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A basic question about DC power conversion

Discussion in 'Electronic Basics' started by aman, Mar 7, 2005.

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  1. aman

    aman Guest

    I have a 15V voltage regulator V which supplies current to a device D.
    This device supplies current at 5V.

    Is it correct to say that for 300 mA supplied by device D at 5V, the
    device D draws 100mA at 15V.

    Does this make sense by power conservation ? (100ma * 15 = 300ma * 5)
     

  2. Assuming that there is no other power flow or
    storage other than needed for the conversion,
    those values represent the limiting case that a
    perfectly efficient converter would impose.
     
  3. Your question confuses me, so I'm assuming you meant:

    I have a 15V power supply which supplies a 5V linear regulator U1.

    Linear regulators are series elements (for the most part) and thus pass
    exactly the same current as the circuit as they draw from the supply
    (there is some leakage for control, but they try to minimize this.)

    Thus, for a linear regulator, the current in (300mA) is equal to the
    current out (300mA), and the regulator itself dissipates I*V = 300mA *
    10V = 3W.

    On the other hand, a switch mode power supply will attempt to use other
    means to minimize the power consumed by the regulator itself. For a
    perfectly efficient regulator (which does not exist) the loss in the
    regulator would be 0W. Thus, the total power drawn from the supply would
    be 300mA * 5V = 1.5W, and you would thus draw 1.5/15 = 100mA.

    However, there is always some inefficiency in the regulator, expressed
    as a percentage. Thus, a 90% efficient regulator would dissipate 10% of
    the total energy drawn from the supply, and deliver 90% to the load.
    Consequently, the power consumed would be

    Pload = 0.300 * 5
    Preg = Pload / 9 = 167mW
    Ptotal = 1.666W

    So the average current from the supply would be

    I = 1.666/15 = 111mA
    Not unless you neglect power used by the regulator.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  4. aman

    aman Guest

    Thanks Larry and Robert. Device D is not a linear regulator. It is a
    board which has 4 output ports at 5V but needs 15V input power.
     
  5. John Fields

    John Fields Guest

    ---
    Here's your situation:

    +-----------+
    +15V>------| PORT1|------------------------+->5V
    | PORT2|-----------------+->5V |
    | PORT3|----------+->5V | |
    GND>--+----| PORT4|---+->5V | | |
    | +-----------+ | | | |
    | [RL4] [RL3] [RL2] [RL1
    | | | | |
    +--------------------+------+------+------+


    You said in your first post that the current drawn by the loads on the
    5V output ports is 300mA, so the total power taken out of the board
    would be:

    P = IE = 300mA * 5V = 1.5 watts.

    Since you'd have to put at least that amount of power into the board
    to get that much out, with a 1.5 watt input requirement the input
    current requirement would be:

    P 1.5W
    I = --- = ------ = 0.1A
    E 15V

    So the answer to your first question would be a guarded "yes", subject
    to the conditions Larry Brasfield explained in his reply to your first
    post.
     
  6. PeteS

    PeteS Guest

    Along with the other asnswers, then provided you have an efficient
    converter (switch mode supply, for instance), then the answer is the 5V
    device draws V5/Vsys (where Vsys is your 15V input) , [5/15 * I(15v)] /
    converter efficiency.
    For this case, that's 100mA / (n < 1), so it's something above 100mA.

    There is literally no such thing as a 100% efficient converter. A
    typical off the shelf converter for this range will typically have an
    efficiency ranging from 80-90%. It is possible to get much higher
    efficiency for a specific system, but off the shelf converters are, by
    definition, not optimised for a particular problem, except 15V in, 5V
    out, I(o) >= some value.

    For a typical converter then, the current drawn at the input to the
    converter to feed device 'D' would probably be about 115mA (typ).

    Cheers
    PeteS
     
  7. Kitchen Man

    Kitchen Man Guest

    So what you are talking about is a product you have bought that performs
    the voltage conversion? If you can find out the efficiency of the
    device, as others have mentioned, you can figure out how much input
    current it will draw from the total of the output current, by using the
    equations given.
     
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