Posted by bode on Saturday, March 26, 2011 at 11:40am.
Address lines are required to address each word. For n lines, the number of possible address is 2^n addresses.
For 128K words, there are 2^7 addresses, so 7 address lines are required.
Each word has a width of 8 bits, so to address each bit individually (and internally), 8 lines are required.
The storage is
128K x 8bits
= 128K * 1 byte
= 128*2^10 bytes
= 131072 bytes
So answer for the first case is 7 address lines, 8 data lines and 131072 bytes.
The remaining cases are similar.
address lines is reqired in 512 k word memory
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