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555 timer chip tone voltage monitor ???

Discussion in 'Electronic Basics' started by twilightzonepinball, Apr 2, 2007.

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  1. Hello,

    My name is Tom and I am new to this group. Please forgive my
    newbieness. I am an Industrial controls tech with limited electronics
    knowledge.


    I was hoping I could get some help designing a simple tone generator
    for a unique application. I have so far built a simple tone generator
    using a 555 timer chip and some resistors and caps with a circuit I
    found online. I am using pin 5 to control it as a VCO astable.
    Everything works great, the input dc voltage is varying the tone, but
    I want the frequency to go up with voltage. Kind of like an audio
    voltage monitor. This input voltage is about 0-14 volts. The problem
    is that the frequency goes down when the voltage goes up. Just the
    opposite of what I need.


    Can anyone suggest a solution?


    Thanks in advance,


    -Tom
     
  2. I think you may be able to achieve what you want by just
    adding another resistor from your voltage signal to pin 7.
    It must not have less than twice the resistance of R1 if you
    want to have a tone at zero volts. It will provide the
    widest range if the 555 is programmed to produce a pulse,
    instead of a nearly square wave, by lowering the value of R2
    to something considerably less than R1.
     

  3. No opamp, but the resistor goes between your input signal
    and pin 7, instead of pin 5. Just leave pin 5 unconnected.
     
  4. John Fields

    John Fields Guest

    ---
    You can do it by using an inverting opamp to drive the 555 control
    voltage pin so that with zero volts into the opamp the control
    voltage input is higher than when the input to the opamp is more
    positive than 14V.

    Can you post a schematic of what you have so far?

    (Not here, since this is a text-only newsgroup, but to
    alt.binaries.schematics.electronic or to a website somewhere)
     
  5. Yes, I used the one on this website.
    http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm
    The tone generator without the transistor on the output. I am just
    using a headphone so not much power needed.

    Thanks,

    -Tom
     
  6. So no need for the inverting opamp? Just another resistor to pin 7
    from pin 5? I am using trimmer pots for all the resistors so I can
    tweak things.-Tom
     
  7. Thanks for the clarification. I will try it and get back.

    -Tom
     
  8. It works!!!!!!!!!!

    You are the man, thanks so much.

    -Tom
     
  9. My circuit is working great except I need to control the volume from
    pin 3. I have a 100uf cap in series with a headphone. I have tried
    using a 10k pot wired to ground, but I don't get the control I need.
    The volume is either too low or full blast at the top of the pot. I
    guess I am using the wrong value pot. How do I choose the correct
    value?

    Thanks,

    -Tom
     
  10. Randy Day

    Randy Day Guest

    twilightzonepinball wrote:

    [snip]
    You probably need to order a 'linear taper' pot.
    It sounds like you have an audio taper one.
     
  11. If you don't need to be able to adjust the volume all the
    way to zero, try disconnecting the ground end of the pot
    element. I don't know what ohms your headphones are, but a
    series variable resistor (pot with only two connections)
    with a total resistance from 10 to 100 times the headphone
    resistance would probably work better.
     
  12. The pot is too large in value, in comparison with the headphone.

    It's not clear how you have it wired, but to begin let's pretend the
    pot is in series with the headphone, one end and the arm being used
    and leaving the other end unconnected. That gives you a variable resistor.
    You've now created a voltage divider, with the variable resistor as one
    of the resistors, and the headphone as the other. When the variable
    resistor is the same value as the headphone, it's halving the voltage.
    Lower the variable resistor more, and the voltage divider will be
    having less and less effect, until the variable resistor is at zero
    ohms, and then there's full volume.

    But when you start raising the resistance of the variable resistor above
    the value of the headphone, it won't take long before the variable resistor
    is much larger in value than the headphone, and thus the voltage is divided
    by a very large amount.

    It's not that it's no longer working, it's that once the variable resistance
    goes above a certain level, you just won't notice any more difference as
    you increase the resistance. Because you've already got the volume below
    a useable level.

    You only see basically "loud" and "no volume" because the useable resistance
    of the pot is on a very small percentage of its travel.

    Use a much lower value pot, like the same value as the impedance of the
    headphones or not more than two or three times the impedance of the
    headphones, and then you'll see a real volume control. Because then
    over all or most of the potentiometer's travel, it will have some effect.

    Michael
     
  13. Does that pose any risk to the output of the chip using a low value
    pot?

    -Tom
     
  14. I mean with it wired with one end to ground and the wiper going to the
    headphone and the other end to pin 3. I assume that is the correct way
    to wire a volume control.

    -Tom
     
  15. "twilightzonepinball" () writes:

    I was going to explain that, but at that point I'd gotten tired.

    It's basically the same concept.

    MIchael
     
  16. The worst case is when the volume is turned all the way up,
    and the headphone and pot resistor are in parallel across
    the output. As long as the pot resistance is several times
    the headphone resistance, it does not add much load to that
    case. If the chip cannot handle the current drawn by a
    direct connection to the headphones, then no pot value will
    be safe at full volume.
     
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