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Looking for simple LC Butterworth Coefficient tables and scaling rules

Discussion in 'Electronic Basics' started by zebedont, Nov 16, 2004.

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  1. zebedont

    zebedont Guest

    I have Googled but get a gazillion (exaggerates) hits and it's all a
    bit complicated. Perhaps my search terms aren't too good.

    This is for low impedance sources so the first component is an
    inductor and the load is resistive. I've had a fiddle with LTspice and
    remembered something about root(2) and 1/root(2) so for a single LC

    L1 = root(2)*R/2piF
    C1 = 1/root(2)2piFR

    And it seems to work. The -3dB point is at the right frequency and the
    response is critically damped.

    Is that right for Butterworth?

    I suppose I don't want to go much above 4th order.

    Any help appreciated.

  2. Your using the wrong software:)

    If you use SuperSpice it will do all of this for your. You use Fantastic
    filter dialog and it will design the filter for you. You can then press
    the button and it will place the filter schematic on the page so you can
    simulate it.

    Kevin Aylward
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
  3. John Larkin

    John Larkin Guest

    No, a Butterworth rings a bit for a step input. Critically damped is
    closer to Bessel.

    It's worth getting a good filter book, like Williams+Taylor, or even
    the old classic "Simplified Modern Filter Design" by Geffe. They have
    scads of normalized filter tables, response curves, and scaling rules.

    Lancaster's Active Filter Cookbook is good to have around, too.

  4. john jardine

    john jardine Guest

    It's a second order filter set for a Q of 0.707. It's just a series resonant
    tuned circuit with the source and load resistances sized to provide
    sufficient 'damping' to knock the Q down to that (optimum) value of
    1/root(2). Easy to sort if you just want a quick filter but awkward if you
    want to go up to higher roll off rates or different response shapes.
    For particular Butterworth, Cheb, Bessel, response shapes it's quite easy
    just to set equal source and load resistances and read from a table of
    standardised values. If it's of any use, I've posted a .GIF page with some
    Butterworth tabled values, onto alt.binaries.schematics.electronics.
    (Unequal source and load resistances need an initial fudge factor
    calculating ).

    By far the easiest filters for calculation are the traditional (but
    seemingly forgotten nowadays) 'prototype K' and 'M-derived' filters. These
    use 2 simple calcs to make a 3rd order section. Sections then stacked to any
    filter order required. Not theoretically perfect but good enough for most
    Of course there are the free filter prog's out there.
  5. zebedont

    zebedont Guest

    Thanks for the software. Unfortunately it calculates values assuming
    equal source and load impedances. I have a voltage source as the input
    to the filter.

    I notice that for a second order Butterworth that if I set the source
    impedance to a low value and halve the value of the capacitor it gives
    me the response I'm looking for.

    I tried a similar thing for higher order filters but the damping
    doesn't work out properly. Is there some other trick I should be

  6. zebedont

    zebedont Guest

    Thanks, is there another source for this? I don't have access to the
    group you mention. Kevins software assumes equal source and load
    impedances as well but I am dealing with a voltage source and
    resistive load. What is the fudge factor?
    I shall go looking for one.
    Thanks again

  7. zebedont

    zebedont Guest

    If I pick F=40KHz and R=8R then I get L=45uH and C=350nF, if I use the
    (guessed)sums I've given above. In LTspice the AC filter response is
    flat to crossover with no peaking and -3dB down at 40KHz.

    If I do a transient response on it then the output overshoots and
    recovers in about one cycle. That's what I thought critically damped

    Overdamped would rise to the final value without overshoot. I think
    I'm getting confused now, maybe critically damped rises to the final
    value in the fastest possible time.

    I started out by making the impedance of the L and the C equal to the
    load resistance at the crossover frequency, 32uH and 500nF, and got a
    peaked response.

    Then I tried making them twice the load resistance, thinking that (in
    ac terms) they appear in parallel to the load. That gave 64uH and
    250nF and the response looked overdamped.

    Then I made them equal to the load and scaled them by that root(2) and
    1/root(2) and the response looked much nicer..... 11.3 ohms, 45uH and

    So I fiddled the values by root(2) and 1/root(2) and it got better.

    Kevins software gives me 45uH and 700nF but uses equal source and load
    resistances. If I make the input resistor small then the response
    peaks at crossover.

    Well, I think I am suitably lost now.

  8. John Larkin

    John Larkin Guest

    It's fine to fiddle with parts on a simulator until you get something
    you like, as long as the result is realizable with practical parts.
    But "Butterworth" and "critically damped" are formal mathematical
    definitions, and a filter can't be both. What you have, something that
    rings a little and looks flat in frequency response, is probably
    intermediate between the classic Butterworth (maximally flat in
    frequency response) and Bessel (no overshoot) forms. Nothing wrong
    with that.

    I often "design" filters, especially digital filters with shiftable
    (power-of-2) coefficients, by fiddling until I like what I see. You
    can dignify the results by call it a "transitional" filter.

  9. john jardine

    john jardine Guest

    I'll email that page and the series-generator that develops those magic
  10. The Phantom

    The Phantom Guest

    These are the (very nearly) exact values for a 2nd order
    Butterworth low-pass with zero ohms source impedance and 8 ohms load.

    The Butterworth response is less than critically damped; the poles
    are complex. Therefore it has overshoot.
    Critically damped means that the poles of the denominator merge at
    the negative real axis into a single, second-order pole. The
    oscillatory component of the time response just disappears for this
    amount of damping. This would be achieved if you keep the same L=45uH
    and C=352 nF but reduce R to 5.657 ohms. R^2 = L/(4C) is the
    condition for critical damping (the discriminant of the denominator is
    zero) for your zero ohm source impedance case.
    The critically damped response has no overshoot, so it rises to the
    final value in the shortest time without overshoot. The overdamped
    case, of course, has no overshoot, but a slower rise.
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