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9v pp3 battery, how to get 5v from this?

Discussion in 'Electronic Basics' started by carl, Feb 14, 2004.

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  1. carl

    carl Guest

    How can i get 5v from a 9v pp3 battery to supply my circuit that requires
  2. David Cole

    David Cole Guest

    How can i get 5v from a 9v pp3 battery to supply my circuit that requires
    You could use a component called a 7805. It is a 5V voltage regulator
    that would give you a nice 5Vs out.

    David Cole
    The farm boy from Saskatchewan.
  3. Or you could make this switching regulator which will not eat up
    batteries as quickly.

  4. Use a component called a 7805

    Is this a homework question
  5. John Fields

    John Fields Guest

  6. carl

    carl Guest

    Well in a way it is a homework question.... im doing a project, at the
    moment im doing all the theory and write ups in college and building a
    prototype of my project at home. The reason for asking how to get 5v from a
    battery is that i don't have a power supply at home and some chips that im
    using require 5v's

    Thanks for pointing out this voltage regulator as i think its will solve my
  7. Byron A Jeff

    Byron A Jeff Guest

    As another posted pointed out, it'll probably create more problems than it
    solves. There are three issues:

    1) 9V batteries have a limited amount of power and...
    2) 7805 regulate voltage by burning/wasting power and...
    3) The 7805 needs a minimum voltage level of 7.5V to regulate properly.

    The end result is that the regulator will go through the batteries like candy.
    Instead of useful power being applied to your chips, it'll just be wasted and
    the 9V size has the least power capacity of all battery types.

    Take another look at the switching regulator suggested in this thread. Instead
    of burning nearly 45% of your power (power efficiency of 55%) it has nearly
    90% efficiency. It'll operate down to a lower battery voltage. Nothing you can
    do about the power issue unless you double the battery feed either 18V or a
    double current 9V to the switching regulator, thereby doubling the potential
    run time.

    I would suggest a few things:

    1) Test with the 7805 and get things working. It's fine for development.
    2) Test the 9V/7805 combo with the expected current of your circuit and see
    how long it takes for a fresh 9V to run down to 7.5V.
    3) After you get your project working look into more efficient ways to use
    batteries such as the switching regulator.
    4) And your best bet is to ditch the 9V. Much better would be to consider
    using 2 1.5V D cells and a step up switching regulator. The D cells have
    10 to 20 times the capacity of the 9V and the step up regulator has about
    the same effiency as the step down. Check out the National Semiconductor
    Simple Switcher part number LM2575 as a possibility. Should be available
    from Digikey.

    Hope this helps.

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