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sample and hold pulse source?

Discussion in 'General Electronics Discussion' started by AFex54, Jun 6, 2015.

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  1. AFex54

    AFex54

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    Apr 10, 2015
    I would like to be able to sample and hold white noise for random voltages but I need a very short pulse for this. I need a 50µs or less pulse with a general frequency range from about 50ms to 3 seconds,
    Unless a 555 can operate at these extreme Im not sure how else to achieve this,
    help is appreciated
     
  2. Laplace

    Laplace

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    Apr 4, 2010
    Have you looked at using mono-stables in tandem? One generates the long delay and triggers the other to generate the short pulse.
     
  3. AFex54

    AFex54

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    Apr 10, 2015
    That would work perfectly, thanks for the tip :)
     
  4. AFex54

    AFex54

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    Apr 10, 2015
    hey I just tried this idea there and im having a problem, the long delay pulse is retriggering the short monostable pulse as soon as it ends until the the astable goes low again which just defeats the purpose.
    im curious if you had any idea how to fix this problem?
    also im actually triggering a transistor with the astable to give the monostable a negative pulse
     
  5. Laplace

    Laplace

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    Apr 4, 2010
    Without a schematic it is difficult to evaluate the problem. Are you using an edge-triggered monostable, or still using the 555? The 555 datasheet contains the following helpful advice:

    "NOTE: In monostable operation, the trigger should be driven high before the end of timing cycle."
     
  6. AFex54

    AFex54

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    Apr 10, 2015
    still using the 555 but I dont have to.
    could you name an alternative edge triggered monostable?
     
  7. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    You can use a 555 to set the long period (50ms to 3s).
    Add a differentiator (high-pass) to generate a short pulse from the edge of the 555's output.
    You may need to add an amplifier with Schmitt-Trigger characteristic or simply a limiter with high gain to create a "good" digital on/off pulse from the differentiated signal.

    This circuit shows the principle:
    [​IMG]
     

    Attached Files:

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  8. AnalogKid

    AnalogKid

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    Jun 10, 2015

    Attached Files:

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  9. AnalogKid

    AnalogKid

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    Jun 10, 2015
    Note that with a 4093, Vcc can be anything from 3 V to 18 V. If you use a 74HC14 (or HCT, AC, ACT, etc.) then Vcc is limited to no greater than 5.5 V. Check data sheets for the exact operating range.

    Also, 50 ms to 3 s is a 60:1 adjustment range. That a lot for a single-turn pot. Consider a 10-turn trimpot Also, if you want the time period to increase linearly with rotation you'll need a logarithmic taper pot. There is a pot configuration where you put a fixed resistor in parallel with the wiper to turn a linear pot into a pseudo-logarithmic pot. This, or a true log pot, might give you better adjustability.

    ak
     
    Last edited: Jul 2, 2015
  10. Arouse1973

    Arouse1973 Adam

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  11. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    I like that idea, never tried it myself with a Log pot.
    Adam
     
  12. AFex54

    AFex54

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    Apr 10, 2015
    Hi again, Ive only gotten to this stage of my project now and I like this circuit because If I add a SPDT that can bypass the differentiator I can 'track and hold' too!
    anyway all im wondering about is making the filtered pulse more digital,
    I have 4069 hex inverters , would these be ok?
     
  13. Old Steve

    Old Steve

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    Jul 23, 2015
    A schmitt trigger inverter might be a better choice, as Harald suggested in post #7. That'll 'square up' the signal nicely. A 74HC14 for voltages up to 5V, or you could use one gate from a CD4093, with the inputs tied together, for higher voltages. A CD40106, too, if you can easily get one. Up to 20V supply with the 40106.
     
  14. AnalogKid

    AnalogKid

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    Jun 10, 2015
    My life is complete.
     
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  15. AFex54

    AFex54

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    Apr 10, 2015
    but you see I already have 4069s , 25 of them!
    from what ive read a Schmitt trigger has a threshold band rather than a centre threshold point like the standard hex inverter , I cant see how this would make any difference when digitalizing a simple 9V spike.
    could you explain why a Schmitt trigger is preferred here?
     
  16. Old Steve

    Old Steve

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    Jul 23, 2015
    Without a Schmitt trigger, the output can oscillate at the threshold with slow-changing or noisy inputs. The hysteresis of a Schmitt trigger stops that. It's ideal for the purpose. But you could try a 4069 and see how it goes.
     
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  17. AnalogKid

    AnalogKid

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    Jun 10, 2015
    An R-C differentiator can turn an edge into a spike. And it can be a pretty narrow, fast spike, fast enough not to need a Schmitt trigger to clean it up. A 50 us spike probably is fast enough. But if you want the added security of a Schmitt trigger using non-Schmitt inverters, connect two of them in series, connect a 47 K resistor from the output of the second gate to the input of the first gate, and drive the first gate through a 10 K resistor. This approximates the hysteresis range of a 40106 or 4093. Note that this circuit now is non-inverting - a positive-going edge will produce a positive pulse.

    ak
     
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  18. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    This is an incorrect interpretation of a Schmitt Trigger's operation. It will turn on when the input crosses the upper threshold from below. Then the lower thresholf becomes active. Thus a small dip in imput voltage (noise on top of Vin) will not reach the lower threshold and the output stays high.
    Conversely, once the input voltage sinks from a high input voltage to one below the lower threshold, the upper threshold becomes active. Again, any small change in input voltage, even if it crosses the lower threshold, will not change the output of the Schmitt Trigger.
    The circuits's noise immunity is greatly enhanced.
    The "band" you mention is a so called "dead band", wheer nothing happens to the output of the Schmitt Trigger. It is not a "threshold band".
     
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  19. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    You can build a Schmitt Trigger from standard inverters quite easily:
    [​IMG]
    The positive feedback via the 100k resistor in cooperation with the 10k input resistor will add a fraction of the ouput voltage to the input of the left inverter, thus effectively creating the effect of two thresholds (or a dead band). Component values can be adjusted to set the width of the dead band.
     

    Attached Files:

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