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Choosing by pass and decoupling capacitor for an optocoupler

Discussion in 'Electronic Design' started by john, Oct 27, 2008.

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  1. john

    john Guest

    Hi all,

    I am trying to figure out the size or capacitance value for the
    following chip ( IL- 715 ),

    http://www.nve.com/Downloads/il71x.pdf

    I know that I can use the following formula to find the capacitor

    C = ( I x N x t ) / V

    Where,
    C = Capacitance
    I = Current needed to switch output from low to high
    N = number of outputs
    t = time required for the capacitor to charge
    V = allowable voltage drop in Vcc.


    The problem is that I can not find the current needed to switch output
    from low to high and also the drop in Vcc in the data sheet. Can any
    body give some suggestions!! Is there another practical method to
    calculate capacitance value?

    I am planning to use all four channels of the IL-715. The signals are
    30MHz frequency clock signals.


    Regards,

    John
     
  2. qrk

    qrk Guest

    Yikes! To use this formula means you need to know more than you could
    ever hope to know about your power supply lines and the part over
    temperature and unit to unit variations.

    You're trying to supply a low impedance power source to the part,
    especially when switching. The job of the decoupling capacitors is to
    handle the current spike on the power supply lines during transitions.
    You must think about the layout of the parts as trace length on your
    bypass capacitors will increase inductance which increases the
    impedance at higher frequencies.

    For this sort of thing, you can use basic rule of thumb design
    guidelines - use 10nF caps real close to the power supply pins
    (minimal loop length between the power and ground pins). One or two
    100nF caps nearby is also a good thing. Somewhere on the board, you
    should have some bulk capacitance in the 100s of uF which takes care
    of the lower frequency garbage. The bulk C is for all your circuitry.

    It's also good to have a solid ground plane and power plane closely
    spaced. The planes will form a good quality capacitor which aids in
    decoupling. At least a solid ground plane. With power and gnd planes,
    you could probably get by with just the 100nF caps close to the power
    supply pins and ditch the 10nF parts.

    After you lay out your circuit, monitor the power supply pins on your
    parts and see if the noise is acceptable (usually less than 50mVpp).
    Be sure to use proper oscilloscope probe grounding procedures for high
    speed signals or you will end up with a higher noise reading. Proper
    grounding technique: don't use the dangly ground lead on your probe,
    connect the probe ground ring to the board system ground. You can use
    a small screw driver to bridge the ground ring to system ground.
     
  3. DJ Delorie

    DJ Delorie Guest

    Why that much? The spec says 47nF low-ESR per Vdd.
     
  4. john

    john Guest

    Hi,

    I know the thumb rules etc. But I am looking for some formula or
    mathematical solution to this problem. I know 47nF valus is given in
    the data sheet. But how did they come up with that?

    I read the following note from texas instrument , Please take a look
    at it,

    http://focus.ti.com/lit/an/scba007a/scba007a.pdf

    The note has the formula which I mention in my post. I am unable to
    find the switching current value.

    Regards,
    John
     
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