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Voltage regulation

Discussion in 'Electronic Basics' started by Chris W, Jan 10, 2008.

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  1. Chris W

    Chris W Guest

    Suppose you have a 12V supply and need to power a 5V device. If that
    device is a simple resistive load like say an incandescent light bulb,
    you can simply use the right size resistor to drop the voltage. If the
    device is more complex and does not draw a constant current, the
    resistor obviously won't work. Something like a 7805 would be the
    easiest solution. In both cases you are wasting a lot of power. If the
    first case over half the power is just generating heat in the resistor.
    I assume the same, or close to the same, amount of power is lost in
    the 7805 regulator.

    If you needed to light two 5V light bulbs then you could hook them in
    series and waste a lot less power by using a smaller resistor.

    Now for my question, can you do something similar by using 2 7805
    regulators hooked up in series to power 2 separate loads? Alternatively
    and more what I want, can you also have the outputs hooked up in
    parallel to power one 5V device?




    --
    Chris W
    KE5GIX

    "Protect your digital freedom and privacy, eliminate DRM,
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  2. default

    default Guest

    Sounds like you already have a grasp of the principles. In theory
    with two regulators in series (and the ground returns isolated from
    ground) each regulator would do its best to keep the voltage constant
    to its load (barring oscillations or other anomalies).

    But if one regulator/load is drawing less current than the other -
    that regulator would have lots of input voltage to play with and the
    other might not. So it would only work with very close unvarying
    loads and enough voltage headroom to keep two regulators happy -
    nothing that suggest higher efficiency.

    Best bet is to use switching regulators to supply the loads from a
    single source. National Semiconductor has a line of "simple
    switchers" (now in their 3d or 4th generation) that are very easy to
    apply and very efficient - and still available in packages you can
    hand solder to without a magnifying glass.
    --
     
  3. Jamie

    Jamie Guest

    The use of a "BUCK" regulator works well in these cases to save
    power.
    Look at the below example.

    http://www.national.com/pf/LM/LM5005.html


    http://webpages.charter.net/jamie_5"
     
  4. redbelly

    redbelly Guest

    As others have said, use of a buck switching regulator will save on
    the power waste.

    Rather than connect two 5V regulators in series, it is possible (and
    simple) to make a variable supply from any regulator. See the
    datasheet for the LM317 regulator (essentially a 1.2V regulator) for
    how to do it.

    And rather than connect two regulators in parallel, a greater current
    capacity can be realized by using a bypass transistor and a single
    regulator. You might find a schematic in a regulator datasheet, or
    perhaps through Google.

    Regards,

    Mark
     
  5. whit3rd

    whit3rd Guest

    There wouldn't be any point in using two regulators for loads like
    light bulbs; a single regulator can power both, either a current
    regulator
    with two lamps in series, or a 10V regulator with two lamps in series,
    or a 5V regulator with two lamps in parallel.

    To get around the heat-generation problem, one can ALSO use a high
    frequency switch (like, 1 kHz). The result, if one controls the
    switch
    duty cycle, is similar to voltage regulation. Because the heat in
    the lamp filament is proportional to V-squared, a 5V lamp
    can be driven by a 12V source with duty cycle 25/144.
    Or a 10V pair of lamps can be driven by 12V chopped at duty cycle
    100/144.
     
  6. ehsjr

    ehsjr Guest

    You got some good answers, but I'm not sure if
    any explained why the series idea won't work.
    You cannot hook up 2 7805 regulators in series:

    Input---7805---7805---output

    The first one will produce 5V output. That output
    is too low to allow the second 7805 to work properly.
    The 7805 needs a minimum of about 8 volts input.

    Ed
     
  7. redbelly

    redbelly Guest

    Perhaps he meant something more along these lines:

    ------
    12V --o--------| 7805 |---- 10V out
    | | |
    | ------
    | |
    | |
    | ------ |
    `--| 7805 |--'
    | |
     
  8. redbelly

    redbelly Guest

    I think this might work too, and they're in series:

    ------
    12V --| 7805 |---- 10V out
    | |
    ------
    |
    |
    | ------
    `--| 7805 |--o
    | |
    ------
    |
    |
    -----
    ---
    -


    BUT here is what is actually suggested in datasheets. Choose R1 and
    R2 so that:

    10V = 5V*( 1 + R2/R1) + Icom*R2

    ------
    12V --| 7805 |--o-- 10V out
    | | |
    ------ \
    |com /
    | \ R1
    | /
    | \
    | |
    `------o
    |
    \
    /
    \ R2
    /
    \
    |
    |
    -----
     
  9. Bob Monsen

    Bob Monsen Guest

    You can do the first thing, connecting two of them (although ehsjr's point
    is a good one, and the 'drop out' of the 7805 regulators will cause a
    problem with 12V).

    However, the second one, connecting them in parallel, does not work. The
    problem is that you need a ground reference, and the reference will be
    different for the two regulators. There is no good way to hook them up in
    parallel in the way you describe.

    On the other hand, the industry has decided that the best way to do this
    kind of thing is with a 'switched mode power supply', or SMPS. That circuit
    uses devices that store energy (inductors) to offer power output at a lower
    voltage using a larger voltage. The power efficiency can be in the 90% range
    with good switchers. There are cheap chips that do most of the heavy lifting
    for you, and some companies specialize in building those chips, like
    www.linear.com. A linear regulator from 12V to 5V can only get around 42%
    efficiency.

    Regards,
    Bob Monsen
     
  10. gearhead

    gearhead Guest

    Right. By switching an incandescent light or a heating element faster
    than its thermal time constant, you get the efficiency of a switching
    power supply. You don't need an inductor.
     
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