MEMORY ORGANIZATION IN 8051
ARCHITECTURE OF 8085
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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How many bit address are used to access the data memory
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16
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8
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64
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32
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Explanation:
Detailed explanation-1: -For example, an 8-bit-byte-addressable machine with a 20-bit address bus (e.g. Intel 8086) can address 220 (1, 048, 576) memory locations, or one MiB of memory, while a 32-bit bus (e.g. Intel 80386) addresses 232 (4, 294, 967, 296) locations, or a 4 GiB address space.
Detailed explanation-2: -because 8 bits can only have 256 combinations, the memory addresses must only have around 256 bytes, considering that 8-bit computers can only have 256 possible memory addresses for each byte of RAM.
Detailed explanation-3: -Also what does 8 bit words mean? You used it indirectly: the total amount of addressable memory is 2 ^ 16 * 8 bits. Since one byte = 8 bits, that’s 2 ^16 bytes, i.e. 65536 bytes, or 64 KiB.
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