moog sonicV diode ladder
Don Tillman
don at till.com
Sat Apr 3 20:11:11 CEST 1999
Date: Fri, 02 Apr 1999 14:52:24 -0800
From: Sean Costello <costello at seanet.com>
Terry Michaels wrote:
> The recent discussions on Moog transistor ladder filters, and diode ladder
> filters, brings up a question: Is there a consensus that these filter
> types sound significantly different?
Oh, gosh yes.
> If so, one reason might be the coupling between filter sections.
Don Tillman posted an analysis of just this very phenomenon (sp?) a few
years back. He did a Spice analysis of pole migration in a diode ladder
filter as resonance is increased. It is in the archives somewhere,
wherever the archives are now.
Hi.
No Spice, I did a pure math analysis with used some software I wrote
to search out the poles of polynomials.
Here are three message I posted on the topic a while ago. The first
is about the Roland diode filter and how it the Q values are
remarkably low, but can start to come up when you add a lot of
feedback.
----------------------------------------------------------------
Date: Tue May 28 1996
To: synth-diy at horus.sara.nl
Subject: Re: butter and pole locations
From: Don Tillman <don at till.com>
From: Haible_Juergen#Tel2743
Date: Mon, 20 May 96 17:39:00 PDT
(BTW, has anybody yet calculated the exact locations / pole spread
for the diode ladders?)
The Roland cascaded SV 4p-filters might be different ... anybody
knows if *they* start on an arc (as in a butterworth)?
Derriving the transfer function isn't all that difficult, but it's a
very mistake-prone process.
The generalized version (any values for the Rs and Cs) is too long for
me to conveniently write out, but if you set all the Rs equal and all
the Cs equal, the transfer function becomes:
Vout 1
---- = ----------------------------
Vin s^4 + 7s^3 + 15s^2 + 10s + 1
(I *think* this is correct. If anybody cares they can do it out
independently and we'll compare notes.)
Which means the poles are at (normalized to 1Hz for clarity):
2.35Hz, 3.53Hz, 1.00 Hz, 0.12 Hz
As expected we have four real poles.
Adding a feedback factor simply increments the 0th coefficient by that
amount, soooo, with a feedback gain of 1 the poles are:
2.62Hz, 3.41Hz, 0.59 Hz, 0.38 Hz
At a feedback gain of about 1.07, the lower two poles meet.
Feedback gain of 2:
3.00Hz, 3.15Hz, double pole at 0.56 Hz (Q= 0.66)
Not too interesting.
Feedback gain of 3:
3.14 Hz (Q= 0.50), 0.64 Hz (Q= 0.84)
(Note the higher two poles have met.)
At a feedback gain of 4:
3.20 Hz (Q= 0.51), 0.70 Hz (Q= 1.04)
The lower pole pair is starting to show some gain.
At a feedback gain of 5:
3.25 Hz (Q= 0.51), 0.75 Hz (Q= 1.25)
At a feedback gain of 10:
3.45 Hz (Q= 0.52), 0.96 Hz (Q= 2.98)
At a feedback gain of 15:
3.60 Hz (Q= 0.52), 1.11 Hz (Q= 9.57)
Here's some serious resonance. And it starts oscillating with a
feedback gain of about 18.3. The higher pole pair never really gets
off the ground
Executive summary: not too exciting.
And yes, in answer to the upcoming question, I have some software I
wrote to do the pole analysis.
-- Don
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A later post mentions the Moog ladder polynomial for comparison.
----------------------------------------------------------------
>From till Thu May 30 1996
To: ftom at netcom.com
CC: synth-diy at horus.sara.nl
Subject: Re: butter and pole locations
From: Don Tillman <don at till.com>
Date: Wed, 29 May 1996 22:51:18 -0700
From: Tom May <ftom at netcom.com>
So the small signal circuit for a fixed control voltage would look
like this, right? Hardly the "all 4 poles the same" circuit that I
had been led to believe it was!
+--R--+--R--+--R--+--R--+---
| | | | |
Vin C C C C Vout
| | | | |
+-----+-----+-----+-----+---
The R*land diode ladder circuit is indeed equivalent to the RC ladder
you drew.
The Moog transistor ladder is equivalent to an RC ladder with a buffer
in between each stage, so that will have four poles the same. The
transfer function for that would be:
Vout 1
---- = ----------------------------
Vin s^4 + 4s^3 + 6s^2 + 4s + 1
The difference between diodes and transistors here is *very*
important.
-- Don
----------------------------------------------------------------
Later I discovered something interesting about the Moog Ladder when
you add feedack.
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>From till Sat Sep 20 1997
To: johns at oei.com
CC: synth-diy at horus.sara.nl
Subject: Re: Moog filter clone question
From: Don Tillman <don at till.com>
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
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