in this paragraph, I do NOT undrestand the sentence \"memory content is one byte
ID: 3664853 • Letter: I
Question
in this paragraph, I do NOT undrestand the sentence "memory content is one byte wide". (I thought16 bits, we can handle up 2^16 addresses. Why do we need one byte wide in memory content?)
MEMORY ADDRESSES The address bus size in processors is usually determined by the number of memory positions it can address. For example, if we have a microprocessor with an address bus of 16 bits, we can handle up to 216 addresses. If the memory content is one byte wide then the processor can handle up to 21 bytes 64KB 8 bits Address 0000 0000 0000 0000: Oxoo00 tentsEytewide thenthe 0 0001: 0x0001 to 2 odreses hememor conn .the 0000 0000 000 Here, we use 16 bits per address, or 4 nibbles. The lowest address 1111 1111 1111 1111; 0xFFFE (in hex) is 0x0000 and highest address (in hex) is 0xFFFF.Explanation / Answer
Primary recollections are ordinarily greater than the measure of a solitary RAM chip. Along these lines to get to a memory word the memory equipment needs to peruse a line in a few RAM chips at the same time and after that connect the returned results from every RAM chip.
The RAM chips that make up a principle memory framework, are ordinarily gathered into banks that are one memory word wide:
The bank divide into 4*4 bits and four banks
Example: Given Main Memory = 1M × 16 bit (word addressable),
RAM chips = 256K × 4 bit
BANK size = RAM chips per memory word = Width of Memory Word / Width of RAM Chip = 16/4 = 4
18 bits are required to address a RAM chip (since 256K = 218 = Length of RAM Chip)
A 1M × 16 bit word-addressed memory requires 20 address bits (since 1M = 220).
Therefore 2 bits (2018) are needed to select a bank.
The Total number of RAM Chips = (1M × 16) / (256K × 4) = 16
Total number of BANKS = Total number of RAM chips / BANK size = 16/4 = 4
One location addresses one byte. Utilizing 16 bits, you can compose 65536 locations from 0 to 65535, that is 65536 unique addresses and address 65536 bytes. 65536 bytes is 64kB. In software engineering, b is bit, B is byte. The byte is the littlest measure of memory you can address. A kilo unit is 2^10=1024 units.
Memory (RAM) is made out of bits, gatherings of 8 which frame bytes, each of which can be tended to ,and henceforth byte addressable memory.
Address Bus stores the area of a byte of memory.
In the event that a location transport is of size 32 bits, that implies it can hold upto 232 numbers and it henceforth can allude upto 232 bytes of memory = 4GB of memory and any memory more noteworthy than that is pointless.
Information transport is utilized to send the quality to be composed to/peruse off the memory. In the event that I have an information transport of size 32 bits, it implies a greatest of 4 bytes can be composed to/peruse off the memory at once. I discover no connection between this size and the most extreme memory size conceivable.
Despite the fact that most frameworks are byte-addressable, it bodes well for the processor to move however much information around as could reasonably be expected. This is finished by the information transport, and the span of the information transport is the place the names 8-bit framework, 16-bit framework, 32-bit framework, 64-bit framework, and so on originated from. At the point when the information transport is 8 bits wide, it can move 8 bits in a solitary memory operation. At the point when the information transport is 32 bits wide (as is most basic at the season of composing), at most, 32 bits can be moved in a solitary memory operation.
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