RC differentiator

Discussion in 'General Electronics Chat' started by kaiosama, May 2, 2013.

  1. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    I am trying to design an RC differentiator that will detect the edges of a square wave.

    The square wave is 1khz (1ms period) with 5v amplitude. I need to select the values of R and C.

    To have a good differentiator, I've read that the impedance of the capacitor must be at least 10 times larger than the resistance of the resistor at the operating frequency, and that the time constant RC must be smaller than the period of the input signal.

    Zc = -j/wC = -j/(2*Pi*f*C)
    and therefore -j/(2*Pi*f*C) > 10*R
    plugging f = 1000 hz and solving for C
    C must then be < 1/(20000*R*Pi)

    We also need RC << 1ms.

    Setting R = 100 ohms,
    C < 0.159 uF
    I choose C = 50nF.
    So RC = 5 us which is << 1ms.

    Here is what I get (no load):
    [​IMG]
    [​IMG]

    Questions:
    1. are my assumptions and constraints for a good RC differentiator corrrect? I read that on the internet.
    2. The differentiated output is from 0.3 to 0.4V, which is too small to drive a diode (0.6V forward drop). What can I do to make the differentiated signal larger? Obviously I need to play with R and C but I don't know which one and how. Id rather understand why and how rather than do it blindly.

    Thank you
     
  2. #12

    Expert

    Nov 30, 2010
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    What's the impedance of the source? Ideal in a simulator?
    Make the resistance high enough to not load the source very much, like 10x the source impedance, then increase the capacitor until you get a good response.

    The more resistance, the smaller the capacitor will be.

    I think your "rule of thumb" is backwards. That's why you're only getting little bumps on the graph.
     
    Last edited: May 2, 2013
  3. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    I have found out that by setting C to a higher value, say 200n, the output voltage becomes higher.
    This is with C= 200nF
    [​IMG]


    yes I think on LtSpice the voltage source is ideal. Is it good practice to put series resistance and/or parallel capacitance with voltage source?
     
  4. #12

    Expert

    Nov 30, 2010
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    I think that's called, "success".
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Simply make RC time constant equal to 1ms/20 = 50us
    For R = 100R C = 50us/100R = 500nF = 470nF
     
  6. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    thank you for your help.

    I read here http://www.daenotes.com/electronics/digital-electronics/rc-high-pass-filter-circuit that the impedance of the capacitor must be at least 10 times larger than the resistance of the resistor at the operating frequency.

    Usually the load must be much higher than the driving impedance to avoid loading effects, but in this case it seems backwards as you said. Maybe because the R is not considered a load and is considered part of the driving circuit?
     
    Last edited: May 2, 2013
  7. #12

    Expert

    Nov 30, 2010
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    Only if you are trying to simulate something that exists in the real world.
     
  8. #12

    Expert

    Nov 30, 2010
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    Now you know what the truth is.

    ps, your link doesn't work for me.
     
  9. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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  10. #12

    Expert

    Nov 30, 2010
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    "There is nothing false on the internet." = Joke

    For everything you can find on the internet, you can also find something to contradict it on the internet. = True
     
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