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
