A certain semiconductor memory chip is specified as 2K times
Solution
part(a):
2K x 8 = 16 K.
1 word size =8 bits
then 16K/8 = 2048 words.
Total bits = 2*1024*8 = 16384 bits.
part(b):
5M x8 = 5*1024*1024*8 = 41,943,040 bits
1M x16 = 1*1024*1024*16 = 16,777,216 bits
Therefore 5M x 8 memory can store most bits.
part(c):
4K x 8 memory:
(a)No. of input/output lines = size of word = 8
(b) Address Lines = N where N is 2^N words in memory
2^N words = 4K words
Therefore N = 12 = Address Lines.
(c) Capacity in bytes = 1 Byte = 8 bits
Therefore 4k bytes = 4096 bytes
part(d) :
A3 A2 A1 A0 address lines corresponds to adress
1 0 1 0 .
where A1 and A0 corresponds to row select therefore A1A0 corresponds to \"10\" which is 2.
where A2 and A3 corresponds to column select therefore A1A0 corresponds to \"10\" which is 2.
Therefore row = 2 and column = 2 intersects at register 10.
I hope the answer satisfies your need and fulfilled your request.