More result, this time with the second board in various slots/configurations
> - in MEM 1 with MEM 2 empty
> - in MEM 2 with MEM 1 empty
> - in MEM 2 with a good board in MEM 1
> - in MEM 1 with a good board in MEM 2
Just a bit of extra information in case it's useful. I tried the first
memory board (the one with the faulty IC at E22) in Slot 2 with Slot 1
empty. I got a checkerboard pattern on the screen. In Slot 1 with Slot 2
empty this showed 00000186 0000 0000 0000 0001 0000 0000 0000 0000 of
course.
Hope this helps. I'll leave off replacing that IC at E22 for now in case
you want me to do any more tests under various configurations with the two
faulty boards in order to figure out a robust faulty chip ID scheme.
Many thanks James. At least your ID procdure looks like it will fixed at
least one memory board problem for me (and hopefully both).
Terry
> James,
Checkerboard pattern. Doesn't go any further.
Error 70
00000186 0000 0000 0000 0200 0000 0000 0000 0000
Error 70
00000186 0000 0000 0000 0200 0000 0000 0000 0000
Error 70
00000186 0000 0000 0000 0000 0200 0000 0000 0000
>
>> Bit 0 maps to column 22 on the memory board, so the suspect chip is the
>> one at coordinates E22. (If that isn't it, I think the only other
>> possibility is B22.)
>
> You were right on the button here! I piggybacked a 4164 over the IC at
> E22 and the board passed the memory check with no problems. It seems
> that's the faulty one allright. Later today I'll extract and resolder a
> replacement in.
>
>> I think the four possible slot combinations might give different results,
>> so try each configuration with the bad board:
>> - in MEM 1 with MEM 2 empty
>> - in MEM 2 with MEM 1 empty
>> - in MEM 2 with a good board in MEM 1
>> - in MEM 1 with a good board in MEM 2
>
> In a few minutes I'll run these tests with the second board and post up
> the results. Stand by..
>
> Terry
>
> ----- Original Message -----
> From: "James MacPhail" <gg__at_email.domain.hidden>
> To: <lisalist_at_email.domain.hidden>
> Sent: Saturday, March 26, 2011 3:19 PM
> Subject: Re: Troubleshooting X/Lisa RAM boards - first draft
>
>
>>
>>>One USED to come
>>>up with a parity error. Now it doesn't even get that far. Instread I
>>>get a
>>>checkerboard pattern so there might be more than one chip faulty on this
>>>one.
>>
>> I think the four possible slot combinations might give different results,
>> so try each configuration with the bad board:
>> - in MEM 1 with MEM 2 empty
>> - in MEM 2 with MEM 1 empty
>> - in MEM 2 with a good board in MEM 1
>> - in MEM 1 with a good board in MEM 2
>>
>>
>>>The second board comes up with a read/write error. Using just that one
>>>board in memory slot one I examined the words at the place indicated and
>>>found..
>>>
>>>00000186 0000 0000 0000 0001 0000 0000 0000 0000
>>>
>>>Using your rule of thumb, this would seem to be Row E, yes?
>>
>> Yes, that's correct (I hope).
>>
>>>translate this to exactly which bit this is and also what column? I do
>>>know what binary is and how it (and hex) translates to things regarding
>>>8-bit but not really at the chip level . Are you able to elaborate (or
>>>point me to a URL) on how this is translated? I assume these words are
>>>in
>>>hex? I assume so if you can get something like 4000.
>>
>> Yes, the words are in hex. Each character of the four in one word is one
>> hex value, from 0 to F, and those correspond to the 16 possible binary
>> values of 4 bits.
>>
>> The bit positions are (sometimes) numbered, with bit 0 being the least
>> significant, and rightmost bit. Your value of 0001 has the rightmost bit
>> set, and that is bit 0.
>>
>> Bit 0 maps to column 22 on the memory board, so the suspect chip is the
>> one at coordinates E22. (If that isn't it, I think the only other
>> possibility is B22.)
>>
>> I know that's not a clear explanation of the bit numbers, but perhaps
>> will suffice.
>>
>> If there is just one bit set (ie. only one bad chip), then there are 16
>> possible values for the word, which map to memory board columns as
>> follows:
>>
>> Word Bit# = Column
>> ---- -- --
>> 0001 0 = 22
>> 0002 1 = 21
>> 0004 2 = 20
>> 0008 3 = 19
>> 0010 4 = 18
>> 0020 5 = 17
>> 0040 6 = 16
>> 0080 7 = 15
>> 0100 8 = 1
>> 0200 9 = 2
>> 0400 10 = 3
>> 0800 11 = 4
>> 1000 12 = 5
>> 2000 13 = 6
>> 4000 14 = 7
>> 8000 15 = 8
>>
>> If there is more than one bad chip, there will be more than one bit set,
>> and you'd get a character other than 1,2,4,8, or more than one in the
>> word.
>>
>> Good luck!
>>
>> James
>>
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Received on 2015-07-15 16:53:21
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