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Resistor vs. voltage regulator question

Discussion in 'Electronic Basics' started by robert gagnon, Jul 21, 2003.

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  1. I have this 12 volt/5000 rpm motor that I want to run from 6 volt at 450
    A 6 ohm resistor connected to the battery does just that and the motor draws
    ..750 amp. from the battery. However I made a simple adjustable voltage
    regulator with a LM317 that draws the same amperage at 450 rpm. My question
    is: will the battery lasts as long with either method or is one method
    better than the other? Also are there other more efficient ways to do that?
    I'm a newcomer in electronics so any somewhat simple answers would be
    appreciated. Thanks , BobG.
  2. Eric Immel

    Eric Immel Guest

    A much more efficient way to do this is to use a pulse width modulator.

    Your current setup simply burns off a bunch of power by dissipating it
    in either the regulator chip or the resistor. You can figure out how
    much power is being used by the formula P=IE. That is, multiply the
    current through the circuit .75A, by the voltage dropped across the
    limiting device 12V(source)-6V(motor)=6V(resistor or regulator). This
    means that the wasted power is 6V*.75A=4.5W this is true with either
    method. Dissipating this much power not only drains your battery, it
    creates a large thermal load on the limiting device, you'd need a 10W
    resistor, or a good sized heat sink on your regulator. You can calculate
    the electrical efficiency of this circuit by dividing the useful power
    by the total power. Total power = 12V*.75A = 9W. Electrical efficiency =
    4.5W/9W = 50%. How to improve this?

    A pulse width modulator (PWM) limits the average power by delivering
    full power for some time, then delivering no power for some time in a
    repating periodic fashion. In other words, the limiting device connects
    the full 12V to the motor for a while, then disconnects it completely
    for a while, then connects, then disconnects, etc. This is repeated
    quickly enough that, to the device being powered, the applpied power
    looks continuous. The ratio of time on to time off determines the
    average power delivered to the load. To see why this is more efficient
    repeat the calculations above for each of the two states of the limiting
    device. Power disipated by the device when connected (assume 1A motor
    current) = 12V(source) - 12V(motor) = 0V(limiting device). Wasted power
    = 0V*1A)=0W. Power disipated by the device when disconnected (no motor
    current) = 12V(source) - 0V(motor) = 12V(limiting device). Wasted power
    = 12V*0A=0W. If the limiting device connects the battery to the motor
    half the time, the average power to the motor will be the same as
    applying 6V. If you calculate the efficiency for this PWM, you come up
    with 100%! Yay! Free energy! Not so fast. This is a description of an
    ideal device. Such things exist only in our dreams. Here in the real
    world, the limiting device will drop some voltage when connected say
    about .2 to 1 volt depending on the type of device. Also the circuitry
    that generates the timing will use some energy. PWM drivers can provide
    very high efficiencies, but there's no such thing as a free lunch.

    For more detail, google PWM motor control. Additional keywords will
    probably be necessary.

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