Hallo Constant!
>First I have pressed them firmly in their sockets, no change.
Sometimes cleaning the chip's pins with a rubber gum (the thing you use
to remove writing from your pencil) helps.
>Reading below the banks I found following info. Bank 0 and 1 -4164/
>41256 Bank 2 and 3 4164 only. As the chips of the 286 where 256 chips
So you could take all chips out and put the 256s in bank 0 and 1 and
leave 2 and 3 open. Possibly there are some jumpers or DIP switches near
the memory banks where you have to disable them or something.
>that the chips I need are 64 chips with any number combination in
>front. Am I right there? What does this 41 before the 64 tell an
>expert?
Generally, yes. But you will have only 256K of total memory then. Oh,
the 41 says, if i remember correctly, "standard DRAM with 1 bit wide
data path" and the 64 / 256 is the number of KBits in the chip.
So my suggestion is to get 18 * 4164 and 18 * 41256 chips to give you a
full 640K of memory for you operating system. If you can actually get
hold of this type of chips. Here they are quite difficult to get if you
want new ones.
>
>In the bank I find the chips numbered from 0 to 7 with the last chip
>lettered P.
That will be one for each bit of the data bus, and a parity bit.
>If I only have 512 kb at booting, does that mean that a
>whole bank (bank 3 ?) is not functioning? Is there a way to spot the
Wild guess: Bank 2, as bank 0 and 1 together should get you 512K.
>misbehaving chip? And if so then what?
Sometimes broken chips get hot. So open the thing, turn it on, and use
your fingers to feel the heat. If no single chip is any hotter than the
others after 5 minutes, you can stop feeling...
>Anyhow, how can you calculate the memory. 16 chips with 256 (?) makes
>4096. 16 with 64 1024, totaling 5120 (what ?)
KiloBits
>If I save time, when do I get it back?
Oh, never, of course.
--
Gunnar Thöle
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