Look at the data sheet again,
It shows an internal sample switch AS WELL AS selector switches.

  *
  |  __O    Thomas C. Sefranek  tcs@cmcorp.com
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-----Original Message-----
From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf Of
Michael Rigby-Jones
Sent: Wednesday, March 01, 2006 3:10 PM
To: Microcontroller discussion list - Public.
Subject: RE: [PIC] Question about a/d converter



>-----Original Message-----
>From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] 
>Sent: 01 March 2006 20:02
>To: 'Microcontroller discussion list - Public.'
>Subject: RE: [PIC] Question about a/d converter
>
>
>
>
>-----Original Message-----
>From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] 
>On Behalf Of Gerhard Fiedler
>Sent: Wednesday, March 01, 2006 1:54 PM
>To: piclist@mit.edu
>Subject: Re: [PIC] Question about a/d converter
>
>>The conversion time is what you always have (and you seem not to be
>>considering it here). The acquisition time is what you only 
>need after you 
>>switched channels. If you are reading only one ADC channel, 
>you usually can
>
>>use an acquisition time of 0.
>
>>Gerhard
>
>
>I disagree. Selection of charge source is NOT the sampling 
>switch! The sampling switch and series impedance dictate a 
>phase/time delay in combination with the internal charge hold 
>capacitor. (120 pF) The holding capacitor is NOT continuously 
>refreshed. 

>From the datasheet diagrams, the capacitor would appear to be continuously
connected to the A/D pin once a channel has been selected (the "sampling
switch" refers to the analog mux according to my FAE).  In fact common sense
tells us this must be the case, as you don't have to do anything to initiate
an acquisiation once a channel has been selected for the first time.  There
is some impedance in the analog switch though, so the cap would undoubtedly
need some recovery time after a sample has been taken.

Regards

Mike

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