FERD20S100SB-TR STMicroelectronics

FERD20S100S 100 V field-effect rectifier diode Datasheet - production data A Description K The device is based on a proprietary technology K A that achieves the best in class V /I trade-off for a F R K given silicon surface. This 100 V rectifier has A been optimized for use in confined casing K A IPAK applications where both efficiency and thermal A K performance matter. With a lower dependency of A TO-220AB leakage current (IR) and forward voltage (VF) in K K function of temperature, the thermal runaway risk A A is reduced. Therefore, it can advantageously A A replace 100 V Schottky diodes. DPAK Table 1: Device summary Symbol Value IF(AV) 20 A Features V 100 V RRM ST advanced rectifier process VF(max.) 0.415 V Stable leakage current over reverse voltage Reduced leakage current I (max.) 110 A R Low forward voltage drop Tj(max.) 175 C High frequency operation ECOPACK 2 compliant component November 2017 DocID029416 Rev 2 1/12 www.st.com This is information on a product in full production. Characteristics FERD20S100S 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 100 V I Forward rms current 40 A F(RMS) IF(AV) Average forward current = 0.5, square wave TC = 150 C 20 A TO-220AB 220 A Surge non repetitive forward tp = 10 ms I FSM current sinusoidal DPAK, IPAK 150 A Tstg Storage temperature range -65 to +175 C (1) Tj Maximum operating junction temperature 175 C Notes: (1) (dP /dT ) < (1/R ) condition to avoid thermal runaway for a diode on its own heatsink. tot j th(j-a) Table 3: Thermal resistance parameters Symbol Parameter Value Unit Junction to case Rth(j-c) 1.3 C/W Table 4: Static electrical characteristics (anode terminals short circuited) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 110 A j V = V R RRM (1) IR Reverse leakage current Tj = 125 C - 6 12 mA T = 125 C V = 70 V - 3 6 j R Tj = 25 C - 0.395 0.445 IF = 2 A T = 125 C - 0.36 0.415 j Tj = 25 C - 0.520 0.585 IF = 5 A (2) V Forward voltage drop T = 125 C - 0.500 0.555 V F j Tj = 25 C - 0.680 0.780 IF = 10 A T = 125 C - 0.600 0.660 j Tj = 125 C IF = 20 A - 0.690 0.760 Notes: (1) Pulse test: t = 5 ms, < 2% p (2) Pulse test: tp = 380 s, < 2% To evaluate the conduction losses use the following equation: 2 P = 0.45 x IF(AV) + 0.021 x IF (RMS) 2/12 DocID029416 Rev 2