STTH806 Turbo 2 ultrafast - high voltage rectifier Main product characteristics A K K I 8 A F(AV) V 600 V RRM A T 175 C j NC 2 V (typ) 1.1 V F D PAK STTH806G t (max) 35 ns rr Features and benefits A A Ultrafast switching K K Low reverse current TO-220AC TO-220AC Ins STTH806D STTH806DIRG Low thermal resistance Reduces conduction and switching losses Insulated package TO-220AC Ins Order codes Insulated voltage: 2500 V RMS Part Number Marking Typical package capacitance: 7 pF STTH806G STTH806G Description STTH806G-TR STTH806G STTH806D STTH806D The STTH806 uses ST Turbo2 600 V technology. STTH806DIRG STTH806DI This device is specially suited for use in switching power supplies, and industrial applications. Table 1. Absolute ratings (limiting values per diode at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 600 V RRM 2 TO-220AC, D PAK 30 A I RMS forward current F(RMS) TO-220 Ins 24 A 2 T = 140 C TO-220AC, D PAK 8 A c I Average forward current, = 0.5 F(AV) T = 120 C TO-220 Ins 8 A c I Surge non repetitive forward current t = 10 ms Sinusoidal 90 A FSM p T Storage temperature range -65 to + 175 C stg (1) T Maximum operating junction temperature 175 C j dP 1. condition to avoid thermal runaway for a diode on its own heatsink tot 1 --------------- < -------------------------- dT R j th()j a August 2006 Rev 2 1/9 www.st.comCharacteristics STTH806 1 Characteristics Table 2. Thermal parameters Symbol Parameter Value Unit 2 TO-220AC, D PAK 2.5 R Junction to case C/W th(j-c) TO-220AC Ins 4 Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ Max. Unit T = 25 C 8 j (1) I Reverse leakage current V = V A R R RRM T = 150 C 20 200 j T = 25 C 1.85 j (2) V Forward voltage drop I = 8 A V F F T = 150 C 1.10 1.40 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 = 1.07 x I + 0.041 I F(AV) F (RMS) Table 4. Dynamic characteristics Test conditions Symbol Parameter Min. Typ Max. Unit I = 0.5 A, I = 0.25 A, I = 1 A, F rr R 35 T = 25 C j t Reverse recovery time ns rr I = 1 A, dI /dt = -50 A/s, F F 40 55 V = 30 V, T = 25 C R j I = 8 A, dI /dt = -100 A/s, F F I Reverse recovery current 4.5 6.5 RM V = 400 V, T = 25 C R j I = 8 A dI /dt = 100 A/s F F t Forward recovery time 200 ns fr V = 1.1 x V , T = 25 C FR Fmax j I = 8 A dI /dt = 100 A/s F F V Forward recovery voltage 3.5 V FP V = 1.1 x V , T = 25 C FR Fmax j 2/9