led at output of integrator never goes off

Discussion in 'The Projects Forum' started by GatorCpE, Jul 20, 2010.

  1. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    I posted here several days ago asking how I could get a triangle wave out of a 555 timer. I am trying to build an LFO for a simple guitar phaser and I need a triangle wave. I decided to try and feed the square wave output from the 555 to an inverting integrator op amp. I have a 100k pot that i am using to control the frequency of the 555's output pulse. I have leds hooked up to the output of both the 555 and the integrator to monitor the outputs.

    The output of the 555 works as expected. It flashes on and completely off with frequency controlled by the pot. The integrator however, will brighten, but then never go completely off. It brightens and then returns to a dim state rather than turning completely off. What could be causing this?

    Another question I have is, how can I calculate what the integrators component values should be? Right now, I have a one microfarad capacitor and two 1 mega ohm resistors(in series). This has caused the output at the integrator to pulse and not blink like the 555. However, the output of the integrator still never completely turns off.

    Thanks
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Post a schematic of your circuit. It eliminates many questions that could take lots of going back and fourth.
     
  3. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    What's the best way to go about doing this? I have a hand drawn circuit that I can post, but it will be later in the day before I can scan it and upload it.
     
  4. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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  5. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Five time constants of the integrator must be less than the period of one-half of your timer's output if you expect the LED to extinguish.

    If you use a variable frequency, you should set the integrator based on the highest frequency.

    You should keep all your project related stuff in one post.
     
  6. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    Yea that was pretty stupid of me. Sorry about that. Anyway, are you saying that if I want the led to extinguish, then 5 time constants of the integrator must be less than half the period of the 555's square wave output? So, I could calculate what the 555's highest output frequency is then set up an equation such that 5*tau = RC < (1/2) period of highest square wave frequency. Am I understanding this correctly?

    I'm assuming if I do this right, this should correct my problem and my integrator should give me a good triangle wave? Will I also be able to control the frequency of the triangle wave with the 100k pot on the 555 side of the circuit?
     
  7. Wendy

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  8. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    I worked through some math, and if i did it right( big if ), I came up with the following:

    Half the period of the square wave oscillator:

    .693*((1000+100000)*10E-6 + (100000*10E-6))/2 = 0.6946465

    Now to find the resistor and capacitor for the integrator i did:

    5*tau = 0.6

    solving for tau yields:

    tau = 0.12

    0.12 = ( 10E-6 )*R

    R = 12000 ohms.

    I built the circuit with these values, and while it is working MUCH better than it was, the led still doesn't fade all the way out. I gets pretty dim, but it doesn't extinguish all the way.

    Any ideas?
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Assuming you're using Bill Marsden's schematic with your calculated values and a bjt 555 (not CMOS),

    1) Add a 1k resistor from pin 5 (CTRL) to Vcc. There is an internal resistive divider consisting of three 5k resistors which establishes the threshold and trigger values at 2/3 Vcc and 1/3 Vcc, respectively. Adding the 1k resistor from pin 5 to Vcc puts it in parallel with the upper 5k resistor, so the threshold voltage is increased considerably. If you're using a 9v battery for a supply, the original threshold and trigger will be at 6v and 3v, respectively. Adding the 1k resistor from pin 5 to Vcc will change them to about 8.31v and 4.15v. That part should allow your LED to turn completely OFF.

    2) Add a 1k resistor from pin 3 (OUT) to Vcc. Standard BJT 555 timers have a Darlington follower on their output, so the output won't go higher than about Vcc-1.2v even with light loading. Adding a pull-up resistor provides a current source. The 555 will have no problem sinking the current to bring it low. If it still doesn't oscillate, decrease that resistor to 470 Ohms.
     
    Last edited: Jul 21, 2010
  10. Audioguru

    New Member

    Dec 20, 2007
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    Like I said in your other thread, you have the (+) input of the opamp connected to ground where it might not work and if it did then the output of the opamp would always be low, instead of biased at half the supply voltage with two series resistors.
     
  11. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    @ SgtWookie:

    I'm not using Bill's Schematic, I'm still trying to get mine working at the moment. Are you referring to turning an led off at the output of the 555 or the integrator? I want to clear up any confusion because I feel I wasn't very clear. My 555 timer seems to be generating a perfect on/off pulse that is variable via the 100k pot. The integrator is where I'm having trouble with the led not fading all the way out. If this is what you had in mind when you made your recommendations, then I will give it a try. I just didn't want anyone to be confused.

    @ Audioguru:

    do you mean that I need to make a voltage divider to divide up the source voltage, feeding half into the non inverting terminal?

    Thanks
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    Yes of course.
    Most opamps have inputs that do not work if they are within about 3V from the opamp's negative supply voltage which is 0V (ground) in your circuit. A few opamps (LM358 and MC33171) have PNP transistors as inputs and they work when the input is at 0V in your circuit but:
    1) The square-wave from the 555 oscillator is positive at the inverting input of the opamp so that forces the opamp's output to always be as low as it can go which might be +2V dimly lighting the LED.
    2) If you bias the non-inverting input of the opamp at half the supply voltage then the square-wave input will go positive and negative compared to the non-inverting input then the output can swing properly as a triangle waveform.
    3) Many opamps are designed to use a dual polarity supply then 0V is the voltage at the inputs which is "half the total supply voltage".

    Two resistors in series must divide your single polarity supply so the non-inverting input is at "half the supply voltage".
     
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