ADA/Flintlock Flanger filter question

[Tall Steve]

There's a Sallen-Key LPF on the ADA flanger that's made up of a 22k/1n5 network and a 68k/510p network (R28/C15 and R29/C16 on this schematic: http://lectric-fx.com/wp-content/uploads/2015/07/FlintlockFlangerV1.2.pdf). With 510p caps being nearly impossible to find, could I sub in 62k/560p?

Also, from a design standpoint, what's the purpose of scaling the latter network? Is it for damping? I looked through the Active Filter Cookbook, but all of the examples Lancaster gives have equal value resistors.

[PRR]

> could I sub in 62k/560p?

Free opinion: Seems reasonable to me. Everything interacts, but this is a small change.

> what's the purpose of scaling the latter network?

There's Equal Component Value (with gain) and Unity Gain (with unequal components). To make the unity-gain form work you have to significantly reduce impedance in one RC. If I understand, by about the square of the gain you would use in Equal Value design. That gain (for flattish filters like we usually want) is 1.7 or so IIRC(*). 1.7^2 is almost 3, which is about the ratio 68/22. Changing 68 to 62 is about 10% change in one RC, but probably 5% change overall, which is half-dB and probably not a killer difference for an effect.

See also: http://sim.okawa-denshi.jp/en/Fkeisan.htm

EDIT (*) 1.6... you used to know this, kid! Old age does something but I ferget what.

[Rob Strand]

FYI:

   Original             Yours
R1   2.20E+004   2.20E+004
R2   6.80E+004   6.20E+004
C1    1.50E-009   1.50E-009
C2   5.10E-010          5.60E-010
      
w0    29559.936          29542.903
f0   4704.610      4701.899
Q   0.737            0.720

I think that translates to 0.2dB drop at f0.
No one will know of your evil doings.

[Tall Steve]

Sweet! Thanks, guys!

Rob--would you mind showing me the equations you used to get w0 and Q?

[PRR]

http://sim.okawa-denshi.jp/en/OPstool.php

22K 68k 1.5n 510p
Cut-off frequency fc = 4704.6098909642[Hz]
Quality factor Q = 0.73702773119009

22K 62k 1.5n 560p
Cut-off frequency fc = 4701.899081237[Hz]
Quality factor Q = 0.71958071421905

Block-out the 13-digit "accuracy" because 2-pole audio filters never have to be better than 5% so not-even 2-digit exactness.

It is 4.7KHz Q=0.73 either way you look at it. It is sub-Butterworth so there's no audible peaking. It is slightly over Q=0.707 so technically it bumps.

For 22K 62k 1.5n 560p, "text data" says -3.01dB at 4786Hz (not 4701); also +0.00455dB at ~1KHz, which is "a bump", but hardly large.

"text data" says provides up to 1 minute.... I finally figured this out. (not minutes of angle!) The data is held 1 minute after Calculate, then it expires. So grab it quick, or re-Calculate.

[Rob Strand]

PRR's link looks pretty cool (it agrees with the values I posted).

I actually use a lot of spreadsheets for stuff (mainly because I've learnt not to trust anything and I've also got a lot of tricks I've come-up with over the years).   The filters are pretty straight forward (until you start getting in to pre-distorted designs):

This link has the equations: (book page 4, PDF page 8 )
http://www.ti.com/lit/an/sloa024b/sloa024b.pdf

They look OK but I can't say I've checked them.
One thing is they call the frequency fc, implying the -3dB point, this is wrong. 
It's actually f0 or fn, the "natural frequency" of the filter.
Only for Butterworth is f(-3dB) = fc = f0.
Given f0 and Q you can calculate f(-3dB) with more big formulas.

[Tall Steve]

Cool, thanks again! I sent an email to Lectric FX with this substitution in case they wanted to add it to the build doc.