# [sdiy] Rick Jansen's Moog ladder filter design question...

Tim Stinchcombe tim102 at timstinchcombe.co.uk
Sun Mar 16 20:20:43 CET 2014

```> I'm about to build the modified Moog ladder filter you posted here:
>
> http://dropmix.xs4all.nl/rick/Emusic/Moog/moogvcf_schematic.gif
>
> I would like to change the resistor R15 2k37 1% for a 1k 1%
> 3300ppm tempco but this will change the overall A3 opamp gain
> and screw up the 1v/oct ratio on input called "KOV" and all
> the other CV inputs. Since I don't know the exact mv/oct
> value this schematic needs at Q1 base for a good oscillation
> tracking, can you help ? I think I'll just have to also
> change both P3 & R17 values right ? But to what values ?

Unfortunately that is going to be tricky, unless you also change all of
R10-R14 too!

The overall gain from input to base of Q1 needs to be about 0.018 (but
probably a little more to compensate for the base current of Q1). The
arrangement of P3+R17 will scale the output of A3 from between 1 and about
0.74 (=(x+700)/(250+700) as x runs from 0 to 250). Thus assuming 'KOV' or
'ECV' are the standard '1V/oct' inputs, with the values shown A3 gives a
gain of 0.0237, so adjusting P3 will give between 0.0237 and 0.0237 x 0.74 =
0.0175, which just encompasses the desired 0.018 value. Swapping R15 for a
1k value will mean the _biggest_ value you can get is 0.01, *whatever*
values of P3 and R17 you chose! Thus you'd need to reduce R10-R14, by a
little under a half to compensate.

Perhaps it would be better to stick the tempco as R17, and increase the
value of P3 a little (double it to 500 say)?

Hopefully with the little algebra you can play with your own sets of values!
(And the '0.018' is 0.026 x log_e 2 by the way!)

Tim
__________________________________________________________
Tim Stinchcombe

Cheltenham, Glos, UK
email: tim102 at timstinchcombe.co.uk
www.timstinchcombe.co.uk

```