Integrator triangular wave

Discussion in 'The Projects Forum' started by thiagofdp, Nov 6, 2014.

  1. thiagofdp

    Thread Starter New Member

    Nov 6, 2014
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    Guys, I need help with this project.

    I have this circuit
    circuits.png


    This is the project
    IMG_20141106_113433.jpg

    Here's what I want from V1
    IMG_20141106_113439.jpg


    Can someone give me a hand? Thanks in advance
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    What frequency, rise time, fall time, are you trying to achieve?

    R3*C1 determines the rise time. R4*C1 determines the fall time. The sum of these determines period = 1/frequency.

    10*R2/R1 determines amplitude.

    I question why the inverting input of the second opamp is connect to +5V instead of ground?
     
    Last edited: Nov 6, 2014
  3. thiagofdp

    Thread Starter New Member

    Nov 6, 2014
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    it's on the third picture..
    rise time 500ms
    fall time 1000ms


    thanks
     
  4. MikeML

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    Oct 2, 2009
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    Reread my previous post.
     
  5. thiagofdp

    Thread Starter New Member

    Nov 6, 2014
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    IMG_20141106_135455.jpg
     
  6. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Looks like your math is good. Note that the sim and your analysis depends on the opamps being rail-to-rail output. Note V(b); see how close it must pull to the rails... Note I used repeated trials to get close to your 1500ms period.



    223.gif
     
  7. thiagofdp

    Thread Starter New Member

    Nov 6, 2014
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    Thanks a bunch!
    But it would be a nice a capacitor of lower value, because the period is too damn big
    I tried 500k, 1M and 10nF, doesn't work that well... Can you show me how you calculated R3, R4 and C1?
    Thans a lot.
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The times are a little difficult to calculate because of the 5V bias on the Schmitt stage. I just played around with R and C until the rise time ~ 1s and fall time ~0.5s. The time constant is roughly R*C = 1e6*1e-6 or 1meg and 1uF gives one second. The input bias current of the opamps puts an upper bound on R. I wouldn't go much over ~1meg. So you are stuck with C.
     
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