Re: Cinematronics > Atari Adapter???

On Tue, 15 Oct 2002 02:17:44 -0500, Rodger Boots <rlboots@cedar-rapids.net>
wrote:

>>You do need 12 bits for the 10 bit vector resolution, using the hardcoded timers
>>built onto the Cine CPU board. Anything less than 12 bits for 10 bit vectors
>>and the intensity difference between the start of the vector and the end,
>>becomes noticeable. That and the timing generator on the CCPU would have to be
>>changed to compensate for the slower drawing rates of the smaller DACs. And the
>>instruction set would need to be changed since the Normalization instruction
>>currently normalizes for 12 bit DACs.
>>
>>
>
>You might be comparing apples to oranges here. The vectors are drawn by
>the integrator, NOT the DAC. The only difference dropping a few bits
>will have is less accuracy setting the start and end points. It doesn't
>affect timing, brightness, or anything else about the vector except the
>start and end points, and they will be real close to being right.

No, I've got my fruits straight on this. ;-) You could be thinking of how the
Atari VG works, and not the Cinematronics approach.

You refer to the R/C generator as an integrator, in the classic sense it isn't.
The capacitor is not connected across the feedback of a op-amp, and does not
integrate current, but holds a voltage as in a sample and hold circuit.

The capacitor is connected to ground. It is driven to a starting position using
a DAC set to the actual starting position voltage (only 10 bits of the DAC are
used to do this), through an analog switch connected directly to the cap. The
Z-axis is then turned and the DAC is set to a voltage way beyond the endpoint,
and through a resistor, (using an analog switch) the capacitor is setup to
charge to the new DAC voltage, which if allowed to go there, would be way off
the edge of the screen. This is done for both axis simultaneously. This need to
send double the needed voltage to cap is why a 12 bit DAC is needed for 10 bit
vectors (you need to double the voltage in both positive and negative directions
which is why you need 2 additional bits.)

Timing is very critical at this point, when the capacitor reaches the position
of the original endpoint (not the one loaded into the DAC, but where we *want*
the end point to be), the analog switch must be opened, and the Z-axis turned
off. All timing is dependent upon the value of the resistor, the capacitor, and
the voltage on the DAC.

If the voltage on the DAC is too low, the CRT trace will not have moved far
enough at the timeout period, and the vector will be too short, if too high, the
vector will be too long when the timer times out.

As far as brightness. An R/C circuit (as opposed to a real integrator), does
not charge at a consistent speed. It starts by charging quickly and slows down
logarithmically as it nears it's final ending voltage. If the DAC were to be
set to the actual end point voltage, the CRT trace would start by moving quickly
and then slow down the closer it got to the ending point. The result is a dim
vector that gets logarithmically brighter as it approaches the end point, and
would take much longer to draw. By setting the DAC's endpoint to way beyond the
original end point, only the starting charge of the R/C circuit is used to
generate the vector, this part of the R/C charge curve is fairly linear and is
used to draw a vector of consistent intensity and speed.

When building a Cine->WG replacement, one has no control over the timing, this
is all done on the CCPU board. Same with the voltage values being sent to the
DACs. The CCPU is sending out 12 bit DAC information, and is expecting to time
the charge of a preset R/C constant. Changing any of the parameters: DAC
voltages, R or C, will result in screwed up timings and vectors of incorrect
lengths.

-Zonn
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Received on Tue Oct 15 15:46:06 2002