DIP Memory Organization

Discussion in 'General Electronics Chat' started by nDever, Jan 13, 2011.

  1. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    Hey Guys,

    I have a memory DIP (one of the HM62256B series chips) and the datasheet says that the memory is organized in a 32-kword x 8 bit grid but the chip only has 15 address inputs. How can the datasheet claim that the chip is organized in words when the packet only has 15-bits of address access?

    http://www.datasheetarchive.com/Indexer/Datasheet-028/DSA00489713.html
     
  2. tom66

    Senior Member

    May 9, 2009
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    A "word" is arbitrary; on an 8-bit processor, it could be 8 bits (synonymous with byte), and on a 16-bit processor, it could be 16 bits. In this case, your memory has an 8 bit word. It only has 32,768 locations, which requires exactly 15 address lines.

    http://en.wikipedia.org/wiki/Word_(computing)
     
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  3. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    Thank you; I understand now.
     
  4. nDever

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    Jan 13, 2011
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  5. hgmjr

    Moderator

    Jan 28, 2005
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    The device in the datasheet is 256K bits organized as 32Kbytes.

    hgmjr
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Some memory capacity is given in bits, so the word size doesn't matter.

    Normally, when memory is specified in bytes, an 8 bit byte is assumed. Otherwise, 4GB of memory would be "twice as much" in a 32 bit system compared to a 64 bit system.
     
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  7. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    256 Kbits = 256,000 bits
    32 Kbytes = 262,144 bits

    How can this configuration be possible?

    I believe that the confusion I am having arose from the ambiguity of whether a kilobit is equal to 1,000 bits or 1,024 bits. I have seen both definitions.
     
  8. hgmjr

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    The use of the prefix kilo is used loosely when specifying bytes. Your bit count calculation is correct.

    1 Kilobyte = 1024 Bytes

    hgmjr
     
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  9. tom66

    Senior Member

    May 9, 2009
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    In electronics a Kbit is 1024 bits, a Mbit is 1048576 bits and so on, binary prefixes are mixed with metric and it all gets confusing.
     
  10. nDever

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    Jan 13, 2011
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    Ah, I see. Thanks guys for the quick responses.
     
  11. thatoneguy

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    It's frustating and confusing. This practice carries over to hard drive and other storage devices as well. Sometimes kilo/mega/etc is a 1024 based number, sometimes kilo/mega/gig/etc is a 1000 based number, and most annoying of all is when they mix the two, such as 1 million bytes. First is the 1000 system, second is the 1024 system. Until you work with it, you can't exactly figure out the number of usable bits, especially with hidden bad sectors, etc.
     
  12. nDever

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    Jan 13, 2011
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    As an experienced person in electronics, would you conclude that the potential loss of 24 bits would affect larger-scaled systems badly? Can they be safely ignored?
     
  13. tom66

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    May 9, 2009
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    One potential problem: An operating system computes drive space in megabytes by using integer division. It divides say 1,000,000,000 bytes by 1024 to get the number of kilobytes = 976562 KB. In this case, it rounds down - but then another operating system decides to round up to 976563 KB. This creates a size incompatibility - for example, duplicating a backup. Now if the sizes were multiples of 1024, there'd be no problem.

    It's not the loss of just 24 bits, either. On a large scale, a 1 GB drive is 1,000,000,000 bytes instead of 1,073,741,824 bytes - a whole 7.3%. A terabyte drive is usually only 931 GB. As you can see, a considerable amount of space is not actually there.
     
  14. thatoneguy

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    It depends on what the 24 bits are.

    If they are a platter on a Terabyte hard drive, it wouldn't matter at all.

    If it is a defective area of RAM in a microcontroller that only has 128 bits of RAM to start with, it can be very bad.
     
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