DC-DC Buck/Boost Converter

Discussion in 'The Projects Forum' started by kfrazie1, Jan 14, 2010.

  1. kfrazie1

    Thread Starter New Member

    Jan 4, 2010
    Using a DC-DC Buck/Boost Converter with an input of 8 - 30 V DC, and an output of 12 V DC and maximum of 10 Amps. The generator that is attached to the input of the converter has an efficiency of 72.5% and the converter has an effiency of 85%. The manufacturer states that the converter can output 100 watts continuous and 120 watts peak.

    so, with this DC-DC Buck/Boost converter is Vin*Iin*generator efficency=Vout*Iout*converter efficiency?

    or can you input less power than you put out?

    or do you always have to put in more power than you put out?

    or is the first equation true, basically Pin=Pout

  2. steinar96

    Active Member

    Apr 18, 2009
    Ask yourself....does "more power out, then in" violate physics ?;)
  3. Wendy


    Mar 24, 2008
    Wattage out will alway be less than wattage in, always. Wattage is another term for power.

    Voltage is a different story, as is current.

    If you use 9V to create 12v then the input current has to be a lot larger to make up the wattage difference. It 20V is used to create 12V then the current on the input is a lot less. Generally this only applies to SMPS, because they convert the voltages.

    Linear can't boost voltage (without a lot of hassle), so the current in is larger than the current out, and they get rid of the excess power as heat. This is why linear regulators, while they work extremely well, are not always the best way to go.
  4. 3ldon

    Active Member

    Jan 9, 2010
    72.5% and the converter has an effiency of 85%...

    61.6% overall

    not bad :p
  5. kfrazie1

    Thread Starter New Member

    Jan 4, 2010
    with the inverter its down to about 55%, preety bad.
  6. ericwertz


    Aug 26, 2009
    certainly violates English;)