buck converter

Discussion in 'General Electronics Chat' started by LaurenceR, Apr 19, 2014.

  1. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    I am building two separate circuit boards, one for driving 12-24 vdc high current and the second to provide 5 volts for the logic circuit and atmel controller (low current). Essentially I have 12 - 24 volts dc available but have to reduce it to 5 volts for the control board. It appears to me voltage regulators would waste a lot of current to heat dropping to voltage that much. Therefore I suspect it would be better to use a step down buck converter. I think I could use a better understanding of the efficient use of the buck converter. Any suggestions which one to use would be helpful. Need to keep it simple as this is driving a pump that works off a solar panel and has to conserve as much energy as possible.

    Thanks,
    LaurenceR
     
  2. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    A 5 V regulator might waste about 200 mW & buck reg not too far different at low current?? Do you know 5 V supply current?
     
  3. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
    2
    I believe the ICs draw on the order of 20-30 milliamps. putting a signal on an opto isolator, another 20ma. Total in the neighborhood of 100ma.
    Atmel attiny26 IC
     
  4. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
    2
    The attiny26L consumes 15 ma at 5 volts. Just looked it up.
    Thanks
     
  5. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Probably not worth the trouble of a switcher. The linear would waste about 300 mw, the switcher 60mw.
     
  6. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
    2
    Thank you for your help. I didn't realize the little difference at lower current draws, not to mention how much easier it is to set up.

    Many thanks.
     
  7. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
    2
    Have to ask this question.
    If I am dropping 24 vdc down to 5 vdc at approx. 25ma, would it work to first use a voltage divider ahead of the linear voltage regulator.
    example:
    voltage divider R1 600 ohms, R2 500 ohms, Vin=24 volts, Vout=9.6 volts with no load. This would draw approximately .0045 amps or (5.0 volt * .0045a=.0255 watts.
    Now if you subtract the 5 volts from the 9.6 volts for the regulator you have 4.6 volts * .050 amp power consumption of the circuit, this would give you another 0.23 watts.
    Adding the .23 watts and the .0255 watts gives a total of .2555 watts which is much less than if you dropped 24 volts down to 5 volts with the regulator alone.
    19 volts * .050 current draw of circuit = .95 watts.
    I am probably missing something here and if so I would appreciate it is someone could explain what.
    Thanks again.
     
  8. BobTPH

    Active Member

    Jun 5, 2013
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    The voltage divider would waste exactly the same amount of power as it would save in the regulator. Your calculation is wrong in that the power in the voltage divider would be 24V times the current, not 5V times the current. Also, the current through the regulator would also be passing through the upper resistor, using more power still.

    Bob
     
  9. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
    2
    Thank you. I can see what you mean about the voltage divider and the regulator. Six of one and half a dozen of the other. Appreciate the explanation.

    Larry
     
  10. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
    2
    Do you think an LM7805 would handle this with minimal wasting of current, or should I take another approach.
    Thanks,
    Larry
     
  11. ScottWang

    Moderator

    Aug 23, 2012
    4,850
    767
    Using a Step Down Switching voltage Regulator LM2576-5 5V/3A output, eff=75% when Vin=12V.
     
  12. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
    2
    Thank you, I'll try it.
     
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