What is the decimal number 75 in binary and hexadecimal?
1 answer:
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Answer:
Required memory size is 16k x 8
16k = 24 x 210 = 214
Hence, No. of address lines = 14
No. of data lines = 8
a) Size of IC 1024 x 1
Total number of ICs required = 16k x 8 / 1024 x 1 = 16 x 8 = 128
b) Size of IC 2k x 4
Total number of ICs required = 16k x 8 / 2k x 4 = 8 x 2 = 16
c) Size of IC 1k x 8
Total number of ICs required = 16k x 8 / 1k x 8 = 16 x 1 = 16
Explanation:
For a, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1024 memory locations present in a IC.
For b, 11 address lines from A0 to A10 are used to select any one of the memory location out of 2k=2048 memory locations present in a IC.
For c, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1k=1024 memory locations present in a IC.
