Buck converter

Discussion in 'General Electronics Chat' started by Dritech, Nov 21, 2015.

  1. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
    756
    5
    Hi all,

    I did the following buck converter simulation which should output 10V 2A. On simulation it worked fine, but will it work in prectice? are the values selected good?

    Any help would be appreciated.

    [​IMG]

    Edit: The voltage on the transistor base is 15V.
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
    3,244
    622
    Don't see how it could work. No circuitry to regulate to 10V.
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,014
    3,234
    Very little increased efficiency is obtained when going from 12V to 10V with a switching regulator.
    A linear regulator is 83% efficient for that conversion, so even using a 90% efficient switching regulator will only gain 7% efficiency or about 2W less dissipation.
    And you'd be hard pressed to get 90% efficiency using a BJT emitter follower for the switch since it has a minimum drop of about 0.7V when conducting.
     
  4. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
    756
    5
    Thanks for the reply. This will be an open loop circuit, so I will connect the output to a multimeter and vary the function generator duty cycle (at the base of the transistor) till I get the disired voltage.
     
  5. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
    756
    5
    Hi crutschow. I am going to implement this circuit just for the fun of testing a switching regulator :)
    But before I buy the components and assemble it, I wish to hear from others with excerience if this would work in real world, or it I need to change something.
     
  6. DickCappels

    Moderator

    Aug 21, 2008
    2,653
    632
    It should work with the right Q1.

    BUT it would work better and be more efficient if you use a PNP transistor so that the collector drives the inductor. If you drive the inductor with Q1's emitter there is a good chance that the free-wheeling current will flow through Q1 and the 12 volt supply instead of through D5. That will increases Q1's dissipation and reduce overall efficiency.
     
    Hypatia's Protege likes this.
  7. crutschow

    Expert

    Mar 14, 2008
    13,014
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    Using a P-MOSFET would work even better.
     
    DickCappels likes this.
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