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Convert 5V to 12V

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by jjimenez101, Aug 1, 2006.

  1. jjimenez101

    jjimenez101 Guest

    Hello all,

    What chip can I use to get an output of about 9V t0 12V from an input of 5V. I believe that these chips are called charge-pump IC's?? A part number would be great, thanks.
     
  2. _nox_

    _nox_ New Member

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    is that possible ?
     
  3. Papabravo

    Papabravo Well-Known Member

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    A charge pump can be used to double a voltage so doubling 5V would get you 10 volts, not 12 volts. They are used for low current applications like RS-232 Transceivers and LCD Bias supplies.

    A switching regulator in the boost configuration is what you need. So.


    1 Go to the Linear Technology Website
    2 Download LTSpice/SwitcherCAD III
    3 Run the program and go to the switch selector
    4 Enter your requirements in terms of input and output voltager and current
    5 Hit the go button and get an instant schematic and bill of materials

    Sorry, it only uses LT parts but you wouldn't expect them to promote other peoples parts -- would you?
     
  4. dave

    Dave New Member

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  5. dknguyen

    dknguyen Well-Known Member

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    Why don't you google step-up switching Dc-DC converters (or regulators?). There are a few solutions, and many are complex- meaning only you can choose the one you want.

    Or...here's a list of them.
    http://www.chipcatalog.com/Cat/479.htm

    Only a few in the list step-up to 12V (most step-up 1.8/3.3 to 5V). Actually, until I googled this stuff just now, I have never come by a steup-up converter that steps up that high. In the list, most of the LM____ devices found on page 1 will stepup to 12V (usually more if adjusted to do so). Just pick the current you need.

    Why don't you just add some extra cells? It's much cheaper and is more efficient with MUCH longer run times, and often the extra cells take up the same space as the converter would (although these are ICs and are much smaller than the tiny converter PCBs I have seen before so the space argument might breakdown now). Remember, that if you draw 2x as much current from a battery it will run LESS than half as long. If you draw 10x current from a battery it will run 1/100th (or a really big number) as long as it normally would. THe more current you draw from a battery, the exponentially less run time you will have. This will happen if you use a step-up converter since power can't be created from nowhere, to boost a lower voltage to high voltage, you must draw more current at lower voltage to produce less current at higher voltage.
     
    Last edited: Aug 1, 2006
  6. William At MyBlueRoom

    William At MyBlueRoom New Member

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  7. oldtimer

    oldtimer New Member

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    Build a simple oscillator using a 555 timer and a 3-1 step-up transformer, the stun gun circuits on this forum will do nicely. Rectify the output and use 78 series regulators to get 9V and 12V.
     
  8. HiTech

    HiTech Well-Known Member

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  9. audioguru

    audioguru Well-Known Member Most Helpful Member

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    It divides a voltage, it doesn't boost a voltage.
     
  10. jjimenez101

    jjimenez101 Guest

    I think I found just the one I need, the Maxim 681.
    Thanks alot people, alot of good stuff here.
     
  11. t.o.

    t.o. New Member

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    Hi all,

    There are some mini DC-DC converter devices out in the market such as NMA0515 from C&D technologies can convert 5v to +/- 15 v.
     
  12. RasCreationKing

    RasCreationKing New Member

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    Hi everyone, there is a good device that can step up voltage from as minimum as 3VDC to 37V DC. this IC can delive maximum current of 1.5Amps.ofcourse you can increase the current using transistors etc.

    the IC is MC34063 DC to DC converter. you can download the datasheet of this device online and make the best out of it. Thanks!

    Ras Creation King
     
  13. OutToLunch

    OutToLunch New Member

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    nox

    i don't think the equation in your sig (i² = j² = k² = ijk = -1) is correct.

    for i² = j² = k² = -1 to be true, i = j = k = sqrt(-1) must be true, but

    for ijk = -1 to be true, the three variables must be one of the following combos: -1,-1,-1 or 1,1,-1 or sqrt(-1),sqrt(-1),1

    don't mean to be picky - i guess i'm a little bored at work today.
     
  14. HiTech

    HiTech Well-Known Member

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    oops! my bad for not looking at it closer before posting it. Yep you're right.... I took a longer look at that circuit. The one I should have posted was on the next page of the web site.
     
  15. Papabravo

    Papabravo Well-Known Member

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    Reaching through my monitor.....

    SLAP-SLAP!
     
  16. poopeater

    poopeater New Member

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    can you post a link?
     
  17. shalbk

    shalbk New Member

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    your requirement is simple with simple solution

    use diode or any voltage doubler double the 5 volts to 10 and 10 to 20v then use LM 317 three pin voltage regulator and get the desired o/p u need by simply putting one register at the controll pin.
    thats all

    but the thing is your current requirment must be low:)
     

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