Moog filter clone question

[Don Tillman]

   From: John Speth <johns at oei.com>
   Date: Fri, 19 Sep 1997 08:52:58 -0700

   I looked over this schematic and, yep, it's a ladder filter.  It's my
   first look at a ladder design using diodes.  First, just a general
   question to those who know: How would a ladder filter using diodes sound
   different from one that uses transistors (assuming optimal designs for
   both)?

A Moog ladder without feedback has four real poles in the same
frequency.  Adding feedback causes those poles to leave the real axis
and spread out in an X pattern (the surprising thing is that it's
*exactly* an X pattern).  So any feedback causes some resonance.

A diode ladder without feeback has four real poles at locations 0.12,
1.00, 2.35 and 3.53 times the tuned frequency.  So the poles are
really spread out and the curve will be much more subtle.  You need to
add much more feedback, a gain of 5 or so, to get resonance.

   Is there any matching requirements for the diodes?  

The top diodes should be matched.  The others don't matter.

   Will Ge diodes work as well or better than Si diodes?

Probably.  They'll certainly distort sooner.  'Might sound pretty
cool. 

   From: "Grant Richter" <grichter at execpc.com>
   Date: Fri, 19 Sep 1997 13:22:53 -0500

   The EMS line uses a diode ladder and the Moogs use a transistor ladder. The
   main difference seems to be how much of the low frequency is still passed
   when the resonance is cranked up. 

Huh?  It's the same.  

Oh wait a second, no, because you have to crank up so much more feedback
before it'll resonate, so in comparison to *that*, yeah.

Let me rephrase that: The low-end response of the diode and Moog
versions is the same for a given amount of feedback.  But since the
diode ladder has so much less resonance you need to apply that much
more feedback to get it resonating, and that'll mean that the diode
ladder has far less low-end response for a given amount of resonance.

Cool, I'd never considered that before.

 -- Don