FERD30SM100DJF Datasheet 100 V, 30 A field effect rectifier Features ST patented rectifier process Stable leakage current over reverse voltage Low forward voltage drop High frequency operation ECOPACK 2 compliant Applications Switching diode Notebook adapter LED lighting DC / DC converter Description The FERD30SM100DJF is based on a proprietary technology that achieves the best in class V / I trade-off for a given silicon surface. F R Packaged in PowerFLAT 5x6, the FERD30SM100DJF is optimized for use in confined applications where both efficiency and thermal performance are key. Product status FERD30SM100DJF Product summary Symbol Value I 30 A F(AV) V 100 V RRM T 175 C j(max.) V 0.665 V F(typ.) DS10782 - Rev 3 - February 2019 www.st.com For further information contact your local STMicroelectronics sales office.FERD30SM100DJF Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified, anode terminals short circuited) Symbol Parameter Value Unit V Repetitive peak reverse voltage 100 V RRM I Forward rms current 45 A F(RMS) I Average forward current, = 0.5, square wave T = 100 C 30 A F(AV) C I Surge non repetitive forward current t = 10 ms sinusoidal 180 A FSM p T Storage temperature range -65 to +175 C stg (1) T Maximum operating junction temperature +175 C j 1. (dP /dT ) < (1/R ) condition to avoid thermal runaway for a diode on its own heatsink. tot j th(j-a) Table 2. Thermal resistance parameter Symbol Parameter Max. value Unit R Junction to case 2.6 C/W th(j-c) For more information, please refer to the following application note : AN5046 : Printed circuit board assembly recommendations for STMicroelectronics PowerFLAT packages Table 3. Static electrical characteristics (anode terminals short-circuited) Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - - 150 A j V = V R RRM (1) T = 125 C I Reverse leakage current - 8 16 R j mA T = 125 C V = 70 V - - 9 j R T = 25 C - 0.480 j I = 5 A F T = 125 C - 0.395 0.435 j T = 25 C - 0.595 j (2) V Forward voltage drop I = 10 A V F F T = 125 C - 0.510 0.555 j T = 25 C - 0.970 j I = 30 A F T = 125 C - 0.665 0.735 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 = 0.562 x I + 0.0057 x I F(AV) F (RMS) For more information, please refer to the following application notes related to the power losses : AN604: Calculation of conduction losses in a power rectifier AN4021: Calculation of reverse losses on a power diode DS10782 - Rev 3 page 2/9