Welcome! I have a theoretical question.
How many bits would be needed to address 14kB of case data memory if one cell is 1 byte in size?
My answer would be, 14kB is 14,336 bytes and would require that many memory cells. And the result should be 114,688 bits... right?
I'm not sure what question you are actually asking, but a 14 bit address can address 16384 individual cells of any size.
Then it would take 14 bits to come out, which would use 14226 cells, where each cell would be 1 byte?
Given the following information .. Each memory cell has 1 byte, need to address 14kB. The question is how many bits are needed to do this?
If English is not your native language, try using a translator program to explain the question you are asking, or explain what you actually want to do.
14 kByte of memory requires 14 address bits (or an address bus that is 14 bits wide). 14 address bits can address 214 addresses (16k).
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