2nd generation VCFs (was: Xpander VCF)

[Haible Juergen]

	>I was looking at OB Xpander's VCF on Grant's page, and while I
	>understood
	>theory of different filter modes, I can't figure out cap values
used.
	>In most of the quad OTA filters (CEM,SSM,13700) caps used are all
the
	>same value and in range 100p-1n (integration cap that is), and
Xpander
	>has 33n caps for first 3 stages (first stage can be switched "off"
with
	>100p cap needed for some modes) which seems too high, and fourth
stage
	>has 100 times smaller 330p cap. Why's that? And why so large caps?


Hi Marjan and list,

I have not seen the filter in question (Can you give me the address ?), but
there 
is something more general that I have discovered and reverse engineered just

lately.

You know, there are several "2nd generation" VCF chips that don't use buffer
stages between their OTA stages.

If you look at 1st generation VCF chips like SSM2040 or CEM3320, you
have the gm cells that need a small input voltage, so there is a resistor
divider from the previous stage, and a resistor divider in the feedback loop
from the same stage's output as well.

At one point the circuit designers must have seen that this is really
unnecessary,
boosting up - dividing down - boosting up - dividing down.
You can just omit the resistor dividers and run the capacitor and buffer
stage
at the same 20mV as the OTA inputs. You need very low noise buffers
of course. And because of the altered internal loop gain, you need much
larger
cap values for the same cutoff frequency.

Next step: Why use these stupid buffer stages at all ?!  Sure, the input of
the
following OTA has a painfully low impedance, but then this impedance is
changing with the control current just like the cutoff frequency. So instead
of
isolating that impedance with a FET or darlington buffer, just connect it
in parallel to the capacitor. And while you're at it, just connect the
negative
input of the stage's own OTA parallel to the cap as well.

If you draw the schematics for that, it's ridiculously simple. Like that:
3080#1_pin3 = input. Connect together 3080#1_pin6, capacitor, 3080#1_pin2
and 3080#2_pin3. Hard wire it all together, and build a chain of 3080's
and capacitors with *no* resistors or buffer transistors whatsoever.

Does it work ? In simulation it does, in certain SSM and CEM and Roland
chips apparently it does as well. (If my interpretation of the data sheets
is right).

Surprised ? I was surprised for sure !

All right, for the last of 4 stages you probably want a buffer stage,
because
you're driving the feedback loop, and a VCA and whatever.
But there would still be no reason for a different cap value and for
resistor
dividers, no ? One could just insert a FET (and source resistor), and then
drive the OTAs for VCA and VC feedback with 20mV as well.

BUT ! (And that's a rare occasion when you'll find me using capital letters)
BUT if you do this for the last stage as well, you get some feedthru from
the
input *backwards* thru the feedback OTA to the output.
The result is that your filter will not sound as smooth and "dark" as it
should.
The rolloff has the 24dB / oct at first, but after a while it becomes flat
and
you get your high frequencies thru with a fixed attenuation of a few dozen
dBs. That's why I think the chip designers used a different scheme for the
last stage. There is even a chip intended for 2pole and 4pole use which has
alternately unbuffered and buffered OTA stages. (Might be in the JP-6,
but I'm not sure.)

I may be completely wrong with my interpretation of course, but then at
least
the explanation fits nicely in retrospect. I'd love to see this confirmed or
denied from SSM and CEM designers - let's see if there will be some
reaction.
Or is there even some publication about it ?

At last, a "caveat": If you're going to try this with 3080's, make sure that
offset voltages won't get you in trouble.

Some food for thought: What does this (possibly) tell us about Moog filters,
and different buffer and feedback stages ??

JH.