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Two questions on voltage regulators

Discussion in 'Electronic Basics' started by Jeff Dege, Jan 12, 2006.

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  1. Jeff Dege

    Jeff Dege Guest

    If have a circuit that needs 5V, and I have a battery that generates 9V, I
    can wire a couple of resisters into a voltage divider, and have a 5V
    supply. But the big "but" is that you're throwing away nearly half the
    energy in the battery, dissipated as heat in the resisters.

    Do voltage regulators do the same, internally? Do they limit the current
    supplied externally by throwing away the excess? Or are they more
    friendly to battery life?

    Second - batteries come with two ratings - Volts and Amp-Hours. (Or
    milliamp-hours in the usual sizes.)

    Strictly by dimensional analysis, the energy stored in the battery is the
    product of the two - Watts = Volts * Amps, so Volts * Amp-Hours =

    Suppose you have a 12V battery with a 60 mAH rating. If you plug it into
    a DC circuit with 600 Ohms of resistance, you'll get a 20 mA draw. And
    you can expect about 3 hours of battery life.

    But suppose I have a circuit that has wants 3V and a 20mA draw, and I use
    a voltage regulator to drop the voltage. It's still a 20mA draw, so
    should I expect 3 hours of battery life? Or could I expect 12?
  2. Dan Hollands

    Dan Hollands Guest

    Voltage regulators waste energy just like a resistor

    AmpHrs is based on current coming from the battery. If 20mA comes out of the
    battery it doesn't matter how much of reaches the load.

    The only exception is if the regulator is dc to dc convertor - a small
    switching power supply - This has the possibility of providing a lower
    voltage with greater efficiency however the gain may be negligable at low
    current levels.

    Dan Hollands
    1120 S Creek Dr
    Webster NY 14580
  3. Rich Grise

    Rich Grise Guest

    There's still 20 mA coming out of the 12V battery, so, yes, you get
    3 hours battery life - 3/4 of the energy is thrown away as heat.

    That's why they invented the "switching regulator" - it turns the
    supply on and off rapidly, storing the energy between pulses to feed
    the load smoothly - that _does_ increase the efficiency, but you still
    won't get 100%, but 80% to 95% are not uncommon.

    Google "switching regulator".

    Good Luck!
  4. You can, if the current through the two resistors is very much more
    than the current needed by what is attached to the divider point.
    Otherwise, not so good. Another often not so good way would be if
    your 5V device current requirements are constant and then you can just
    use a single dropping resistor.
    You'd be throwing away a lot more than nearly half, using a divider.
    Let's say your 5V device needs 10mA at 5V and that this 10mA is _very_
    stable and doesn't fluctuate. If you simply use a dropping resistor,
    you could get away with "nearly half" by using a (9v-5v)/10mA or 400
    ohm resistor. But with a divider, and you might use any quiescent
    current through it you want and calculate from there, you might pick a
    divider current of 20mA, knowing that half of it will go through your
    device. Somthing like this:

    20mA / R1
    \ 4V/20mA
    / =200
    | 5V
    | |
    | |
    \ \
    10mA / R2 / 10mA
    \ 5V/10mA \ Device
    / =500 /
    | |
    | |
    gnd gnd

    In this case, the power is 9V*20mA or 180mW. Your device needs
    5V*10mA or 50mW. As you can see, 130mW is wasted. Way more than
    half. It's only in the limit-case where R2 is infinity, that less
    than 1/2 of the power is wasted.
    They effectively have a dynamic R1 that changes its value according to
    the time-dependent device current demands. Roughly speaking, R2 is
    nearly infinite in linear regulator cases (almost.) So they do waste
    power and they would, in the case you cite, waste almost half.
    Not the excess current, as their is no excess current. But they throw
    away the unneeded voltage. And to do that, they have to dissipate
    Nope. Not unless you look at switchers, which can put some of the
    energy into magnetic or electric fields for repeated short times
    instead of tossing it away as heat.
    Energy is the watt-hours thing. Yes.
    If that's close to the draw they used in coming up with the 60mAh
    rating. Batteries do not have the same Ah-rating at all current draws
    you might propose and manufacturers pick the better figure (the peak
    or near-peak) of the curve over current loading. So if your draw is
    much different, expect that rating to be optimistic.
    With a linear regulator, yes. With a switcher, you'd probably get
    better life than with the linear regulator. As always, mileage

  5. Noway2

    Noway2 Guest

    You have some good advice here. When you search for switching
    regulators, specifically look for DC-DC converters. A quick searrch on
    google should take you to a number of companies that make devices that
    will accomplish your goal for you. Based on your input, output, and
    current requirements, I would be surprised if you can't find a device
    in the < $10 (US) range.
  6. Guest

    That's a class of devices I'd never heard of before, and they provide
    some interesting options.

    Consider the MAX619CPA - puts out a regulated 4.8-5.2V, from a 2.0-3.6V
    input. Supplies 20mA, with an input of 2.0V, or 50mA with an input
    And with a shutdown pin.

    Quite a neat little device - lets you run TTL logic off of a pair of
    primary cells, instead of those clunking 9-Volts. The next time I try
    to build something in a palm-sized box, I'll very much keep them in

  7. Try dropping the voltage across a string of diodes or LEDs.

    I knew of someone who runs a 7 volt electric drill or similar tool off a 12v
    car battery by dropping the voltage through a string of diodes.

  8. What about pulse width modulation? You will waste no battery power that way
    if the device will tolerate a square wave current supply.

  9. Jasen Betts

    Jasen Betts Guest

    eg 7805, LM317: some of the same, they don't pass a significant waste current to
    ground, but they do waste the energy in the extra 4 volts.
    You'll get three hours with a series regulator like the LM317
    a switching regulator will get you more than three (maybe 9 or 10) but they
    cost a bit more.

  10. The advantage is that this drops a relatively more precise voltage (unlike
    resistors, it drops about the same voltage regardless of current draw).

    But it does still waste the excess as heat. The only way to lower a DC
    voltage without wasting the excess is a switcher of some sort.
  11. Jeff Dege

    Jeff Dege Guest

    Batteries don't provide precise voltages. And while electric motors are
    fairly tolerant of imprecise voltages, digital circuitry isn't.
  12. Byron A Jeff

    Byron A Jeff Guest

    You will have a 5V supply if and only if you draw current at some specific and
    constant value. If the amount of current draw varies, then the voltage will
    vary also. That's one of the major reasons why resistor voltage dividers
    are not used as voltage regulators.
    Linear voltage regulators work like that. They present a varaible resistance
    and dissapate the excess energy as heat.
    Not current.
    Not linear regulators.
    Be aware that a battery does have an internal resistance. This resistance
    has an effect on the amount of power one can extract from the battery.
    No necessarily. the AH rating is at a specific current draw. Many battery
    manufacturers have a 20 hour draw. So you can expect to get 60 mAH if you
    drew 3 ma for 20 hours.

    However, when you change the current draw, the amount of energy you can
    extract is vary based in the internal resistance of the battery. So if you
    draw more power, more energy is dissapated as heat, and less is delivered to
    the load. So in your example the battery would be exhausted well before the
    theoretical 3 hours if you drew 20 mA from it.
    But at a different voltage, so the amount of energy is different.
    Nope. See above.
    Definitely not!

    If you are using a linear regulator then the other 9V is being dissapated as
    heat. You can only draw 12V from the battery. So if your circuit is using 3V
    @ 20 mA then 9V at 20 mA is being converted to heat.


    And again because the 20 mA draw is nearly 7 times more than the nominal 3 mA
    draw for the AH rating, you can expect your battery to peter out well before
    3 hours.

    So the real question is how to improve your performance. The best way is to
    use a switching regulator, which delivers upwards of 95% of the energy to the
    load instead of burning it up. The current draw will then change as 3V @ 20 mA
    is the same as 12V @ 5 mA (Note that a switching regulator does not burn excess
    energy as heat). So due to two factors (less heat dissapated and lower current
    draw against the battery) your circuit will run longer. But it still won't be
    3 hours.

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