DC/DC Converter Auto Boost Buck 80W Power Supply CC CV

 Specification:

• Input Voltage:   DC 8 ~ 30 V ( input Change, Output not Change ) 
• Output Voltage:  DC 2 ~ 16 V ( auto Buck & Boost; Input change, output constant voltage, adjustable )
• Output Current:  0.5~6A ( adjustable, Constant current )  
• Maximum Power:  80W Natural cooling
• Efficiency:  94 % ( high output voltage, high efficiency )
• Module Properties:  Non-isolated step-down/UP module (BUCK & BOOST, CC & CV)
• Full Load temperature rise:   40 ℃
• No load current:   45 mA
• Output ripple:   20M bandwidth, 20mV
• Load Regulation:   ± 2%
• Voltage Regulation rate:  ± 0.5%
• Dynamic response speed:  5% 200uS 
• Short-circuit Protection :   YES, Output current limit protection
• Input voltage Protection :   7.7V
• Indicator :   Yes, the work of normal (Green), the work is extremely or protection ( RED )
• Input Reverse Protection:  YES
• Operating Temperature:   Industrial (-40 ℃ to +85 ℃)
• Size: 72 x 42 x 23 mm ( L*W*H    Installation location: 90mm  )

Module introduction:

• All core components are imported
• Imports chip capacitors and ceramic capacitors
• High-grade magnet core
• Effectively reduces the output ripple
• Improve the reliability of the module
• In order to improve heat dissipation, PCB board using thick copper plate, switch independent heat sink design

Application:

• Adjustable Output Car Power
• Boost Charger
• Power for your electronic equipment
• Solar battery charger for electric vehicles

Description:

• Use four pillars, fixed directly above the machine.
• The module efficiency in non-synchronous rectification in very high.
• Not short-circuit protection, if you need,please install the fuse or circuit protection.
• With no reverse polarity protection,if need, install a diode in series to protect.
• Boost module, the output voltage is always higher than the input voltage.
• Output direct parallel is not possible, they will form a large circulation among.
• When the temperature exceeds 40 degrees, lower power use, or to enhance heat dissipation.
• Potentiometer is damaged, the output voltage will suddenly increase it ? ------ NOT
• P=U*I   (Power = voltage * current )