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What's the best way to drop a voltage, a bit?

Discussion in 'Electronic Basics' started by Jeffrey C. Dege, Apr 9, 2004.

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  1. What's the best way to drop a voltage, a bit?

    Or do I need to?


    I'm playing with a small robot, made from FischerTechnik components,
    that I'm driving with a Basic Stamp, through a Pololu Micro Dual Serial
    Motor Controller.

    Both Stamp and Motor controller are powered by an 8-pack of AA NiMHD
    batteries, providing c. 10.5 V.

    This is well within the capability of the Basic Stamp's onboard voltage
    regulator, and the Stamp's regulated 5V output is within the 3.0-5.5V
    capacity of the Motor Controller's logic supply.

    The question is regarding the Motor Controller's motor supply. This is
    spec'ed at 1.8-9.0V, with currents of up to 1 A per motor.

    I'm using FischerTechnik's S-motors, which run at 9V, and have a .7A
    stall current, so we're fine, there.

    The question is whether I should reduce the 10.5V for input to the Motor
    Controller, and if so, how?

    I see four alternatives:

    1. Let it run at 10.5V.

    2. Stick in a couple of resistors to create a voltage-divider.

    3. Use a voltage regulator.

    4. How about a pair of diodes?


    The problem is I don't know why which would be preferred over the others.

    Any suggestions?

    --
    Only justice, and not safety, is consistent with liberty, because safety
    can be secured only by prior restraint and punishment of the innocent,
    while justice begins with liberty and the concomitant presumption of
    innocence and imposes punishment only after the fact.
    - Jeffrey Snyder
     
  2. CFoley1064

    CFoley1064 Guest

    Subject: What's the best way to drop a voltage, a bit?
    1) Only the manufacturer knows for sure. You might want to contact them. But
    it does sound like 9V is the upper limit.

    2) A voltage divider will only work if the load current is small in relation
    to the voltage divider -- in other words, just pour current down the drain for
    regulation. Probably not a good idea. Using a series resistor is a somewhat
    better idea, and will not waste current. Let's say your peak current is 0.7A,
    and you want to drop 1.4V. Using Ohm's Law, that would mean you would pick a 2
    ohm resistor, at least 2 watts -- I^2*R = (0.7A)^2 * 2 ohms = 1 Watt -- pick 2
    Watts for margin. The thing is, the supply voltage will vary linearly as
    current varies, between 9V (0.7A) and 10.5V (0A). This is a better idea, but
    probably not where you want to go if you want to stay below 9V for any load.

    3) All voltage regulators have a specification called "dropout voltage", which
    is the required minimum difference between the input and output voltage
    necessary for the regulator to work as advertised. Standard regulators have a
    dropout voltage spec between 1.5V and 3.5V., which means you'd probably need
    what's called a LDO/low dropout regulator if you wanted 9.0V. If you were
    willing to go down to 6V or so, you might want to use an LM317 adjustable
    voltage regulator. If you had very tight specs for your load voltage, this
    would be the right way to go if you had to use the 10.5VDC source. However, it
    sounds like a little much, considering the wide specification for the
    controller. Also, all regulators waste some current to run themselves. That
    will reduce battery life.

    4) How about 3 series diodes instead of 2? Forward voltage drop on 1N400X
    diodes will be between 0.55V (for a couple of mils) to 0.9V or so (for 1A),
    depending on manufacturer. A good rule of thumb average is about 0.7V for
    reasonable currents. Since they can handle up to 1A forward current, you
    should be O.K. there. So, you can expect a voltage drop of between 1.55V
    (minimum current) to 2.7V (high current, maximum voltage drop across diodes).
    You should only get about 0.4V change in load voltage between minimum current
    and maximum current, which sure beats a series resistor. That will put you in
    the safe range for all eventualities. The KISS principle (Keep It Safe &
    Simple) applies here, too. As an added bonus, you're not wasting current for
    the regulator. All current going through the diodes will be used by the load.

    By the way, you can use any 1N400X diode -- whatever you can scrounge. The
    difference between the diodes is mostly how much reverse voltage it can hold
    before breaking down -- not a consideration here.

    Have fun, and good luck
    Chris
     
  3. Doug

    Doug Guest

    The way I would do it is either;

    Create a dummy battery with a piece of metal rod to make the power pack a 7
    cell, or Tap the regulator from some battery besides the end one.
     
  4. If the specs are 1.8-9V then 10.5V is overkill. May be fatal.
    A voltage divider is not an option. A series resistor may help but it
    requires a minimum load to keep the the voltage <=9V and lowers the voltage
    below 9V when the load increases. It has its use with known, fixed voltages
    and a known, fixed load. Otherwise it's bad practice from a technical point
    of view .
    A voltage regulator should be an option if the available and the required
    voltages differ for more then 1.5V. Some regulators will do the job but
    require at least 1V between in- and output. This will work fine with a fresh
    charged battery but will give up pretty fast when the battery voltage
    decreases. I see two other options.
    - As you spec says 1.8-9V you may go down to 8V or less. FAIK low drop
    regulators for 8V/1A are available (from Linear?) Disadvantage: Your motor
    get less then full power.
    - Use a switched mode (buck) regulator. The one that (almost) shorts input
    to output when the input voltage becomes too low.
    Looks like the most simple method. You'll have to look at the voltage drop
    at the minimum and at the maximum current. If two normal Si rectifiers let's
    say 1N4007 have not enough voltage drop at he minimum current you may add a
    third diode. It may be usefull to take a Schottky type for that third one as
    its voltage drop is (less then) half the drop of a normal SI rectifier.

    petrus
     
  5. Seth Koster

    Seth Koster Guest

    Why dont you limit it with a zener?
     
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