LM2907 TACH CIRCUIT .. NEED HELP BADLY

Discussion in 'General Electronics Chat' started by ivan_the_red, Dec 16, 2008.

  1. ivan_the_red

    Thread Starter Member

    Dec 7, 2008
    11
    0
    I have been trying for some time now to create a circuit that will read a pulse frequency from my distributor and convert it to a voltage that will change with the engines r.p.m. (like a tachometer). I have done all of the calculations according to the chip makers data sheet but still cant figure out why this circuit wont work correctly. I'm a newbie to electronics and still have alot to learn, but I really need some help at this point. I have attached a schematic along with the calculations for someone to pick apart. Any help surely would make my day...
    Thanks
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You probably fried the IC.

    That's easy to do in automotive environments. You need to ensure that the IC gets clean, regulated voltage - with no spikes. A single spike will kill the IC.

    You just show 12V in. I'll tell you what though, it's a terrible hash of noise that's on that input. If you try to run it from "car voltage" (the vehicles' electrical system) without regulation and protection, you're in for a world of hurt (the chip will get fried, and your circuit won't work.)

    Just FYI.

    Why don't you give some more detail about how you're supplying voltage to this IC?
     
  3. ivan_the_red

    Thread Starter Member

    Dec 7, 2008
    11
    0
    The IC is powered directly from the "+" and "-" terminal of the car's battery without any additional circuitry. The frequency input is tapped from a lead that exists on the distributor. Does this circuit and its calculations look o.k. ? , if they do then how can I protect the circuit from the input spikes from the battery, can you help me out with a schematic, because I probably dont know enough to construct one just by an explaination? or do you suggest using a seperate regulated supply from another battery (ie 9v)?.

    B.T.W. Thanks for replying, I really do appreciate it.
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Use a LM7805 voltage regulator to regulate the battery's voltage down to 9V. Also, use capacitor filters on the input/output of the voltage regulator and a transient suppressor on the input of the voltage regulator.
     
  5. ivan_the_red

    Thread Starter Member

    Dec 7, 2008
    11
    0
    Thanks Mik3,

    Did you happen to look at the circuit? Nobody has been able to tell me if it looks correct yet.
     
  6. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    You have pin 10 connected to ground and pin 11 connected to the output. They are connected backwards.
    Pin 10 should connect to the output and pin 11 should connect to ground.

    You should have used the LM2917N that has a zener diode voltage regulator but you can use your LM2907N with an external zener diode and current-limiting resistor or use a voltage regulator IC.
     
  7. ivan_the_red

    Thread Starter Member

    Dec 7, 2008
    11
    0
    Thanks for taking a look at my schematic. Do those calculations look correct? Before I recalculate for using a 9v (LM7809 Circuit) I want to make sure that everything else is being figured out correctly. Can you offer any assistance on figuring out the value for C2 (Ripple capacitor)? I dont understand what "pk-pk" means in the attached formula.
     
  8. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    I didn't look at the frequency calculation.
    Vripple is the amount of unsmoothed AC from the input that occurs at the output.

    p-p is peak-to-peak. It is the amplitude of ripple measured.
     
  9. kathapurushan

    Member

    Dec 17, 2008
    14
    0
    try using a schmitt trigger at the output with p.s viltage as +5 volts. u will get square wave. try a differentiator to conver the square to pulse . :)
     
  10. ivan_the_red

    Thread Starter Member

    Dec 7, 2008
    11
    0
    Do these calculations look correct, I'm concerned about the value for the C1 capacitor. (see attached image)
     
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