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please help...with circuit

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

  1. dakiller

    dakiller (Oscillating & Impeding)

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    Doesn't work like that at all, I don't even know where to start explaining how that wont work
    For starters, diodes only conduct when they forward biased
    On a normal computer, say you are sourcing 10A to run it normally, and sinking 1A in total running fans at 7V, then the total current on the 5V rail will be 9A, as long as you source more than you sink you're ok
     
  2. kizzap

    kizzap Member

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    yeah just forget everything i said here...i fscked up bad...time to reread millman's, etc...

    what voltage are you sourcing that 10A from?

    wait...nvm...it clicked.
     
  3. dakiller

    dakiller (Oscillating & Impeding)

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    Yea, I should have said the 5V rail, I didn't re-read what i was typing well enough and was in too much of a rush
     
  4. kizzap

    kizzap Member

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    So would it be safe to say that the following circuit is the circuit that the OP will want to produce?

    [​IMG]
    Click to view full-sized image!
    Hosted by UGBox Image Store

    with a 25 ohm max pot, 3W value (i know the power rating is low but at least it is a possible value).

    going on my calculations this should give a voltage drop between 0 and 10 volts. Correct?
     
  5. nux

    nux Member

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    How can you have posted so much with so little understand of how electrical circuits work?

    As has been said, on the original circuit when the wiper is turned one way there will be 12V to the fan, when it is the other way, there will be 5V to the fan.

    There will always be a current through the pot of (12V-5V)/R amps. The problem is probably that the pot is rated too low.
    The voltage at the middle will depend on where the pot is positioned, it will vary between 5V and 12V.
     
  6. kizzap

    kizzap Member

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    http://forums.overclockers.com.au/showpost.php?p=6602168&postcount=35

    that is all
     
  7. nux

    nux Member

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    It works perfectly fine as a voltage divider, it just doesn't give a linear output.

    You said it would not function as a voltage divider and then posted a whole heap of incorrect information about how you think circuits work. I don't think you understand voltage potential and how it is referenced in circuits.

    A better way to make a fan controller would be either an adjustable voltage regulator like an LM317, or a multipole switch with diodes connected to drop multiples of 0.7 volts.
     
  8. dakiller

    dakiller (Oscillating & Impeding)

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    Close, it will be between 12v and Xv where X is going to depend on the response of the fan when it has the full resistance of the pot in series with it

    To analyse these circuits, draw them up with the pots at their upper and lower limits to see what they do, it makes it much more clearer
     
  9. kizzap

    kizzap Member

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    I never posted that it wouldn't function as a voltage divider. I in fact said it was not a voltage divider. google voltage divider and find me a website that shows a voltage divider with two positive voltages entering and both going through a different path, being grounded at another point completely. Let me tell you now, you wont find one - because that circuit is not a voltage divider.

    There is no point in the OP trying to use the circuit he wired up, as it will cause the circuit to not power the fan at points, this not really being the ideal function of the circuit.

    We also have acknowledged the fact that a LM317 circuit or even PWM would be a better option, but these would be too complex for the OP.
     
  10. nux

    nux Member

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    Yes it actually is a voltage divider. Every single link in google shows it. It doesn't matter what the voltages are, it all depends on the reference you use. You really do not understand voltages at all. Hopefully someone else can explain why better than my 'just because' answer.

    As long as the fan operates at 5V (the minimum possible voltage going to the fan) then it will always power the fan with an appropriate pot.
     
    Last edited: Nov 11, 2006
  11. kizzap

    kizzap Member

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    it isnt a voltage divider, simply because of the fact that the fan doesnt ground itself back to either the 12v or 5v at the pot.

    did you read Goth's posting at all? He tried the circuit, and it showed that at around the mid point of resistance, the wiper voltage was actually only around 2 volts. i am fairly sure this is not enough to power the fans. the only real thing that will change with the difference in pot value will be the current flowing in the circuit.
     
    Last edited: Nov 11, 2006
  12. dakiller

    dakiller (Oscillating & Impeding)

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    Actually it is, while not being your bog standard type, the only thing different is the voltage levels are a little mixed, but it will still give you a fraction of the input voltage which is all a voltage divider needs to do

    If we could ignore the impedance of the fan connected, it would output a linear voltage from 12 to 5V along the pot's travel, WRT ground
     
  13. nux

    nux Member

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    That in no way means it isn't a voltage divider. I have just made a PSPICE model of the circuit to prove it. First with the given 10 ohm pot, and sweeping from turned fully to 5V and then fully to 12V. If you increase the pot value, you get strange things happening as per the second graph.

    [​IMG]
    Click to view full-sized image!

    10 ohm pot
    [​IMG]
    Click to view full-sized image!

    1k ohm pot
    [​IMG]
    Click to view full-sized image!

    If you remove the fan from the circuit, with any value pot you will get a nice voltage divider. Just referenced at 5V-12V instead of ground:
    [​IMG]
    Click to view full-sized image!

    This models the fan as a purely resistive load which is wrong, but should be approximately good enough.
     
    Last edited: Nov 11, 2006
  14. kizzap

    kizzap Member

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    i understand the voltages being mixed, however if the fan isnt grounded to the same supplies...wait...would that mean the voltage will go up around 5V or something in this case?
     
  15. nux

    nux Member

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    The 5V and 12V are with respect to ground. You could just as well call 5V 'ground', then 12V would become 7V, ground would become -5V etc.

    I don't know what you mean by the second part. Have you done any sort of electrical study at Uni?
     
    Last edited: Nov 11, 2006
  16. kizzap

    kizzap Member

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    doing some electrical stuff atm in Uni...got exam next thursday too...looks like there may be a great chance i am going to fail it.

    i am trying to understand why the fan is only getting 2 volts, just doesnt make sence, to me it would show that the fan is getting like a -3 V going into it?
     
  17. nux

    nux Member

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    No, it means that at pin 2 of the pot (the middle) its 2 volts. The voltage drop across the fan is thus 2 volts. If you draw the pot as two resistors in series with the middle point being pin 2 then it makes more sense.
     
  18. Goth

    Goth Grumpy Member

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    Clearly you need to choose a value for the pot that's in that 'sweet spot' such that power dissipation in the pot is low, yet not so high so you're not getting this non-linear response.

    P < 500mW would let you use a standard pot such as 24mm Jaycar ones, rather than an expensive wirewound.

    I think that something like 100R might be a good bet - nuxie1, care to plug that into the SPICE model?

    P = (7^2)/100 = 490mW
     
  19. nux

    nux Member

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    Last edited: Nov 11, 2006
  20. Goth

    Goth Grumpy Member

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    Ah yes, my bad there.

    The degree non-linearity that you're seeing with 100R would be perfectly acceptable for this application, but it's a shame because an expensive wirewound pot seems like overkill.

    If you use say 200R or 250R, what's the maximum power dissipation, and what does the voltage-resistance waveform look like?
     

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