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# NEC Ram Module

Posted on 2000-04-18
Medium Priority
399 Views
I have as NEC memory module with the following markings on each of the two chips: 4218160-70 and 9451AU600.

the back of the board contins the following:P7460 and P10-14B74600.

Can anyone advise me as to the capacity of the module and whether it is parity or not.
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Question by:rlpp
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frache earned 200 total points
ID: 2727281

4218160 : 1Mb *16 soit 2Mo each chip
Fast page mode ( no EDO ram).

2 chips = 4 Mo

70 = 70 ns : access time
9451 : 1994 - week 51 ( december)

2 chips : no parity ( no bits for parity )

Each module = 4Mo / 70 ns
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LVL 7

Expert Comment

ID: 2727330

More details :

1 Mb = 1 000 000 bit
1 octet = 8 bits
16 * 1 M = 16 Mb = 16/8 Mo = 2 Mo

1 module = 32 * 1 Mb = 32 Mb = 32/8 Mo= 4 Mo.

Parity : you need one bit more per octet
-> 4 (8+1) * 1 Mb = 36 Mb ... you have only 32 Mb
ECC    : you need two bits more per octet
-> 4 (8+2) * 1 Mb = 40 Mb ...

0

Author Comment

ID: 2727463
Thanks for your help. This is information which NEC refused to answer.
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Author Comment

ID: 2730327
Thanks Frache.

Excuse my ignorance, but what is an Mo.
0

LVL 7

Expert Comment

ID: 2733602

1 Mo = 8 Mb
1 Mb = 1024 bits

-> 1 Mo = 8 * 1024 * 1024 bits.
-> 1 bit = 1 binary data ( 0 or 1).

0

Author Comment

ID: 2734487
Thanks for the clarification.
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LVL 7

Expert Comment

ID: 2734636
Ok, have a nice day.
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