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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Number of address locations that can be accessed in internal RAM are
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32
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8
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64
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256
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Explanation:
Detailed explanation-1: -So to address 2048 (or 2K, where K means 2^10 or 1024), you need 11 bits, so 11 address lines. To express in very easy terms, without any bus-multiplexing, the number of bits required to address a memory is the number of lines (address or data) required to access that memory.
Detailed explanation-2: -The correct answer is 9. The number of address input lines a 512X4 ROM circuit is: If we have n bit address and m bit words then our RAM size will be 2n x m. Thus, 9 bits are required.
Detailed explanation-3: -The memory address space is 128 MB, which means 227. However, each word is 8 (23) bytes, which means that you have 224 words. This means you need log2 224 or 24 bits, to address each word.
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