[sdiy] Why do we need the buffers? - gm-C filter question

[Aaron Lanterman]

If you see a four-pole LM13700 type filter, or CA3080 type filter, you 
often see a series four of stages like:

  +----------------------------+
  |                            |
  |  |\                        |
  +--|-\                       |
     |  \                      |
     |OTA-----+------[buffer]--+--vout
vi--|+ /     |
     | /      |
     |/    -------
             cap
           -------
              |
              |
             gnd

When I look in IC design books, they'll often write what looks to me to be 
an equivalent circuit (assuming you're driving something with a high 
impendence input, say another OTA...)

  +-----------+
  |           |
  |  |\       |
  +--|-\      |
     |  \     |
     |OTA-----+-- vout
vi--|+ /     |
     | /      |
     |/    -------
             cap
           -------
              |
              |
             gnd

Of course, there's the issue of having to put a resistive divider in front 
of the whole thing to get the input range low; in the first diagram above, 
you wind up with a similar divider after each buffer.

Why not just put one resistive divider at the beginning, cascade four 
stages like what's shown in #3, and then put a buffer on the final output? 
I see stuff like that in integrated circuit design books, where everything 
is just OTAs and caps...

Conjectures:

1) Maybe the OTAs we typically use, 3080, 13700, discrete 4-BJT, whatever, 
don't have high enough input impedance to avoid having the next stage load 
down the previous one? (One thing that makes me think this is the IC books 
are all CMOS, and a CMOS-based OTA would probably have higher input 
impedance, so maybe they could get away with it)

2) To keep the input voltage on the next stage low enough to avoid 
distortion, you'd have to keep the current coming out of the previous 
stage pretty small; maybe this gets into noise issues (although constantly 
running through all those resistive dividers and op-amp stages can't be 
good), and also, it may severely limit the range of useful currents at the 
OTA input pin.

Am I right (or partially right and partially wrong) on (1) and/or (2)? 
Other reasons?

- Aaron

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Dr. Aaron Lanterman, Asst. Prof.       Voice:  404-385-2548
School of Electrical and Comp. Eng.    Fax:    404-894-8363
Georgia Institute of Technology        E-mail: lanterma at ece.gatech.edu
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