STPS20L120CFP STMicroelectronics

STPS20L120C Power Schottky rectifier Features High junction temperature capability Avalanche capability specified A1 K Low forward voltage drop current A2 High frequency operation Insulated package K TO-220FPAB Insulating voltage = 1500 V rms Typical package capacitance 12 pF A2 A2 K K A1 A1 Description TO-220AB TO-220FPAB Dual center tap Schottky rectifier suited for high STPS20L120CT STPS20L120CFP frequency switch mode power supplies. Packaged in TO-220AB and TO-220FPAB, this device provides the adaptor designers with an optimized price-performance ratio. Table 1. Device summary I 2 x 10 A F(AV) V 120 V RRM T 150 C j(max) V 0.55 V F(typ) May 2009 Doc ID 15670 Rev 1 1/8 www.st.com 8 Characteristics STPS20L120C 1 Characteristics Table 2. Absolute ratings (limiting values, per diode) Symbol Parameter Value Unit V Repetitive peak reverse voltage 120 V RRM I Forward rms current 20 A F(RMS) I Average forward current, = 0.5 Total package 20 A F(AV) I Surge non repetitive forward current t = 10 ms Sinusoidal 200 A FSM p P Repetitive peak avalanche power t = 1 s T = 25 C 8000 W ARM p j T Storage temperature range -65 to + 175 C stg (1) T Maximum operating junction temperature 150 C j dPtot 1 --------------- -------------------------- 1. < condition to avoid thermal runaway for a diode on its own heatsink dTj Rth()j a Table 3. Thermal parameters Symbol Parameter Value Unit Per diode 2 TO-220AB Total 1.1 R Junction to case th(j-c) Per diode 4.9 TO-220FPAB C/W Total 4.1 TO-220AB 0.2 R Coupling Total th(c) TO-220FPAB 3.2 When the diodes 1 and 2 are used simultaneously : T (diode 1) = P(diode 1) x R (per diode) + P(diode 2) x R j th(j-c) th(c) Table 4. Static electrical characteristics (per diode) Symbol Test conditions Min. Typ. Max. Unit T = 25 C - - 120 A j (1) I Reverse leakage current V = V R R RRM T = 125 C - 8 25 mA j T = 25 C - - 0.74 j I = 5 A F T = 125 C - 0.55 0.605 j T = 25 C - - 0.86 j (2) V Forward voltage drop I = 10 A V F F T = 125 C - 0.63 0.69 j T = 25 C -- 1 j I = 20 A F T = 125 C - 0.72 0.785 j 1. Pulse test : tp = 5 ms, < 2% 2. Pulse test : tp = 380 s, < 2% To evaluate the maximum conduction losses use the following equation : 2 P = 0.595 x I + 0.0095 I F(AV) F (RMS) 2/8 Doc ID 15670 Rev 1