The trimpot in the feedback would allow you to calibrate
exactly ONE input resistor perfectly... as would having the
pot in series with exactly ONE input resistor.
In either case... absolute precison requires you to use the
very same resistor (input) each time.
.1% resistors can be bought... or again I submit that you
can match them easily with the average multimeter. It would not
matter if they were all 99.7K or 103.5K... the pot would calibrate
them out as long as they are ratio matched.
H^) harry
Grant Richter wrote:
>
> > Lots of jacks - not sure what the difference is between the
> calibrated
> > and uncalibrated inputs.
> >
>
> In the summing section of a typical VCO or synth module, you use 1%
> resistors. That is 1
> part in 100 accuracy.
>
> The input A/D is 10 bits or 1 part in 1024. So using any two 1%
> resistors, the input
> reading could differ by 20 "counts" (one res at +1%, one res at -1%).
> The simplest way to
> trim that precisely is with a trimpot.
>
> We want an input that is 0 counts at 0 volts and exactly 1023 counts
> at 5.000 volts (with
> 0.00499 volts per count). We can convert that to floating point and
> actually calculate in
> real world voltages. Since the ouput is calibrated, you can work in
> real world voltages also.
>
> The non-trimpot input is close but not precise. Unless you match the
> resistors by hand to
> 0.1% or buy them matched for 80 cents apiece.
>
> ----------------------------------------------------------------------
> YAHOO! GROUPS LINKS
>
> + Visit your group "ComputerVoltageSources" on the web.
>
> + To unsubscribe from this group, send an email to:
> ComputerVoltageSources-unsubscribe@yahoogroups.com
>
> + Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> Service.
>
> ----------------------------------------------------------------------