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Circuit to control the pulse width of digital signals

  1. Feb 26, 2015 #1
    Hi all,

    I am looking for circuits that control the pulse widths of digital signals and their applications. Specifically, I am looking for circuits that map and a single input pulse width x to a single output pulse width y through function f(x)=y. Thoughts?
     
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  3. Feb 26, 2015 #2

    Svein

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  4. Feb 26, 2015 #3
    Aren't one-shots and monostable multivibrators used for outputing pulses of fixed widths, so that the function above would be y = f(x) = c, where c is constant. What about something that doubles every input pulse width like y = f(x) = 2x?
     
  5. Feb 26, 2015 #4

    Svein

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    Was that what you were after? I did not get that from your original question. I know how to make such a circuit, but I do not know whether it comes prepackaged.
    433px-Basic_integrating_adc.svg.png Start out with something like this. Assume Vin=-Vref and use a comparator on Vout. Let the incoming pulse flip the switch to Vin. The integrator will integrate -Vin for the time the incoming pulse is active. The switch will then flip to Vref and the integrator will integrate back to 0. At that point, clamp the integrator.
    http://www.hardwaresecrets.com/imageview.php?image=3741 [Broken]
     
    Last edited by a moderator: May 7, 2017
  6. Feb 26, 2015 #5
    Very nice! I agree that this will make f(x) = 2x! Did you just think this circuit up or is it used for some application?

    I imagine you can stack these to make f(x) = 4x, etc. I wonder how to do f(x) = x^2....
     
  7. Feb 26, 2015 #6

    Svein

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    This is a variation on the "dual slope A/D converter" design. The way I designed it, f(x) = (1+Vin/Vref). You are, of course, free to create your own versions. Look up "dual slope" on the web.
     
  8. Feb 26, 2015 #7

    Svein

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    This is straining the analog circuits, but two integrators in series create an amplitude proportional to x2
     
  9. Feb 26, 2015 #8
    Awesome, this was a great vocabulary lesson for me. Now I know what to search for in google and such. I will be on the lookout for applications of using pulse width functions for f(x) = m*x, where m is a constant slope.
     
  10. Feb 26, 2015 #9
    By cascading integrators, would the amplitude still rise and fall to create an output width that follows f(x) = x^2?
     
  11. Feb 26, 2015 #10

    Svein

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    Yes, since [itex]\int_{0}^{x}\int_{0}^{x}1dt=\frac{1}{2}x^{2} [/itex].
     
  12. Feb 26, 2015 #11
    You mentioned that this can strain the analog circuits, is that because it would be slow?
     
  13. Feb 26, 2015 #12

    Svein

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    No, because the amplitude is limited to what the analog circuitry can handle. I expect a range of 1:10 or thereabout. If you need a larger range, go digital.
     
  14. Feb 26, 2015 #13

    donpacino

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    Im requesting clarification. By controlling the pulse width of a digital signal, do you mean you want to control the pulse width or duty cycle of a PWM signal?


    If you need the amplitude of the signal to stay constant, and the pulse width to change then the integrator circuit will not work. The integrator circuit will simply find the 'average' value of the input.
     
  15. Feb 26, 2015 #14

    donpacino

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    now if you include a comparator or some other sort of voltage buffer you will be fine
     
  16. Feb 26, 2015 #15

    By controlling the pulse width, I mean that I have a single digital pulse of width x. For example, x = 15.33 ns, and I want a circuit that creates output pulses f(x) = 2x = y so that if I inject a pulse of width x, I get out a pulse of y = 30.66 ns.
     
  17. Feb 26, 2015 #16

    I do not know if this will work for output widths x^2. The amplitude A will be A^2, but if I put a comparator on the output of the rise and fall of the integrators, it will not give me a width that is proportional to x^2.
     
  18. Feb 26, 2015 #17
    For the case of a linear integrator, I can put a comparator on the output and raise and lower the threshold of the comparator to get different output width functions f(x) = m*x, where m is a constant multiplicative factor.
     
  19. Feb 26, 2015 #18

    donpacino

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    not true
     
  20. Feb 26, 2015 #19

    donpacino

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    you are not integrating the pulse width, you are multiplying it by 2. therefore 2 integrators would result in 4 times the pulse width
     
  21. Feb 26, 2015 #20

    hmmm, why? I just did it in LTspice to check.
     
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