STPS50U100CT STMicroelectronics

STPS50U100C ULVF power Schottky rectifier Features A1 K ultralow forward voltage drop A2 high current capability high frequency operation Description A2 A2 K A1 K A1 The STPS50U100C is a dual power Schottky 2 TO-220AB I PAK diode rectifier, suited for high frequency switch STPS50U100CT STPS50U100CR mode power supplies. Featuring an ultralow forward voltage (ULVF) Table 1. Device summary drop, this device, packaged in TO-220AB and 2 I PAK, is intended to be used in notebook, game Symbol Value station and desktop adaptors as well as server I 2 x 25 A F(AV) SMPS. It has been especially designed to help power supply manufacturers meet the recently V 100 V RRM introduced worldwide efficiency standards. V (typ) (25 A 125 C) 0.64 V F T (max) 150 C j TM: ULVF is a trademark of STMicroelectronics November 2010 Doc ID 16603 Rev 2 1/8 www.st.com 8 Obsolete Product(s) - Obsolete Product(s)Characteristics STPS50U100C 1 Characteristics Table 2. Absolute ratings (limiting values per diode at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 100 V RRM I Forward rms current 50 A F(RMS) T = 120 C Per diode 25 C I Average forward current, = 0.5 A F(AV) T = 105 C Per device 50 C I Surge non repetitive forward current t = 10 ms, half sine-wave 250 A FSM p T Storage temperature range -65 to + 150 C stg (1) T Maximum operating junction temperature 150 C j 1 dPtot < 1. condition to avoid thermal runaway for a diode on its own heatsink Rth(j-a) dTj Table 3. Thermal resistance Symbol Parameter Value Unit Per diode 1.3 R Junction to case C/W th (j-c) Per device 0.9 R Coupling 0.45 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 Symbol Parameter Tests conditions Min. Typ. Max. Unit T = 25 C -15 - A j V = 70 V R T = 125 C - 10 - mA j I Reverse leakage current R T = 25 C -30 200 A j V = V R RRM T = 125 C - 15 40 mA j T = 25 C -0.48- j I = 5 A F T = 125 C - 0.38 - j T = 25 C -0.58- j V Forward voltage drop I = 15 A V F F T = 125 C - 0.54 - j T = 25 C -0.67 0.73 j I = 25 A F T = 125 C - 0.64 0.7 j 2 To evaluate the conduction losses use the following equation: P = 0.475 x I + 0.009 I F(AV) F (RMS) 2/8 Doc ID 16603 Rev 2 Obsolete Product(s) - Obsolete Product(s)