Help with address lines

Discussion in 'General Electronics Chat' started by gammaman, Mar 3, 2009.

  1. gammaman

    Thread Starter Member

    Feb 14, 2009
    29
    0
    How do I figure out the number of address lines?

    Say I have 2K x 16. How many address lines would this be?
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    That would be 2048 X 16. 11 lines.
     
  3. gammaman

    Thread Starter Member

    Feb 14, 2009
    29
    0
    Thanks! Would you mind showing me how that is calculated?
     
  4. RiJoRI

    Well-Known Member

    Aug 15, 2007
    536
    26
    1K needs 10 address lines (2^10 - 1024). Every time you double the "K" you add one address line.

    1K = 10 lines.
    2K = 11 lines.
    4K = 12 lines.

    1M = 1K*1K = 20 lines.

    Yeah, it can be done mathematically, but I find this method easier.

    --Rich
     
  5. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    2Kx19 means that you have a data bus 16 wide and 2000 memory locations and each memory location can store a 16bit value.

    Thus you need 2000 different addresses to access each memory location.

    In binary 2048=2^(11)

    Thus to achieve 2000 different addresses you need 11 lines.
     
  6. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    Logarithms to the base 2 is all there is to it.
     
  7. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Yes, but it is not obvious for all people.
     
  8. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    I never claimed it was obvious.
     
  9. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    gammaman,

    2048 = 2^x ===> x = log10(2048)/log10(2) = ln(2048)/ln(2) = 11

    Use the log of any base you want and get the same answer.

    Ratch
     
  10. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    That's the ticket -- You 'da man!
     
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