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please help...with circuit

Discussion in 'Electronics & Electrics' started by effekt26, Nov 10, 2006.

  1. effekt26

    effekt26 Member

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    hello all,

    below is a dodgy (but accurate) diagram of the fan controller im trying to build.

    basically, when theres just straight 12 volts, it works...when the 5 volt lead is attached to the pot, all voltage across the circuit drops to zero...and nothing is shorting...

    [​IMG]

    any help would be greatly appreciated.

    Thanks, Justin

    Note, pot is drawn looking at back, as in, the bar that holds the knob is facing away...
     
  2. nux

    nux Member

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    What do you mean 'all voltages'? Where are you measuring? Have you tried turning the knob half way and measuring the voltage out from the middle of the pot with no fan connected?
     
  3. OP
    OP
    effekt26

    effekt26 Member

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    sorry, i meant all voltages that go through to the fan and to the led, and the go into the pot...

    so basically and of the yellow, red or the common from middle of pot to fan (+) drop to about .02 volts when both red and yellow cables are plugged in...regardless of the position of the pot...

    i have tried it with no fan, and same thing happens, led only lights up when red is not connected, regardless of pot position...

    Thanks, Justin
     
    Last edited: Nov 10, 2006
  4. nux

    nux Member

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    Measure the resistance between all the sets of pins on the pot.
     
  5. OP
    OP
    effekt26

    effekt26 Member

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    ok, i set my multimeter to the 2000 ohm setting. when no red wire is attached, i get a 1, but i moved the red probe off the terminals, and all 3 range between 1450 - 1600, so i think this is just because the probe has left the metal. i think..

    when the red wire is plugged in, i get around the 120 mark...again, set at the 2000 ohm level...

    Thanks, and those figures are roughly what all three get. i can give u table if u wish...

    Thanks, Justin
     
    Last edited: Nov 10, 2006
  6. nux

    nux Member

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    Measure it with nothing connected and nothing plugged in.

    The resistance from the left to right pin should be 10Kohms. The resistance from the left to middle plus the resistance from the middle to right should be 10Kohms as well.
     
  7. OP
    OP
    effekt26

    effekt26 Member

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    ok, with no power, i get 12v: 113, middle: 106, 5v: 104

    just to clarify im doing it right, the multimeter is set on the 2000 ohm level, the red probe is what i touch each three terminals with, and the black pin goes to an earth?

    Thanks, Justin
     
  8. nux

    nux Member

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    No, you need to make a complete circuit.
    Black to the left pin, red to the right pin. Measure
    Black to the left pin, red to the middle pin. Measure
    Black to the middle pin, red to the right pin. Measure

    It sounds like you may have a 100ohm pot instead.

    Can you take a picture of what you have and how you have it setup?
     
  9. OP
    OP
    effekt26

    effekt26 Member

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    Black to the left pin, red to the right pin. 11
    Black to the left pin, red to the middle pin. 11
    Black to the middle pin, red to the right pin. 06

    again set at 2000 ohms

    pics in a couple of mins, there off a phone, so a bit dodgy
     
  10. wabbit

    wabbit Member

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    Sounds more like a ten ohm pot. You're probably overloading the power supply and causing it to shut down. Hopefully nothing is damaged.
     
  11. OP
    OP
    effekt26

    effekt26 Member

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    psu is fine, because when no red is connected to other side of pot, works fine, so thats a good thing...

    written on the side of the pot is 1RH AUST AW 10R K 145

    its from dicksmiths...guy said it was 10k pot...

    hope this helps..

    Cheers, Justin
     
  12. wabbit

    wabbit Member

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    Guy was wrong. It's 10 Ohms.
    You wanted an R6925 (10K) but you got an R6903 (10 Ohms).
    I find this all the time at DSE. Wrong parts in bins/drawers.
    Best you take your meter with you when shopping to make sure you get the right part because many of the assistants have limited knowledge of parts.
     
  13. kizzap

    kizzap Member

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    um am i the only who has noticed that he has TWO power sources going into that pot...

    wont really work that well doing it that way, just cut off the 5 volt supply and it should work fine.

    EDIT: refer to my edited post in your other thread
     
    Last edited: Nov 10, 2006
  14. nux

    nux Member

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    Yes it will, the pot is acting as a voltage divider. Just instead of going between 12V and ground it is going between 12V and 5V.
     
  15. OP
    OP
    effekt26

    effekt26 Member

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    right, well, back down to dicksmiths then.

    thanks for all the help,


    Cheers, Justin
     
  16. kizzap

    kizzap Member

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    um no it wont, there is actually a 12volt difference, a 5 volt difference, and a 7 volt difference, the voltage divider circuit wont work under those circumstances...

    if you remove the 5 it should hopefully run the fan between 12 and 5 volts at most.
     
  17. kizzap

    kizzap Member

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    if you do change it go for a 25R K pot

    it is what i have in my machine, and it works perfectly fine
     
  18. Rt!

    Rt! Member

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    you're trying to run a 12v fan on a supply rail which can only reach 7V?
    why?
     
  19. Goth

    Goth Grumpy Member

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    It's being used as a potentiometer, like a voltage divider, so you can use any resistance value you want and it will function exactly the same. However, if the resistance is too low you'll have excess current flowing, possibly burning up the pot and/or tripping the PSU overcurrent protection.

    5K or 10K or something will be fine. You want a linear pot, by the way.

    Moral of the story? Jaycar or Altronics or similar are your friend.
     
    Last edited: Nov 10, 2006
  20. Goth

    Goth Grumpy Member

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    No, as drawn it will have a voltage of between 5V and 12V across the fan.
     

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