> Note that as the A/D is switched between channels, you actually have a
> charge pump from one channel to the next.  A newly selected channel is
> connected to the same internal capacitor that the previous channel was
> connected to.  This actually creates an equivalent resistance from the
> previous channel voltage to the current channel pin.  This equivalent
> resistance is proportional to the period between selections of the same
> channel.  None of this should matter, however, if each channel's source
> impedence is 10Kohms or less and you are waiting the proper acquisition
> time.

Oops.  I meant to say that the equivalent resistance is *inversely*
proportional to the channel selection period (proportional to frequency of
the channel being selected).


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