Question

How many address bits are required to address a 4096 bit memory organized as a 512 x 8 memory?

Answer 1

Answer:

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.

Answer 2

Answer:

9 address bits are required to address a 4096 bit memory organized as a 512 x 8 memory.

Explanation:

To address n bytes, you'll need ㏒(2)n bits.

An 8 bit number, for example, can store 256 different values, so 8 bits can address 256 bytes. Because 210 = 1024, every byte in a kilobyte requires 10 bits of address. Similarly, every byte in a megabyte requires 20 bits, and every byte in a gigabyte requires 30 bits. For 4 GB of memory, 232 = 4294967296, which is the number of bytes in 4 gigabytes, you'll need a 32 bit address.

2^ Address line * Data line = Memory Location

That is,

$2^{9}$ * 8 = 4096

So, 9 address bits are required.

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