fan connection?

Discussion in 'Electronics Resources' started by logmode, Jun 4, 2012.

  1. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Hi, can you give me a hand please? I can’t find the information I need for this case fan. If you could tell me the simplest way to wire this fan at full speed with a 15 VDC power supply only, would be greatly appreciated, or a place I can go to read about it, diagram would be great too!
    This computer case cooling fan is # DF126025BM, Dynatron, 7000 RPM, 12 VDC, .45 A, and 3 pin (red white and black).

    My guess is red low, black high, and white com.


    Thank you
    Logmode
     
    Last edited: Jun 4, 2012
  2. sbixby

    Active Member

    May 8, 2010
    57
    10
    I assemble my own PCs, so i have a wee bit of experience with these.

    I could be wrong about this particular fan, but I think a 3-wire fan just has a hall sensor to provide information back to the motherboard to indicate the fan's RPM; then software can inform you when a fan isn't spinning, or isn't spinning fast enough.

    So, these three-wire fans run at full RPM always.

    I'm going to guess that red is positive, black is ground, and white is the sensor signal. In which case you can leave the white wire unconnected.
     
  3. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Thank you sbixby, but if your guess is wrong would the ouput become trash? Also if i could just feed 12 VDC to the input and it will run fine if I was sure what the inputs are?

    thank you
     
  4. sbixby

    Active Member

    May 8, 2010
    57
    10
    As long as you're not pumping wall current through it... :) - I doubt you'll hurt it if something hooks up backwards.

    90% sure that red is +12V, black is GND. Try those two on a 12v supply and it should spin merrily.

    Look at the connector body - if the wires are Black, Red, White in that order, then white is the hall sensor. Other combos are Black, Yellow Green, etc - but the three-wire connector standard is for black to be on one end, +12V in the middle, hall sensor on the other end.

    There's a bazillion pages out there with this info, btw, just google "cooling fan 3-wire pinout" or some such.
     
  5. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    sbixby, right on!
    thanks
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,000
    3,229
    If you run the fan on 15V you will need something to reduce to voltage to 12V such as an LM317 regulator.
     
  7. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Hi,
    Ok I am interested in the LM317. Fan current is .5 A and lets drop the volts by 3V. I look for how to find the resistance values for the LM317 but only found a formula with more than one missing value. How do I find these values, just find some that will work, or what?
    Thanks
    logmode
     
  8. crutschow

    Expert

    Mar 14, 2008
    13,000
    3,229
    Here's a calculator to help you with that task.

    P.S. Don't forget the capacitors.
     
  9. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Thank you so much!
    I read: "Typically R1 is 220 ohms or 240 ohms, but it could be some other value."

    So I put in whatever works.220, 240, 6000m, whatever. Then just work it out, ok. Can you get me so info on what ADJ is looking for? For voltage reg.

    unwanted voltage goes in. wanted voltages goes out. But what happens to ADJ? It’s not voltage only. Please look at the diagram above the calculator.
    I did some calculations on the calculator, and found that ADJ is not voltage sensitive, voltage only. R1 will be 240 OHMS. If R2 is 2064 OHMS, then OUT is 12 VOLTS. But, if R2 is 4560 OHMS, then OUT is 25 VOLTS. Now if ADJ is voltage sensing only, then current through R1 and R2 is the same with both voltage outputs 12 volts and 25 volts, which means that 1.25 volts was at ADJ with both outputs 12 V and 25 V. So is ADJ controlled by current?
    Thank you,
    Logmode
     
  10. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Bought some regulators. And found out that the current stays the same at ADJ in both cases
    58 UA @ 12 V and 58 UA @ 28 v.
    I’m tiered, later I am going to look at it again, be back.
    logmode
     
  11. crutschow

    Expert

    Mar 14, 2008
    13,000
    3,229
    R1 cannot be "whatever". You should use about 240Ω. That allows the regulator to maintain regulation even with no load.

    The regulator regulates by maintaining 1.25V across R1. It's related to current since it's the current through the resistors that determines the resistor voltage drop.

    For example for 1.25V across R1 = 240Ω, the resistor current is 1.25/240 =5.208mA. The regulator will then adjust the output voltage so that the same current flows through R2. Thus 5.208mA through R2 = 2064Ω gives a voltage drop of 5.208mA x 2064Ω = 10.75V. Add that to the 1.25V drop across R1 gives an output voltage of 12.0V
     
  12. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Hi,
    Thank you for your correction of my error.

    ADJ is not a variable MA, or V input proportional to the output. But must have 1.25 volts at ADJ or the regulator will change the output till it get 1.25 V at ADJ. Also if the voltage changes at ADJ, then the output voltage will have a reverse action.

    If, R1 = 240 ohms, R2 = 150 ohms, output is 2.03 V, 2.03 V / 390 ohms = 5205.13 UA. Also if R1 = 240 ohms, R2 = 820 ohms output is 5.52 V, 5.52V / 1060 ohm = 5207.55 UA. Then, 5207.55 UA - 5205.13 UA = 2.42 UA. And then, 5.52 V - 2.03 V = 3.49 V. So then 3.49 V / 2.42 = 1.44 v output change for every 1 UA change through R1 and R2.

    Is this correct?

    I said to myself, the current load through ADJ is too small to think about, but it is greater than the current change in R1 and R2. If I am correct, there is very little current change in R1 and R2 for a voltage change in the regulator output. So how does ADJ’s current play in this?

    Should I start a new thread?
    Thank you
     
    Last edited: Jun 17, 2012
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