How do I get 50+V from 12V battery supply

Discussion in 'General Electronics Chat' started by kevin0228ca, Jun 14, 2015.

  1. kevin0228ca

    Thread Starter Member

    Jun 5, 2015
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    I am developing a device which is supplied by battery so less than 12V.
    But I require a voltage of at least 50+V.
    So I think I need a charge pump?
    I found MAX1044/ICL7660
    http://datasheets.maximintegrated.com/en/ds/ICL7660-MAX1044.pdf

    But I am a bit confused.
    datasheet says Vsupply of IC is 1.5-10V, and is a voltage doubler, so I can only get max 20V?
    P.10 shows cascading MAX1044 to get more voltage, but is negative, how do I get positive?
    Also says cascading will increase impedance, what does that mean exactly, can I still get 10mA current?

    Or is there easier ways of getting 50+V from battery?

    Thank you.

    //more answered information
    50+V at most 10mA.
    I have to use only one battery because system has to be portable
     
    Last edited: Jun 14, 2015
  2. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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    50V at how much current?

    Ken
     
  3. kevin0228ca

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    Jun 5, 2015
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    at most 10mA
     
  4. shteii01

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    Feb 19, 2010
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    Four 12 volt batteries in series will give you 4*12=48 volts. Five 12 volt batteries in series will give you 5*12=60 volts. Do you detect the pattern?
     
  5. kevin0228ca

    Thread Starter Member

    Jun 5, 2015
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    Do you mean I just use more batteries?
    I have to use as little battery as I can because system has to be portable.
     
  6. shteii01

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    In project management there is this stage where requirements and constraints are finalized. It is a good thing that you have those.
     
  7. Papabravo

    Expert

    Feb 24, 2006
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    A charge pump can be used for low current applications, but the thing you need to keep in mind is that the power out of a device will always be less than, sometimes much less than, the power in. In you example we have 50V @ 10 mA for a power of 500 mW or half a watt. Assuming you conversion scheme is 80 % efficient the battery will need to supply:

    0.5 Watt / 0.8 = 0.625 Watts
    and
    0.625 Watts @ 12 v requires ≈ 52 mA

    As long as your battery can supply that current draw for however long between recharge or replacement you should be OK.
     
  8. dl324

    Distinguished Member

    Mar 30, 2015
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    From a single 12V battery, you'll need a boost converter. I like MC34063A.

    Seems like that Maxim part would require multiple cascaded doublers to get to 50V. Don't care for the way they draw "schematics".
     
    Roderick Young likes this.
  9. #12

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  10. AnalogKid

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    Aug 1, 2013
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    For how long does the power system have to deliver 10 mA? What is your maximum size limit? One possible answer is six 9-V radio batteries. If that is too large, four of the little 12 V batteries sometimes used in garage door openers would be smaller, but those puppies are expensive.

    ak
     
  11. #12

    Expert

    Nov 30, 2010
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    The little AA size 12V are called A123.
    I don't think you can still buy the 22V that were the size of 9V batteries.
     
  12. dl324

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    OP said single 12V battery.

    Here's an example using MC34063A:
    stepUp.jpg
    Change R1/R2 ratio to get 50V (R1 is 2.2K). For 10mA, inductor, diode, and output cap values aren't critical. Regulator is $0.50 in qty 1.
     
    Last edited: Jun 15, 2015
  13. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Or, If you like charge pumps....
     
  14. Bordodynov

    Active Member

    May 20, 2015
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    The converter supply voltage 12V to + 50V and -50V. Stab+-50V.png
     
  15. AnalogKid

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    Used to sell those back in my TV shop days; that and the 90 V "B" battery. Best not to bump one of those with the soft part of the forearm...

    Back on topic, if the requirement is for a 50 V, 10 mA pulse rather than a continuous output, an inductive kick into a zener might be enough. 555, MOSFET, inductor, zener.

    ak
     
  16. ian field

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    Oct 27, 2012
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    The basic Joule thief circuit that is often mentioned on this forum could form the basis of a voltage converter - but the usual blocking oscillator design is usually rigged for operation on a single 1.5V cell. With careful design that alone could get you up to 50V at such low current, just rectifying the collector flyback pulses.

    A typical 1.5V blocking oscillator might have about 15T for the collector winding, then twist out a tapping point for Vcc and carry on for another 12T - the end of that goes to the base via about 15 - 100R.

    At 12V on the Vcc tap, the biasing arrangement gets a little more complicated. The base winding has to be ground referenced (and correctly phased) and fed to the base via a DC blocking capacitor. Then of course the transistor base needs a resistor to Vcc to get it started - try about 470k and keep reducing it till the oscillator starts reliably.

    You'll probably need at least 4x as many turns on the collector winding, if the collector flyback pulses aren't big enough, you can always wind on a step up secondary.
     
  17. cornishlad

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    Jul 31, 2013
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  18. kevin0228ca

    Thread Starter Member

    Jun 5, 2015
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    Thank you for your answers.

    I am looking through answers, trying to figure everything.

    A few clarification about my system.
    My system has to be wearable by wrist, ideally close to a watch.
    Most of time system will be idle, I only need 50+V 10mA for less than one minute.
     
  19. ian field

    Distinguished Member

    Oct 27, 2012
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    The blocking oscillator is popular in mains/rechargeable shavers, even at mains voltage the transformer is sometimes no more than 1cm^3.

    For the size you're talking about, you may have to resort to using a tiny ferrite toroid - and it'll be a pig to wind enough collector turns for 12V.

    Usually you can salvage a small toroid from most CFLs - but in some makes the toroid transformer is potted!
     
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