High efficiency step down converter

Discussion in 'The Projects Forum' started by TheOtherGuy, Apr 4, 2015.

  1. TheOtherGuy

    Thread Starter New Member

    Apr 4, 2015
    Hi everyone, what I'm looking for is how to make a step down dc to dc "transformer". No, I do not want a buck converter or any switching power source, because I need a large amount of current from it ( around 50A) meaning I need current step up. What I have in mind is making an inverter, passing the ac current through a transformer and then rectifying it. Problem is, this would make it rather bulky, and I only want to use it as a last resort. Does anyone have any idea how to do this without converting the current to ac and back?

    input is a 13.3V, 10A battery

    Sorry if this matches with another topic, I'm new :/
  2. Papabravo


    Feb 24, 2006
    What you want seems to be of dubious utility because you did not specify an output voltage.
    Your input power is:
    13.3V * 10 Amperes = 133 Watts.
    Assuming a conversion process that is 90% efficient, your output power will be:
    133 Watts * 0.9 = 119.7 Watts
    Now 50 Amperes from 119.7 Watts ≈ 2.4 Volts

    Note I did not say anything about a buck converter or any type of switching power supply. It is just basic physics.
    Note also that transformers are exclusively and inherently AC devices. It makes little to no sense to speak of a DC to DC "transformer".

    There is an old joke about retirement.
    Q: What is the best way to end up in Palm Beach with a million dollars?
    A: Bring two.

    It is kinda the same way with power.
    In the early days of fooling around with TTL, a 5V 50A power supply was not uncommon, and it took a big piece of heavy iron to make it happen.
  3. Dodgydave

    Distinguished Member

    Jun 22, 2012
    what output voltage
    input wattage =133W ( 10A x 13.3V)

    so output voltage = 133W/50A = 2.66V
  4. Papabravo


    Feb 24, 2006
    Only if the process is 100% efficient.....chaa, and monkeys will....never mind.
  5. Roderick Young


    Feb 22, 2015
    Is your input source a lead-acid battery? If it's rated at 10 amp-hours, you might be able to draw 50 amps from it for a short time.

    What the others said about conservation of power is quite true, but in the event that you actually did want 50 amps of current at a low voltage, a buck converter can actually increase current. I don't know of any more efficient way to do things, short of redesigning your load to use exactly the battery voltage directly.
  6. #12


    Nov 30, 2010
    Do you understand that these are contradictory statements?
    Do you understand that a switching converter is the smallest possible way to do this?
  7. crutschow


    Mar 14, 2008
    You may not think you do, but you do want a buck switching converter.
    It does indeed act as a transformer to increase (step-up) the output current in proportion to the voltage reduction that it performs (and they can be designed to do that with high efficiency).
  8. DickCappels


    Aug 21, 2008
    You can do with a dynamotor like this and a rectifier, but a buck converter as others are trying to help you understand, would be the most efficient solution available.

  9. TheOtherGuy

    Thread Starter New Member

    Apr 4, 2015
    Thanks for clearing that up wally(crutschow).
    I was a little confused about converters, cause the one I made was very low efficiency, and my duty cycle was only 2/3, so I guess I made assumptions
    thanks for replying everyone!
    Last edited: Apr 5, 2015