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Question about accessing memory in 6264

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abhaybigbang

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could anybody tell me whn thr are 13 address lines in 6264 how can v access 64 kb of memory while 2^13 is only 8kb.
 

M!k

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Re: 6264

this memory has a capacity of 64kbit, organized as 8bit (1 byte) * 8k = 8kbyte
2^13 = 8192 bytes = 8kbytes
So don't confuse 'kb' = kilobit with 'kB' = kilobyte
 

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