Half wave rectification question

Discussion in 'The Projects Forum' started by Hk606, Mar 9, 2015.

  1. Hk606

    Thread Starter New Member

    Oct 19, 2014
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    My project requires 12 volts DC 29mA. The input voltage to the project is 24 volts Ac 2.0 amps this is a must.
    As my load is so small, For simplicity purposes I'd like to use a simple half wave but I don't know what size filter cap to
    use across on the output. to control any ripple. I would be grateful for any help. Thanks
     
  2. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    Any size from 470 to 1000uF , the voltage after rectifying will be approx 16V open load.
     
    Last edited: Mar 9, 2015
  3. alfacliff

    Well-Known Member

    Dec 13, 2013
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    he input current at 2 amps is decieving, just because a power supply or wall wart is rated 12 volts at 2 amps does not mean that 2 amps is necessary, the current depends on the load;. the current rating just tells you tghe maxumum it will put out. your 24 volt ac transformer will be putting out 29 ma RMS, the load current.
     
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  4. #12

    Expert

    Nov 30, 2010
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    (Square root of 2) C Er(peak to peak) F = I
     
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  5. wayneh

    Expert

    Sep 9, 2010
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    You need to define what you mean by "control". For 100% ripple, do nothing. For 0% ripple, give up now. For something in between, there are options.
     
    ebeowulf17 and #12 like this.
  6. Hk606

    Thread Starter New Member

    Oct 19, 2014
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    Yea I just wanted to give all the info on the project. Thanks for getting back
     
  7. Hk606

    Thread Starter New Member

    Oct 19, 2014
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    Thanks for the help. If the 24 volts weren't needed for other relay control it would have been easier, not that what I'm doing is complicated for sure!
     
  8. RamaD

    Active Member

    Dec 4, 2009
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    What is the tolerance of your project requirement voltage of 12V, 29mA? The half wave rectified voltage will have a peak of around 33V.
     
  9. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here is how I would do it.

    Note the voltages at the three named nodes. Note the little bit of ripple left in the load current...

    171.gif
     
  10. wayneh

    Expert

    Sep 9, 2010
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    R1 limits the current into the cap?
     
  11. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Yes, and greatly reduces the wasted power...
     
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  12. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    I screwed up the load current. Here is revised circuit that will make 29mA in the load.

    171a.gif
     
  13. #12

    Expert

    Nov 30, 2010
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    I calculate 285 uf to keep the ripple voltage down to 10% of 12 volts, but there is no good reason to mess with Mike's design unless you have some difficulty with the amount of ripple he has presented.
     
  14. alfacliff

    Well-Known Member

    Dec 13, 2013
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    what is r5 for?
     
  15. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    It is his existing ~2A load.
     
  16. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Actually, the ripple helps keep the power dissipation in R1, R2 and D2 a bit lower.
     
  17. #12

    Expert

    Nov 30, 2010
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    I know.
     
  18. Hk606

    Thread Starter New Member

    Oct 19, 2014
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    Just for clarity purposes the circuit is to operate a miniature air solenoid SMC model: vqz 100 series. Question: Is the R3 420 ohm resistor shown the resistance
    of the load or is this resistor added across the solenoid terminals. By the way. Thanks for the efforts on this!
     
  19. #12

    Expert

    Nov 30, 2010
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    Mikes R3 is pretending to be your load. It's only there for the simulation program.
     
  20. Hk606

    Thread Starter New Member

    Oct 19, 2014
    17
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    gotcha! Thanks.
     
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