STTH60RQ06CWL Datasheet 600 V, 2 X 30 A ultrafast high voltage rectifier Features A1 High junction temperature capability K Ultrafast with soft recovery behavior A2 Low reverse current Low thermal resistance Reduced switching and conduction losses ECOPACK2 compliant component Applications Solar boost diode Output rectification A2 K PFC TO-247 LL A1 UPS Air conditioning Charging station OBC in EV-HEV Description The STTH60RQ06CWL has been developed for applications requiring a high-voltage (HV) capability such as in secondary rectification in HV LLC full bridge topology or in high voltage boost function. Product status link It is ideal for switching power supplies and industrial applications, as rectification STTH60RQ06CWL function, or even freewheeling and clamping diode. Product summary Symbol Value I F(AV) 2 X 30 A V 600 V RRM V (max.) 1.45 V F t (max.) 30 ns rr T (max.) 175 C j DS13257 - Rev 2 - March 2020 www.st.com For further information contact your local STMicroelectronics sales office.STTH60RQ06CWL Characteristics 1 Characteristics Table 1. Absolute ratings (limiting values at 25 C, unless otherwise specified) Symbol Parameter Value Unit V Repetitive peak reverse voltage 600 V RRM I Forward rms current 50 A F(RMS) Per diode 30 I Average forward current T = 103 C, = 0.5 square A F(AV) C Per device 60 I t = 10 ms sinusoidal Surge non repetitive forward current 200 A FSM p -65 to T Storage temperature range C stg +175 T Maximum operating junction temperature 175 C j Table 2. Thermal resistance parameters Symbol Parameter Max. Unit Per diode 0.9 R Junction to case C/W th(j-c) Per device 0.45 For more information, please refer to the following application note: AN5088: Rectifiers thermal management, handling and mounting recommendations Table 3. Static electrical characteristics Symbol Parameter Test conditions Min. Typ. Max. Unit T = 25 C - 40 j (1) I Reverse leakage current V = 600 V A R R T = 150 C - 80 800 j T = 25 C - 2.45 j I = 15 A F T = 150 C - 1.15 1.45 j (2) V Forward voltage drop V F T = 25 C - 2.95 j I = 30 A F T = 150 C - 1.45 1.85 j 1. Pulse test: tp = 5 ms, < 2% 2. Pulse test: tp = 380 s, < 2% To evaluate the conduction losses, use the following equation: 2 P = 1.05 x I + 0.026 x I F(AV) F (RMS) For more information, please refer to the following application notes related to the power losses: AN604: Calculation of conduction losses in a power rectifier AN4021: Calculation of reverse losses on a power diode AN5028: Calculation of turn-off power losses generated by an ultrafast diode DS13257 - Rev 2 page 2/11