Help with splitting a 12V PC fan tacho signal

Discussion in 'The Projects Forum' started by joe brick, May 18, 2013.

  1. joe brick

    Thread Starter New Member

    May 18, 2013
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    Hello,

    As I'm quite new to electronics, I would greatly appreciate a little help with a project I'm working on in relation to PC fans.

    In a nutshell, what I would like to do is to split the tacho output signal from a 12V PC fan. The purpose of this is to be able to connect one tacho output to a fan header on the PC's motherboard (for continuous reading of the fan's speed), and the other tacho output to a fan failure alarm circuit I'm currently working on.

    If I understand correctly, the tacho output of PC fans is of the "open collector" type and that therefore it requires a pull-up resistor.

    I've also been told that the said tacho signal split could be achieved with a couple of small transistors and a few resistors, but no specifics.

    Could anyone help me with info on how to construct this "tacho splitting" circuit please? (a simple diagram would be of enormous help :D)

    Thank you,

    Joe
     
    Last edited: Jun 3, 2013
  2. #12

    Expert

    Nov 30, 2010
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    There are a hundred ways to do this. First glance says try a voltage follower, make that into a frequency to voltage converter, compare that to some DC level, and let the second stage indicate when the fan stops sending tachometer pulses. I'm sure it can be done with less complication, but that requires knowing things like the resistance being used to generate the tach pulses.

    Here's an idea:
     
    Last edited: May 18, 2013
  3. joe brick

    Thread Starter New Member

    May 18, 2013
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    Hey #12,

    Thank you very much for the exceptionally quick response!

    I have to admit I've lost you there, though. You seem to be talking about the fan alarm circuit (or possibly that combined with the tacho splitting circuit?)

    However, what I'm looking for is just a simple tacho splitting circuit (I've already got the alarm circuit covered).

    Also, you mention needing to know the resistance used to generate the tacho signals, but isn't that just a question of what (external) pull-up resistor is used in the receiving circuit?

    thanks again,

    Joe
     
  4. #12

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    I think I made more than one mistake. First, the open collector transistor is inside the fan, not the mother board? Yes, knowing what value the pull up resistor is will lead to knowing how to connect to it. Apparently the pull-up resistor is on the mother board? And finally, I seem to be lacking what you mean by "splitting" the signal. Get it to drive 2 circuits? Change it from 0 to 12 volts into maybe 0 to 5 volts? Divide the frequency by 2?
     
  5. joe brick

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    May 18, 2013
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    I probably didn't explain things as well as I should have. A better term for what I'm looking for might be "signal duplicator". The idea is to take the tacho output of the fan and use it with 2 different receiving devices (motherboard header & independent fan failure alarm circuit).

    Here's a simple diagram that will hopefully make what I have in mind clearer:

    [​IMG]


    Apparently this "signal duplicator" circuit (or "splitter" as I originally referred to it) could be built with a few transistors and resistors, but I haven't got a clue as to how and this is what I need help with.

    Also, If I'm not mistaken, the open collector transistor for generating the tacho signal is indeed in the fan while the pull-up resistor is located in whatever receiving device the fan's tacho line is connected to (for exmaple, the motherboard).

    Suggestions?
     
  6. Ron H

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    Apr 14, 2005
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    You haven't given enough information.
    The circuit you need depends on the input requirements (voltage and current at logic high and logic low levels) of the two destinations.
     
  7. #12

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    We are having a communication problem. The first amplifier in the drawing I posted is a signal duplicator. (Another way to get 2 equal signals from one wire is to twist another wire on to the first one, but that could cause impedance problems as long as I don't know what size the pull-up resistor is.) The fan is designed to be plugged into the mother board. Presto, one signal. The signal duplicator I posted makes an exact copy without loading down the first signal. Presto, a second equal signal.

    Can you work from this and come up with another way to say what you need?
     
  8. Ron H

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    I don't understand the reasons for the peak detector, the pot, and the comparator.:confused:
     
  9. #12

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    One tachometer signal goes to a fan failure alarm. I described a frequency to voltage converter, then botched the drawing. Do you think this will work better?

    It could also be done with a missing pulse detector, but I'm not very good with those, especially if the fan is SUPPOSED to run at different speeds. So I just did a circuit that will fall to zero if the tachometer signal stops, compare that to some non-zero voltage, and there's a signal for, "if the fan is running".
     
    Last edited: May 19, 2013
  10. joe brick

    Thread Starter New Member

    May 18, 2013
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    Many thanks for the revised diagram and clarifications, #12! :D

    Any chance you could please fill it in with component names and suggested values (like I said, I'm a newbie to electronics :confused: if I know what parts I'm looking for, I can get them and test them on the breadboard, but from the "naked" diagram I don't know what resistors, caps, etc. are needed).


    Hey, Ron H. Thanks for the comment. I'm guessing they were aimed for #12 rather than me, right?
     
  11. Ron H

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    Yep. You had said this:
    I assumed from this that all you need is a couple of logic signals developed from the tach output. Hence my post #6:
     
  12. #12

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    Try this. All the functions of the concept circuit jammed into a single transistor.
     
  13. THE_RB

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    Feb 11, 2008
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    That's really a fan good indicator, rather than a fan failure indicator?

    You could reverse the operation by biasing the darlington on by default, and use the incoming AC pulses to keep the darlington turned off. It needs a couple more parts, a cap from base-emitter and maybe a second diode.
     
  14. joe brick

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    May 18, 2013
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    Thank you very much for the updated diagram, #12, and I very much appreciate your patience & time in this, but unfortunately I still can't make sense of your suggestion...

    The circuit I'm looking for ought to have 1 input & 2 outputs (tacho signal in, and 2 identical tacho signals out). However, in your diagram I can only identify what (I think) is the input (the line going in from the right?) , but were are the outputs?

    I realize that the reason I'm not seeing it might very well be my lack of knowledge/understanding in electronics, but like I said before, I'm new to all this.
     
  15. joe brick

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    May 18, 2013
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    Hey THE_RB. Thank you for the comment, but what you suggest seems to be way over my league. I'm trying to read & learn as much as I can about all this, but there are too many concepts here I'm not familiar with. Regardless, thanks for joining in the conversation :)
     
  16. joe brick

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    May 18, 2013
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    Hey Ron H, thanks for the clarification. To your point, I'm sorry, but I don't know what those values are, except that the circuit is based on 12VDC.
     
  17. #12

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    Here are 2 more drawings. The easy way to get an open collector output to pull down two loads and the hard way.
    If I still don't have it right, somebody else on this site is going to have to learn how to make drawings, because today, I have to cut two by fours until a guy arrives with a computer for me to repair.
     
    Last edited: May 20, 2013
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  18. Ron H

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    I was designing these while #12 was posting. Here are some more alternatives.
    If your outputs need to be open collector, leave off the pullup resistors.

    Note that the logic "0" levels on option A are ≈0.7V. This is generally low enough for most logic inputs. If you need logic "0" to be very close to 0V, use one of the other schemes.
    #12's first circuit bring up a good question: Why can't you just send the tach signal to both destinations?
    His second circuit needs a resistor in series with the base of each NPN, and a base resistor for the PNP. See my option C.
     
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  19. #12

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    Geez! I am so scatter brained when I try to do electronics and carpentry at the same time! My "good" drawing is short 2 resistors, as Ron said.

    I was doing that yesterday and came up just as unpredictable then, too.
    Must be like talking on a cell phone and driving. I just can't do two things at once and do either of them well. I'm going to quit logging to AAC in before I lose a thumb!
     
  20. joe brick

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    May 18, 2013
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    many thanks, #12! I very much appreciate you taking the taking the time draw them up in detail :)


    thank you very much, Ron H! these diagrams are absolutely awesome and the explanation is very clear & helpful :) also, many thanks for double-checking #12's suggested circuits and pointing out the missing resistors!

    I will give the suggested circuits a go and see how goes!

    to your question: as I understand it, the reason I can't just send the tach signal to both destinations is that different receiving devices (say, a motherboard or an external fan controller with rpm reader built-in) may use various voltages to generate the tach signal from the open collector circuit within the fan.

    Therefore, if both receiving devices happen to use the same voltage to generate the tach signal, then there's no problem.

    But if you want to be able to use more that one receiving device (in my case: the motherboard header & the fan failure alarm circuit), irrespective of the specific voltage each device might be using to generate the tach signal, then the signal needs to be duplicated while still leaving the output as an open collector.

    makes sense?
     
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