How many 256X8 ROMs are needed to make a 4096×8 ROM and what kind of decoder should be used

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QUESTION 4 4. How many 256X8 ROMs are needed to make a 4096×8 ROM and what kind of decoder should be used? (Use only 256X8 ROMs) A # of 256×8 ROMs : 4, type of decoder : 2×4 OB t of 256X8 ROMs : 8, type of decoder: 3X8 c, # of 256×8 ROMs : 16, type of decoder: 3X8 D # of 256X8 ROMs : 16, type of decoder : 4×16 QUESTION 5 5. To compose a 4 096×1 mux from 8×1 muxes only, how many select inputs are needed and how many 8X1 muxes are needed? @ A select inputs: 3-bits, # of 8X1 muxes : 512 select inputs: 4-bits, # of 8X1 muxes : 585 “select inputs: 12-bits, # of 8×1 muxes : 512 select inputs: 12-bits, # of 8X1 muxes : 585 B D QUESTION 6 6. To compose a 2048X64 ROM using only 256×8 ROMs, how many 256X8 ROMs are needed and what kind of decoder should be used? A # of 256×8 ROMs : 32, type of decoder: 2×4 ROMs : 64, type of decoder : 3X8 cht of 256X8 ROMs : 32, type of decoder: 3X8 D # of 256×8 ROMs : 64, type of decoder: 6×64 ® B. # of 256X8 Click Save and Submit to save and submit. Clhick Save All Answers to save all ansuens! search

The Correct Answer and Explanation is:

QUESTION 4

Question: How many 256×8 ROMs are needed to make a 4096×8 ROM, and what kind of decoder should be used?

  • A 256×8 ROM has 256 addresses, each storing 8 bits.
  • A 4096×8 ROM requires 4096 addresses, each storing 8 bits.

So, we need: 4096256=16 ROMs\frac{4096}{256} = 16 \text{ ROMs}

To select among 16 ROMs, we need a decoder that takes 4 input lines and activates one of the 16 output lines. That means a 4-to-16 decoder is required.

✅ Correct Answer: D — 16 ROMs and a 4×16 decoder

QUESTION 5

Question: To compose a 4096×1 multiplexer using only 8×1 multiplexers, how many select inputs are needed and how many 8×1 muxes are needed?

  • A 4096×1 mux selects 1 of 4096 inputs.

\log_2(4096) = 12 \Rightarrow \text{12 select lines}
]

To build a 4096×1 mux from 8×1 muxes:

  • Each 8×1 mux selects from 8 inputs using 3 select lines.
  • You build a multilevel tree structure: Total muxes=4096/8+512/8+64/8+8/8+1=512+64+8+1=∗∗585∗∗\text{Total muxes} = 4096 / 8 + 512 / 8 + 64 / 8 + 8 / 8 + 1 = 512 + 64 + 8 + 1 = **585**

✅ Correct Answer: D — 12 select inputs, 585 8×1 muxes

QUESTION 6

Question: To compose a 2048×64 ROM using only 256×8 ROMs, how many are needed and what decoder is used?

  • A 256×8 ROM has 256 addresses and 8 bits per word.
  • We need 2048 addresses, each with 64 bits.
    So total required storage = 2048 × 64 bits.

Each 256×8 ROM provides: 256 × 8 = 2048 bits.

Total bits needed: 2048×64=131072 bits2048 × 64 = 131072 \text{ bits} 1310722048=64 ROMs\frac{131072}{2048} = 64 \text{ ROMs}

To address 2048 words, you need a decoder to select among 8 blocks (2048/256 = 8): log⁡2(8)=3⇒3×8 decoder\log_2(8) = 3 \Rightarrow \text{3×8 decoder}

✅ Correct Answer: B — 64 ROMs and a 3×8 decoder

Final Answers Summary:

  • Q4: D — 16 ROMs, 4×16 decoder
  • Q5: D — 12 select inputs, 585 8×1 muxes
  • Q6: B — 64 ROMs, 3×8 decoder

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