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

[harrybissell]

OK I'm ignorant enough... I'll take a shot :^P

I think the input impedance of the OTA is NOT fixed...
its high impedance with small standing currents in the
core (small Gm bias) but gets much lower as the core currents
increase.

What say ye, experts ???  Have I 'got it' ???

H^) harry


Aaron Lanterman wrote:
> 
> 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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