STTH1212 Ultrafast recovery - 1200 V diode Main product characteristics A K I 12 A F(AV) V 1200 V RRM T 175 C j A V (typ) 1.25 V F t (typ) 50 ns K rr TO-220AC STTH1212D Features and benefits Ultrafast, soft recovery K Very low conduction and switching losses High frequency and/or high pulsed current operation A High reverse voltage capability NC High junction temperature 2 D PAK STTH1212G 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 STTH1212D STTH1212D reliability. STTH1212G STTH1212G Such demanding applications include industrial STTH1212G-TR STTH1212G 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 STTH1212 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 30 A F(RMS) I Average forward current, = 0.5 T = 130 C 12 A F(AV) c I Repetitive peak forward current t = 5 s, F = 5 kHz square 160 A FRM p I Surge non repetitive forward current t = 10 ms Sinusoidal 100 A FSM p T Storage temperature range -65 to + 175 C stg T Maximum operating junction temperature 175 C j Table 2. Thermal parameter Symbol Parameter Value Unit R Junction to case 1.6 C/W th(j-c) Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ Max. Unit T = 25 C 10 j (1) I Reverse leakage current V = V A R R RRM T = 125 C 7 70 j T = 25 C 2.2 j (2) V Forward voltage drop T = 125 C I = 12 A 1.30 2.0 V F j F T = 150 C 1.25 1.9 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 = 1.5 x I + 0.033 I F(AV) F (RMS) 2/9