Asked by bode
The following RAM chips are specified by the number of words times the number of bits per word. For each, specify:
a. The number of address lines needed for each memory chip
b. The number of data lines needed for each memory chip
c. The number of bytes that can be stored in each memory chip (note that there are 8 bits/byte)
128 KB words x 8 bits/word
512 KB words x 16 bits/word
512 MB words x 32 bits/word
8 GB words x 64 bits/word
a. The number of address lines needed for each memory chip
b. The number of data lines needed for each memory chip
c. The number of bytes that can be stored in each memory chip (note that there are 8 bits/byte)
128 KB words x 8 bits/word
512 KB words x 16 bits/word
512 MB words x 32 bits/word
8 GB words x 64 bits/word
Answers
Answered by
MathMate
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.
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.
Answered by
samrti
address lines is reqired in 512 k word memory
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