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What resistor to put on 12V to get it down to 10V?

Discussion in 'Electronics & Electrics' started by Omarko, May 3, 2014.

  1. Omarko

    Omarko Member

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    I want to reduce the voltage that is going to my pump DDC1T ( LINK ).

    Its bit too noisy so I want to slow it down. Question is, what resistor do I need to use to achieve that? Its running at 12V so 10V would probably do it.

    Hope that makes sense.

    cheers
     
  2. RobRoySyd

    RobRoySyd Member

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    I'd try a 1.0 or 1.2 ohm 5 watt resistor.
     
  3. OP
    OP
    Omarko

    Omarko Member

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  4. RobRoySyd

    RobRoySyd Member

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    That should do it. Of course just how much that'll slow down your pump I cannot say. If I was you seeing as how cheap these things are I'd pick up a 0.47 ohm and a 1.8 or 2.2 ohm 5 watter while I was there.
     
  5. OP
    OP
    Omarko

    Omarko Member

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    ok thanks, will do that and measure with volt meter before hooking up the pump.
     
  6. RobRoySyd

    RobRoySyd Member

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    The volts will not change until you hook up the pump. The current to the pump flows through the resistor, that creates a voltage drop across the resistor hence the pump gets less volts.
     
  7. rainwulf

    rainwulf Member

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    It depends ENTIRELY on the current you are drawing. Resistors dont resist voltage, they drop voltage as a function of current, IE, no current, no drop, lots of current, lots of drop.

    What this means is that you cant define a voltage drop until you know your current, and if that current changes at all, the voltage drop will change.

    Your best bet is to use a few diodes in series, as they have a fairly constant voltage drop no matter what current (within limits and sanity)

    The other option is an adjustable regulator, IE.. these things:

    http://www.lightobject.com/LM2596-p...36V-Ideal-for-solar-panel-regulator-P862.aspx


    Minimum drop is 1volt, you need at least 2, so it would be perfect. No heat, high efficiency, and output voltage will stay the same regardless of load (within sane limits)
     
  8. OP
    OP
    Omarko

    Omarko Member

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    Thats cool!

    How do i hook it up though?

    edit: if you imagine a simple molex connector which goes to the pump : http://cdn.overclock.net/c/c0/350x700px-LL-c05d3fba__49023_2.jpeg

    how do I hook up this regulator? would the yellow cable be simply cut and I connect one end on the regulator at IN point and the other side to the OUT ?

    what about something like this : http://www.ebay.com.au/itm/LM2596-D...cal_Test_Equipment&hash=item3cdd5fd3c6&_uhb=1

    would the black and yellow cable simply plug in at each (+/-) end and would that work?
     
    Last edited: May 3, 2014
  9. rainwulf

    rainwulf Member

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    Yep that would work. 12 volts in, custom voltage out.
     
  10. OP
    OP
    Omarko

    Omarko Member

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    how do I know which one is + and - ?

    yellow +
    black -

    ???
     
  11. Supplanter

    Supplanter Member

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    Yellow is 12V and black is ground, so attach the yellow wire to '+' and the black wire to '-'.
     
  12. starkers92

    starkers92 Member

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    You could also go down the route of an all in one switching regulator - linear tech. and texas instruments make a few which, only require inductor + caps and a feedback resistor.. Probably not needed if your only dropping a few volts across the linear regulator, but is always an option if you want a challenge..
     
    Last edited: May 3, 2014
  13. OP
    OP
    Omarko

    Omarko Member

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    I just noticed that it says it outputs 10W only but the pump is 18W. Do Ineed a unit which can handle more watts? Also its max 3A (with heatsink), how do I know how many amps the pump is drawing. I imagine it wouldnt be much but just to be sure.

    Also, how do I know how many Volts is the device set to when I get it ? I would not want to pump 30V into the pump when I plug it in and burn it out. Maybe I should try it with something like a fan first.


    can you pls link me up to a product? sorry Im learning as I go !
     
    Last edited: May 4, 2014
  14. RobRoySyd

    RobRoySyd Member

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    Yes, sort of . I think the 10W refers to how much power the buck regulator can dissipate not how much power it can output. As your pump draws under 2A and you're very unlikely to set the output under say 6V ((12-6) * 1.5) = 9 watts you're pretty safe.

    The pump is claimed to draw 18W at 12V, 18 div 12 = 1.5 amps

    These are buck regulators, the output voltage cannot exceed the input voltage.
    That device also has a voltmeter built in.

    If it was me I'd stick with a 5W resistor. They're very cheap and hard to damage by accident. Motors can draw more than their rated current / power at startup or when stalled. A 5W resistor can tolerate dissipating many times more than its rated power for long enough for you to realise something is wrong and turn the power off.
     
  15. OP
    OP
    Omarko

    Omarko Member

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    thank you, very helpful.

    I ordered some resistors but also the voltage regulator. Its cheap and looks like fun to play with.

    will update results once I get it and set it all up.

    many thanks again.
     
  16. 2xCPU

    2xCPU Member

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    Quite simply, you don't. Although the ones I've received all seem to be set to 12V, I'd never assume this without testing first.

    A multimeter is the single most useful tool you can own. Sure you can pay many hundreds for a lab grade meter, but a sub $10 meter will serve you well.
    Don't agonize over which model, just get one.

    2.
     
  17. OP
    OP
    Omarko

    Omarko Member

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    yeah I ordered the one with the LCD. I wish Jaycar sold them as I wanted to play with it today. Will have to wait now to get it in mail, prob mid to end of the week.
     
  18. 2xCPU

    2xCPU Member

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    Jaycar have quite a range starting from $5.
    Not brilliant perhaps, but it's a start.

    2.
    Edit: just realised you meant the reg, not a meter.
     
    Last edited: May 5, 2014
  19. Bravs

    Bravs Member

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    Would it not be easier to just add a diode in there to drop the voltage?
     
  20. RobRoySyd

    RobRoySyd Member

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    Perhaps except the pump draws 1.5A.
    Therefore 1.5 x 0.7 = 1 watt dissipation. That'll probably mean a heatsink is needed for the diode and as the OP feels he'll need to drop around 2V we'll need 2 or 3 diodes.
     

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