Looking for simple LC Butterworth Coefficient tables and scaling rules

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.

John

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 snipped-for-privacy@anasoft.co.uk

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

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 applications. Of course there are the free filter prog's out there. regards john

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 using?

Zebedee

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

Zebedee

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 meant.

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 350nF.

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.

Zebedee

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.

John

I'll email that page and the series-generator that develops those magic numbers.

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.