Digital Electronics - Memory and Storage - Discussion
Discussion Forum : Memory and Storage - General Questions (Q.No. 57)
57.
How many address bits are required for a 4096-bit memory organized as a 512 × 8 memory?
Discussion:
4 comments Page 1 of 1.
Velkumargn said:
1 decade ago
2^ Adderss line * Data line = Memory Location.
2^9*8 = 512*8.
Answer is 9.
2^9*8 = 512*8.
Answer is 9.
Vallivel said:
1 decade ago
Can you explain little bit briefly.. Little bit confused with your answer.
4096 --> 4k So we can use 8 * 512k => 4k right???
4096 --> 4k So we can use 8 * 512k => 4k right???
Jitendra yadav said:
1 decade ago
16,384 byte= 2 the power 14 therefore need 14 address line to access each byte
same as 2 the power 9 will equal to 512
same as 2 the power 9 will equal to 512
Bharath said:
1 decade ago
How do we need 9 bits. Can anyone explain me?
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