Re: Troubleshooting X/Lisa RAM boards - first draft

Hi James,

Great! I'll try those combinations on the checkerboard memory board tomorrow and let you know the result.

I'll also replace the IC at E22 and see what happens.

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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