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please help...with circuit

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

  1. kizzap

    kizzap Member

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    ok the image attached is what his circuit actually is...

    [​IMG]
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    Now as you can see the 5 v and the 12 v are both going into the Pot. Because there is a greater voltage on one side, it means that the voltage at MAXIMUM is going to be 7 V (12 - 5), now the actual applied voltage will be less because of losses in the pot. So in retrospec, i am supprised that the fan is even spinning at all. I can only guess that the fan is just getting enough voltage.

    this is actually a good example of a case where Millman's, Kirchhoff's or the superposition theorems should be applied to work out the actual voltages applied to the fan.

    All i can say is that in NO way will the circuit run anywhere near 12volts into the fan. Like i said above, the maximum will be somewhere around 7 volts.

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    is the proper voltage divider circuit that the OP should be aiming to build. It will supply the correct voltage range to the fans to get a good silent to full speed range of voltages.

    Most other circuits (including the OP's circuit) will cause a greater draw from the power supply, and place the PS to have incorrect loadings placed upon it.

    Hope that clears this up.

    I personally have a slightly advanced version of the second circuit in my computer (I have switches wired in with LEDs running from 5 volts).
     
    Last edited: Nov 10, 2006
  2. dakiller

    dakiller (Oscillating & Impeding)

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    Actually kizzap, your proposed circuit is going to have the greater current flow compared to the original, and the original has 5W over the entire pot without even hooking a fan up to it (7^2/10), with yours 14.5W (12^2/10)
    turning the pot on the original circuit to one end should place 12v over the fan and all the way to the other end should have 5v on the fan, and in between a somewhat complex function depending on the fan

    You have most probably killed the pot to start with, it is rated for 3W and that has been significantly exceeded here, you'll need to get a new one and probably a different value as well and hook it up in the same manner that kizzap proposed just without connecting the ground to the bottom end of the pot
     
  3. kizzap

    kizzap Member

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    ok i can understand that there is going to most likely be a lower voltage and lower current in the first example, shown by the fan not working, but i dont agree with your formulae...

    The fans (two connected in Parallel IIRC) will draw .4 Amps regardless, voltages will change in respect to the resistances.

    I dont agree on your statement that the fan will recieve 12 V when the wiper is at the extreme of travel towards the 12 Volt supply. This is because there is still a path for the voltage to travel towards the 5 V supply. (And it will because of the voltage difference). Thus the Fan will not receive the full 12 V or 5 V, and there is no way for the voltage to reach over 7 V.

    You are correct here on removing the link between the Pot and ground. I was only demonstrating the proper circuit for a voltage divider. By removing that link between the pot and ground, all that is happening is a resistor is added in series with the fan, placing a voltage drop across the resistor, causing the fan to have a lower voltage drop upon it which in turn is causing the fan to spin slower.

    I hate the fact that one connection in an electronic circuit can always cause so much trouble....
     
  4. wabbit

    wabbit Member

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    When the pot is fully counter-clockwise, the fan is directly connected between the +5 output of the power supply and the negative return so it gets 5 volts. When the pot fully clockwise, the fan is directly connected between the +12 supply and the negative return so it gets 12 volts.
    As dakiller said, what it gets in between depends on the pot and fan resistances. The path from +12 to +5 via the pot is just another load on the 12 volt supply.
    The better way to do this is with LM317 adjustable voltage regulators which allow you to use low power pots and gives a voltage range of around 1.2 to 10.
     
    Last edited: Nov 10, 2006
  5. kizzap

    kizzap Member

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    Incorrect, there will be an amount of voltage that will want to go from the 12 V to the 5 V. this will be true in ALL CASES. This is because there is a connection between the 5 and 12 volt supplies. so in no case will there be a 5 or 12 volt supply. the only time that the fan will get the 5 or 12 volts is if the resitor is wired so that there is a 12 V in on only one of the side inputs, the fan is connected to the wiper, and the other input is left open...ie no voltage.

    However, now the size of the maximum resistance of the Pot will indicate what the lowest voltage supplied to the fan will be. Current for two fans will be around 400mA at maximum. Using V = IR where I is current in circuit (400mA or .4A) and V is the voltage voltage drop at the resistor, 7V if you want a range between 5 and 12 volts supplied to the fan, it is possible to calculate a value of only 17.5 Ohms maximum resistance needed to supply this change.

    the minumum wattage needed is equal to VI = 12 x 0.4 = 4.8 W. So it would be ideal to use a 5W Pot. I realise that this may not be an easy task to get a pot with values that could handle this though.

    LM317 voltage regulators are a whole different kettle of fish however, something that is way beyond what the OP is after no doubt.
     
  6. dakiller

    dakiller (Oscillating & Impeding)

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    You shouldn't use a voltage divider to power a low impedance load, and if you do the end to end resistance of that divider should be at least 20-100X smaller than the load it is driving otherwise the impedance of the load is going to have a substantial effect on the dividers output voltage. The 10 ohm pot as a divider nearly meets these requirements but the 3ohm rating is grossly inadequate and the idea of using a voltage divider here is going to waste significant amounts of energy for no good reason when there is much more suitable circuits

    kizzap - my calculations are perfectly correct, V = IR and P = VI, so therefor P = V^2/R, the voltage across each divider is 7v and 12v for each of them

    effekt26 - the best results you are going to obtain in building a fan controller is going to be a pulse width modulation type, these are quite complex though and not recommended for beginners and the next step down from that is a linear voltage regulator, have a read up on them here - http://www.cpemma.co.uk/reg.html i strongly suggest you go the lm317 route
     
  7. dakiller

    dakiller (Oscillating & Impeding)

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    kizzap - you keep speaking about voltage where it wants to flow from one point to another. This is not how it works, voltage is only the potential difference between 2 points, then current flows through a resistance because of this difference
     
  8. kizzap

    kizzap Member

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  9. kizzap

    kizzap Member

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    It still goes to say though that the original circuit will not work in any way with both the 12 and 5 V supplied into the circuit. simply put, the fan will never really go because there is little to no activity going on in that part of the circuit. The only way for the fan to go is to put diodes in place before the pot, which also means that fan is going to be supercharged.
     
  10. wabbit

    wabbit Member

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    Puts Pauline Hanson hat on - "Please explain".
     
  11. dakiller

    dakiller (Oscillating & Impeding)

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    The original circuit should work, from 12V to 5V with a complex relationship in between and the only reason it doesn't is that the power rating on the pot has been exceeded
    [​IMG]
     
  12. kizzap

    kizzap Member

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  13. wabbit

    wabbit Member

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    And the "supercharging diodes"?
    Anyway, current cannot flow into the 5 volt supply because there are diode rectifiers in there that will prevent it.
     
    Last edited: Nov 10, 2006
  14. dakiller

    dakiller (Oscillating & Impeding)

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    Remember, V = IR, I = V/R

    With the pot having a value of 10ohm and lets say the fan has a constant resistance of 100ohm (real life fans are a little more complex than this, but this approximation is good enough for this example) with the first setup I drew, the pot will draw (12v-5v)/10r = 700mA and the fan, 12/100 = 120mA and in the second setup the pot wont change and the fan now has 5/100 = 50mA and you'd expect that as it is running slower on 5v than on 12v
     
  15. Goth

    Goth Grumpy Member

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    OK, i tried lashing up such a circuit just to see exactly what it does do.

    A 60mm stock HSF fan was used, along with a standard 24mm carbon 5k linear pot.

    As expected, at either end the voltage across the fan approaches 12V or 5V, exactly as expected, but in between there is quite a complex function coming in to play, as we expect from the standard circuit analysis, and you see the voltage across the fan actually drop right down, to 2.5V or so, where no 12V fan will operate. As the pot is moved a bit more, the voltage climbs back up, between 2.5V and 12V.

    If you run a full algebraic formulation, what i expect you'll see is a quadratic curve, which is basically being clipped due to practical constraints of the SMPS.

    0R 5.38V
    1000R 2.09V
    2000R 2.07V
    3000R 2.11V
    5000R 12.24V

    So, it perhaps isn't the best way to build a fan controller.

    Either use a simple rheostat configuration in series with the fan, or a voltage divider, using only 12V.
     
  16. kizzap

    kizzap Member

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    HA WRONG TOO

    How do people run a fan connected between 5 and 12 volt then?

    This is exactly the point i was trying to get across.
     
    Last edited: Nov 11, 2006
  17. dakiller

    dakiller (Oscillating & Impeding)

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    The voltage will be over 12V? Where did you pull that one from?
    the current sourced from the 5V rail is going to be significantly larger than the current you are trying to sink into it from the 12v rail, therefor from kirtchovs current law, the total current on the 5V rail is still positive
    You didn't do a very good job at expressing it at all, and if you read back, I explained just about exactly what Goth just built, just that without know the proper response of the fan and the pot used, 10ohm or 10k it was a bit hard to make it more specific
     
  18. wabbit

    wabbit Member

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    How do you think the diodes are going to add voltage to the circuit?
    At best they will reduce it by around 0.7 volts each, giving a range of 4.3 to 11.3 volts.
    Anyway, Robocop's on. I'm off to watch.
     
  19. kizzap

    kizzap Member

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    well 5 + 12 = 17 volts..., less the losses in the resistance of the pot and drops in the diodes...
    I also would have thought that the 5 volt rail will actually sink the current, not be the source, as it is the lower voltage, with less force behind it.
    THIS TEXT IS WRONG

    yeah i know i am crap at explaining things via engrish. im more a physical, lets show what is happening kinda guy.
     
    Last edited: Nov 11, 2006
  20. Goth

    Goth Grumpy Member

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    I guess that the net current through the rectifier diodes has to be such that they're forward biased.

    This fan is drawing 30.2mA, connected in that fashion.

    My bench PSU (A slightly modified computer SMPS) has no other load other than a LED on the 5V rail, drawing perhaps 29mA tops. That could be wrong, i can't remember the resistor used.
     

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