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Cooling fan question

Discussion in 'Beginner Electronics' started by [email protected], Jul 24, 2006.

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  1. Guest

    I've got a plain 12V DC cooling fan that I took out of an old computer
    that I'd like to reuse in another place. If i got an ordinary AC
    adapter that's rated at 12V, I can wire the adapter to the fan without
    any problems correct?

    If I have the female plug that accepts the adapter's male part, it's a
    straight connection is that right?

    If let's say I don't have the female plug, I can cut off the male part
    of the adapter and solder the adapter wires directly to the fan?

    Lastly, if I want to hook up two fans together, can I use a 12V adapter
    and hook up the fans in parallel as described above?
  2. Guest

    Excellent, this is just what I thought. As for the polarity you
    mentioned, that is something that i will keep in mind of course. I know
    that on most adapters I've seen, they label which part is positive and
    which is negative.

    Thanks again!
  3. default

    default Guest

    If it is rated for 12 volts DC (not AC) it should work.

    The wall wart - supplies pulsating DC with an average RMS value of 12
    volts when under load - your fan may not like the pulsating DC or may
    try to synchronize to it - speed will fluctuate - in that case you'd
    need to put a capacitor in there to smooth it out. Wall warts, as a
    rule, don't contain filter caps. A few do.
    DC fans are polarity sensitive and have to have the correct polarity
    to work.
    Should work. You may notice speed fluctuations as result of one fan
    trying to sync to the other - usually only with two identical fans -
    and if the slight noise doesn't bother you it won't hurt anything.

    Fan speed and noise can be controlled, over some range, by lowering
    the voltage too.
  4. default

    default Guest

    Well got one here and measured DC out with a scope and has 18.6 VDC
    zero to peak - unloaded - rated 12VDC 200 ma.Damn I must be the exception then - three out of three had 100% ripple
    - but I have seen caps in some of the ones I took apart.
  5. default

    default Guest

    Fuse on my ammeter is dead but with a 75 ohm 5 watt resistor I see
    16.4 zero to peak. May be a poor choice, I think this particular wall
    wart was to float charge a battery.
    No telling where some these came from, just acquired them in my junk
    box (room). Nothing on the label suggests a product manufacturer.
  6. Jamie

    Jamie Guest

    My own experiments with walwarts concur with yours, they seem to have very
    strange voltages. (particularly new ones w/out a load) I've suspected it
    was the newer switched minimal-transformer transformers. (I hate those..)

    For whatever reasons, (suspect each one is different) it's always a good idea
    to make sure there is a small load on them if you want the voltage right.
    I've seen it too, (the ripple) I usually put a capacitor on them when I power
    something sensitive with them, though, I doubt it matters much with a motor.
    (I have a walwart that makes a mess of a radio if I don't)

  7. Ernie Werbel

    Ernie Werbel Guest

    I've got a plain 12V DC cooling fan that I took out of an old computer

    AC != DC. If the fan runs on 12 volts DC then you can't connect AC to it or
    you will destroy it. You will have to feed it through a rectifier first.
    If the 12VAC is actually an RMS value, the real value (RMS * SQRT(2)) is
    about 17V peak. So you will need to drop the extra 5V somewhere (a 5.1
    Zener could do the job). But I would say if you can find an adapter that
    outputs 12 volts DC, that would be the best way to go.

    It doesn't matter how you connect it physically, as long as the voltage is

    Please see above.
    Parallel can work so long as the voltage is the same. Remember DC != AC
    (not equal). Just make sure that the adapter can put out enough current
    (amperage) to drive all of the fans. You add up the currents of all the
    fans in parallel. Your power adapter has to be able to put out at least
    this much current.
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