STTH25M06FP STMicroelectronics

STTH25M06 Datasheet 600 V, 25 A ultrafast high voltage diode A K Features K K Ultrafast recovery, soft recovery Low power losses at high switching frequency operations A A NC Low leakage current NC DPAK High junction temperature High overcurrent capability ECOPACK2 compliant A Applications K TO220FPAC PFC Boost diode LLC clamping diode Description The STTH25M06 is an ultrafast recovery power rectifier especially suited for boost or LLC clamping circuits working at high switching frequencies in heavy duty applications such as air conditioning equipment or telecom power supplies. Designed with the latest STs ultrafast technology, this 600 V 25 A diode in DPAK and TO-220FPAC has a robust behavior against electrostatic discharge and high overcurrent capability. Product status STTH25M06 Product summary Symbol Value I F(AV) 25 A V 600 V RRM t 25 ns rr(typ.) T 175 C j(max.) V 1.6 V F(typ.) DS13169 - Rev 2 - February 2020 www.st.com For further information contact your local STMicroelectronics sales office.STTH25M06 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 Average forward current 25 A F(AV) I t = 10 ms sinusoidal Surge non repetitive forward current 170 A FSM p T Storage temperature range -65 to +175 C stg T Maximum operating junction temperature +175 C j Table 2. Thermal resistance parameter Typ. Symbol Parameter Unit value DPAK 0.5 R Junction to case C/W th(j-c) TO-220FPAC 2.45 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 - 60 j (1) V = 600 V I Reverse leakage current A R R T = 125 C - 70 800 j T = 25 C - 2.1 j I = 15 A F T = 150 C - 1.3 j (2) V Forward voltage drop V F T = 25 C - 2.5 3.4 j I = 25 A F T = 150 C - 1.6 2.0 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 = 1.04 x I + 0.0385 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 AN4058: Calculation of turn-off power losses generated by an ultrafast diode DS13169 - Rev 2 page 2/13