Address bus sizing – 512 × 8 organization from 4096 total bits How many address lines are required for a 4096-bit memory organized as 512 words × 8 bits per word?
Digital Electronics
Memory and Storage
Difficulty: Easy
Choose an option
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A2
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B4
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C8
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D9
Answer
Correct Answer: 9
Explanation
Introduction / Context:Address line count is a fundamental design decision when interfacing memories. It depends on the number of addressable words (depth), not the number of bits per word (width).
Given Data / Assumptions:
- Total memory: 4096 bits.
- Organization: 512 × 8 → depth = 512 words, width = 8 bits.
- Each unique address selects one of the 512 words.
Concept / Approach:The required number of address lines A satisfies 2^A = number of addressable words. Here, words = 512. Since 512 = 2^9, we need 9 binary address lines. Width (8 bits) determines data bus size, not address count.
Step-by-Step Solution:
Compute words: 4096 bits / 8 bits per word = 512 words.Find A: 2^A = 512 → A = 9.Therefore, 9 address lines are required.Verification / Alternative check:Recall powers of two: 2^8 = 256, 2^9 = 512. Hence A = 9 is minimal and sufficient.
Why Other Options Are Wrong:
- 2, 4, 8: 2^2 = 4, 2^4 = 16, 2^8 = 256, all too small for 512 addresses.
Common Pitfalls:
- Multiplying by width when sizing address lines; address count depends only on depth.
Final Answer:9