features V 60 V RRM ST advanced rectifier process VF (typ.) 0.365 V Stable leakage current over reverse voltage T (max.) 150 C j Reduced leakage current Low forward voltage drop High frequency operation September 2017 DocID030962 Rev 1 1/12 www.st.com This is information on a product in full production. Characteristics FERD20L60C 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 30 A F(RMS) Per diode 10 Average forward current = 0.5, IF(AV) TC = 130 C A square wave Per device 20 I Surge non repetitive forward current t = 10 ms sinusoidal 140 A FSM p 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 2.2 R Junction to case th(j-c) Total 1.3 C/W Coupling Rth(c) 0.4 Table 4: Static electrical characteristics, per diode Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 970 A j V = V R RRM (1) IR Reverse leakage current Tj = 125 C - 30 60 mA Tj = 125 C VR = 45 V - 17 34 Tj = 25 C - 0.305 0.35 IF = 2 A T = 125 C - 0.25 0.29 j Tj = 25 C - 0.38 0.425 (2) VF Forward voltage drop IF = 5 A V T = 125 C - 0.365 0.415 j Tj = 25 C - 0.48 0.535 IF = 10 A Tj = 125 C - 0.51 0.575 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 = 0.255 x I + 0.032 x I F(AV) F (RMS) 2/12 DocID030962 Rev 1