STTH8006W STMicroelectronics

STTH8006 Turbo 2 ultrafast high voltage rectifier Main product characteristics I 80 A F(AV) V 600 V RRM T 175 C j V (typ) 1.02 V F A K t (max) 70 ns rr DO-247 STTH8006W Features and benefits Ultrafast switching Low reverse current Low thermal resistance Reduces switching and conduction losses Description The STTH8006, which is using ST Turbo 2 600V technology, is specially suited for use in switching power supplies, and industrial applications, as Order Code rectification and discontinuous mode PFC boost diode. Thanks to its low V characteristics, this F Part number Marking device exhibits high performances in free- STTH8006W STTH8006W wheeling applications. Table 1. Absolute ratings (limiting values, at T = 25 C, unless otherwise specified) amb Symbol Parameter Value Unit V Repetitive peak reverse voltage 600 V RRM I RMS forward voltage 113 A F(RMS) I Average forward current T = 75 C = 0.5 80 A F(AV) c I Surge non repetitive forward current t = 10 ms sinusoidal 500 A FSM p T Storage temperature range -65 to + 175 C stg T Maximum operating junction temperature 175 C j December 2006 Rev 1 1/6 www.st.com 6Characteristics STTH8006 1 Characteristics Table 2. Thermal resistance Symbol Parameter Value (max). Unit R Junction to case 0.75 C/W th(j-c) Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ Max. Unit T = 25 C 50 j (1) I Reverse leakage current V = V A R R RRM T = 150 C 160 1600 j T = 25 C 1.60 j (2) V Forward voltage drop I = 80 A V F F T = 150 C 1.02 1.30 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 = 0.98 x I + 0.004 I F(AV) F (RMS) Table 4. Dynamic characterstics Symbol Parameter Test conditions Min. Typ Max. Unit I = 0.5 A Irr = 0.25 A I =1 A 70 Reverse recovery F R t T = 25C ns rr j time I = 1 A dI /dt = 50 A/s V =30 V 75 105 F F R Reverse recovery I = 80 A V = 400 V F R I T = 125C 14 19 A RM j current dI /dt = 100 A/s F Forward recovery I = 80 A dI /dt = 200 A/s F F t T = 25C 600 ns fr j time V = 1.1 x V FR Fmax Forward recovery I = 80 A dI /dt = 200 A/s F F V T = 25C 3.7 V FP j voltage V = 1.1 x V FR Fmax Figure 1. Conduction losses versus average Figure 2. Forward voltage drop versus forward current forward current I (A) P(W) FM 140 200 =0.5 =1 180 120 =0.2 160 TT =150C=150C jj =0.1 (Maxim(Maximum vum values)alues) 100 140 =0.05 120 80 T =25C TT =150C=150C j jj 100 (Maximum values) (T(Typicalypical vvalues)alues) 60 80 60 40 T 40 20 20 =tp/T tp I (A) F(AV) V (V) FM 0 0 0 1020 3040 5060 7080 90 100 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2/6