RE: [sdiy] 1-pole LPF using a differential integrator?

[rdrake]

how about that center-tapped pot on the output?  would love to see an alternative way of implementing that...

b

>----- ------- Original Message ------- -----
>From: lanterma at ece.gatech.edu
>To: synth-diy at dropmix.xs4all.nl
>Sent: Wed, 30 Apr 2008 01:08:32
>
>One project idea that I tossed out to my students,
>which one student  
>is trying to get work, is to use the idea of an NPN
>differential amp  
>driving a "differential integrator" to build a
>1-pole LPF, which you	
>could then do four times to get a usual 4-pole LPF.
>
>
>I got the idea from the EML-101 filter schematic,
>as expertly redrawn  
>by Marjan Urekar:
>
>http://members.tripod.com/urekarm/synth/eml101vcf.p
>df
>
>The EML-101 is a state variable filter, and said
>diff amp/diff	
>integrator configurations form the variable gain
>integrators.
>
>My thought was that we could take the output of
>such an integrator,  
>and feed it into the opposite input terminal of the
>diff pair, and  
>that feedback should form a 1-pole LPF, as follows:
>
>
>The NPN pair driving the differential integrator
>forms a variable gain	
>integrator with transfer function A/RCs, where A is
>the voltage gain  
>of the diff pair and R and C are the resistor
>values used in the  
>integrator. Let's denote wo = A/RC. Now imagine we
>take the output of  
>this integrator and feed it back into the input
>with a minus sign. You  
>get something like this:
>
>Y(s) = (wo/s) [X(s) - Y(s)]
>
>Y(s)[1 + wo/s] = (wo/s) X(s)
>
>Y(s)/X(s) = (wo/s) / (1+wo/s) = wo/(s+wo)
>
>Which is the transfer function of a single-pole LPF
>with cutoff wo  
>with unity gain at DC. (This is the same idea as a
>usual OTA-C filter  
>1-pole LPF).
>
>Three questions:
>
>1) I sent my student, Justin, this sketch to get
>him started:
>http://users.ece.gatech.edu/~lanterma/eml_inspired_
>vcf_sketch.jpg 
>, where I just blindly ganked the resistor and cap
>values from the EML  
>circuit.
>
>Anyway, I just realized I drew that wrong - the EML
>integrating stage  
>is _inverting_, so we should actually either:
>
>A) Swap the inputs on the differential integrator,
>to make a  
>noninverting stage, or
>
>B) Take the feedback to the left NPN instead of the
>right NPN
>
>Does (A) or (B) sound like the safer option?
>
>2) Given the way the "differential integrator" is
>going to load the  
>differential amp, should I even be thinking of this
>as a differential  
>amp followed by a differential integrator? Or are
>they interacting so  
>much such thinking breaks down?
>
>3) Does this whole crazy idea of mine sound like a
>decent idea at all  
>or is it doomed to insanity?
>
>- Aaron
>
>P.S. This got me wondering if there's an
>"integrating version" of the  
>standard 3-op-amp instrumentation amplifier? A
>differential integrator  
>with high input impedance, but something more
>elegant than just  
>slapping noninverting op amp buffers at the inputs
>to the above  
>differential integrator?
>
>http://en.wikipedia.org/wiki/Image:Opampinstrumenta
>tion.svg
>
>
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