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12V to 5V

Discussion in 'Electronic Design' started by Constant Meiring, Jun 11, 2006.

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  1. Hi there.

    I'm pretty stupid concerning electronics, but I want to play around a

    I have two 6Volt UPS batteries that can give 4A/Hour (It's written on
    the battery). When I connect these two in series I get 12V. I probably
    will still have 4A/Hour (I don't know please correct me if I'm wrong).
    So now I have 12V. How would I get that 12V to 5V so I have a 12V and a
    5V output from the batteries.

    Another thing, how does positive and negative DC work? The two UPS
    batteries when connected together and measured with a mulitmeter (+
    with + and - with -), it says 11.something volts, but when I connect it
    (- with + and - with +) I get -14.something. Why is the negative
    voltage higher than the positive. How does this work?

    Thank you in advance,
    Constant Meiring
  2. mc

    mc Guest

    I have two 6Volt UPS batteries that can give 4A/Hour (It's written on
    What will you be powering with it?
    Those results are puzzling. Connected in parallel, the two 6-volt batteries
    should give 6 volts (or slightly more, up to about 7, if they are fully
    charged). Why you are getting 11, I don't know. Connected in series they
    should (and do) give 12 volts or a little more (up to 14, wich is what you

    What kind of voltmeter are you using?

    As for basic questions, it's time to go to the library and get a basic
    electricity book.
  3. John Fields

    John Fields Guest

    It's not 4A/Hour, it's 4 Ampere-Hours, and that's called the
    capacity (C) of the battery.

    What that means is that if the battery is fully charged and you take
    4 amps out of it for an hour the battery will be discharged. Or if
    you take 2 amps out of it for two hours, or if you take 1 amp out of
    it for four hours, or any combination of amperes and hours where the
    product of the two is 4. Big "but", though, that "but" being that
    you can only get 4AH out of the battery at C/10 or C/20, depending
    on the manufacturer. C/10 would be 0.4A for 10 hours, and C/20
    would be 0.2A for 20 hours.
  4. redbelly

    redbelly Guest

    Try replacing the battery in your meter, and measure it again.

  5. redbelly

    redbelly Guest

    He has the batteries in series for both measurements. He is referring
    to the METER'S + and - terminals connecting the two batteries' + and -.
    The readings should have the same absolute value, just differing by a
    +/- sign.

  6. mc

    mc Guest

    Those results are puzzling. Connected in parallel, the two 6-volt
    AH!! Sorry.

    Then it's a digital multimeter whose DC offset is WAY out of adjustment. A
    correctly adjusted multimeter should read the same voltage if you swap +
    and -, except that you get a - in front of the number.
  7. redbelly

    redbelly Guest

    It could just be the battery is low. The readings on a perfectly good
    DVM can get rather wacko when this happens.

  8. Ok, thanks with all the meter related help. But I still don't know how
    to change that 12V to 5V... :)

    Is there a simple circuit I can use to do something like that?
  9. HKJ

    HKJ Guest

  10. redbelly

    redbelly Guest

    The circuit posted by HKJ (on 17 June 2006) is typically used for this.
    You might find the parts at Radio Shack ... if not they can be ordered
    from places like . Get 2 or 3 of the 7805's, so
    you have a spare in case one burns out. (part # 497-1441-5-ND at

  11. John Fields

    John Fields Guest

    Yes; I already posted a solution for you, that being to use a 12V to
    5V DC to DC converter connected across both batteries in series.
    That combination will give you better efficiency and longer battery
    life than if you use a linear regulator with a 5V output connected
    to 12V or, possibly, with a low-dropout regulator connected to +6V.

    If you want to use a linear regulator you'll need to let us know
    what the maximum load current will be and then we'll be able to
    advise you further.
  12. PeteS

    PeteS Guest

    The IC is a 7805 (which comes in a bewildering array of packaging
    options for different requirements such as output current and thermal
    dissipation). As it has a dropout voltage (the minimum difference
    between input and output) of typically 2V, you could not use it from
    paralleled 6V batteries. [You need a minimum input of about 7V or so
    for it to work].

    Typical datasheet here:

    (now made by Fairchild, National, OnSemi [the old Mot SPS], TI and

    If, however, you use it from 12V, you will have a large amount of
    thermal dissipation, as John noted. At very low currents that may be ok
    (although it's very inefficient, but that may be ok for you too).

    As John said - tell us the load and we'll be better equipped to advise


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