It appears that I have made a math error!
The memory map of the 2465B, as seen by the 6800, puts the
RAM in two places.
The first 2K block is right at the beginning of the address
space, 0000 to 07FF.
The full 8K of the RAM is located at 8000 to 9FFF.
So, to get to the constants in the RAM chip, the M6800 needs
to add 8000 to the address where the constants are found in
the 8K RAM chip.
That should be 9E00 to 9FFF If I did my math right this time.
Please check my reasoning!
-Chuck Harris
On Wed, 10 Apr 2024 12:31:11 -0400 "Chuck Harris" <cfharris@...>
wrote:
Hi Chris,
256 words, or 512 bytes (0x1e00 to 0x1fff) are always set
aside for the calibration data.
There are only actually something like 170 16 bit words.
The 2465(no suffix letter) used an EAROM chip, and it was
was something like 128 14 bit words. The 2465[A|B] stayed
with the same calibration size, but bumped it up to 16 bit
because it was convenient in a byte oriented world.
The stack pointer was set a polite distance below the
calibration data for safety.
-Chuck Harris
On Wed, 10 Apr 2024 09:42:42 -0500 "Chris Elmquist" <chrise@...>
wrote:
On Tuesday (04/09/2024 at 11:47PM -0400), Chuck Harris wrote:
Hi Chris,
It really doesn't matter where you do your development. I choose
linux. I stopped using MS products in the 1990's, after paying a
ton of money for support on an MS development package. Their
support people repeatedly lied to me, I missed several important
deadlines, and lost a very important customer.
Absolutely agree. I just figured for the OPs goal, it might be a
long slog to get Linux up and running before he ever got to the
task at hand, assembling some 68xx code. I didn't want him to
believe that Linux was required. It's a better choice, but not
required for this exercise.
The memory map for the 2465 is shown in figure 3-2 in the
2465B/2467B Service manual.
Excellent. I have not had to look at said manual in many years so I
have likely forgotten it was there. This info will be needed to
know where to copy to and from and the execution address for where
the little bit of code needs to reside.
The calibration constants are the last 256 words in RAM... (eg.
0x1e00 to 0x1fff)
All of the rest of the RAM gets initialized every time the scope
is power cycled... So it just doesn't matter.
So, does that mean there are always 256 bytes of calibration data?
No fewer? The entire 256 byte block is what would be copied?
Chris
