STTH3012 Ultrafast recovery - 1200 V diode Main product characteristics A K I 30 A F(AV) V 1200 V RRM T 175 C j A V (typ) 1.30 V F K t (typ) 57 ns rr DO-247 STTH3012W Features and benefits Ultrafast, soft recovery Very low conduction and switching losses A High frequency and/or high pulsed current operation K TO-220AC High reverse voltage capability STTH3012D High junction temperature Description The high quality design of this diode has Order codes produced a device with low leakage current, regularly reproducible characteristics and intrinsic Part Number Marking ruggedness. These characteristics make it ideal for heavy duty applications that demand long term STTH3012D STTH3012D reliability. STTH3012W STTH3012W Such demanding applications include industrial power supplies, motor control, and similar mission-critical systems that require rectification and freewheeling. These diodes also fit into auxiliary functions such as snubber, bootstrap, and demagnetization applications. The improved performance in low leakage current, and therefore thermal runaway guard band, is an immediate competitive advantage for this device. March 2006 Rev 1 1/9 www.st.com 9Characteristics STTH3012 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 1200 V RRM I RMS forward current 50 A F(RMS) I Average forward current, = 0.5 T = 105 C 30 A F(AV) c I Repetitive peak forward current t = 5 s, F = 5 kHz square 300 A FRM p I Surge non repetitive forward current t = 10 ms Sinusoidal 210 A FSM p T Storage temperature range -65 to + 175 C stg T Maximum operating junction temperature 175 C j Table 2. Thermal parameters Symbol Parameter Value Unit R Junction to case 0.95 C/W th(j-c) Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ Max. Unit T = 25 C 20 j (1) I Reverse leakage current V = V A R R RRM T = 125 C 15 150 j T = 25 C 2.1 j T = 125 C I = 25 A 1.25 1.9 j F T = 150 C 1.20 1.8 j (2) V Forward voltage drop V F T = 25 C 2.25 j T = 125 C I = 30 A 1.35 2.05 j F T = 150 C 1.30 1.95 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 = 1.60 x I + 0.012 I F(AV) F (RMS) 2/9