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16 March, 22:46

With 64 KB of memory and 8 bits in each memory location, how wide should the address bus be to access all 64 KB of memory? (k = kilo = 1000, whereas K = 2^10 = 1024)

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  1. 16 March, 23:15
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    16-bit wide

    Explanation:

    In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.

    If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:

    64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.

    In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.
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Home » Engineering » With 64 KB of memory and 8 bits in each memory location, how wide should the address bus be to access all 64 KB of memory? (k = kilo = 1000, whereas K = 2^10 = 1024)
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