From: Pasi Ojala (albert_at_cs.tut.fi)
Date: 2007-04-07 10:56:30
> I have a variable voltage of 0 to 5V, and I want to
> translate this in to a current to simulate the variable resistance of a
> potentiometer connected to the SID's ADC
I'm almost certain the SID input isn't a current-input, and
the AD would work just fine by simply connecting the varible
voltage directly. (My hazy recollection is that I once tried to
use the SID ADC's for audio sampling, but obviously they are
just too slow.)
To make the actual conversion, SID samples the input by charging a
capacitor, and then discharges it through a resistor.
The amount of time it took gives the conversion result.
At least that's what I remember reading (from Programmer's Ref.Guide,
probably) 20 years ago.
Because the input impedance limits the draw current, and sampling
really takes place (the sample capacitor is not charged continuously),
you really should not be forced to emulate a variable-current drive.
To think about it in another way, the input impedance provides the
voltage division even if the potentiometer does not have the center
pin connected.
Or am I mistaken?
-Pasi
--
"Khrm."
"It's .. animal magnetism, what can I say?"
-- Londo and G'Kar in Babylon 5:"A Tragedy of Telepaths"
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