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GATE 2016 EC (EC2) – Question 52

Digital Circuits · Computer Organization · 2 marks · Multiple choice

An 8 Kbyte ROM with active-low chip select $\overline{CS}$ is used with an 8085. It must occupy addresses 1000H–2FFFH. Address bits run from $A_{15}$ (most significant) to $A_0$. Which expression generates the correct select signal?

  1. $A_{15}+A_{14}+(A_{13}A_{12}+\bar A_{13}\bar A_{12})$
  2. $A_{15}A_{14}(A_{13}+A_{12})$
  3. $\bar A_{15}\bar A_{14}(A_{13}\bar A_{12}+\bar A_{13}A_{12})$
  4. $\bar A_{15}+\bar A_{14}+A_{13}A_{12}$

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Correct answer: (A) $A_{15}+A_{14}+(A_{13}A_{12}+\bar A_{13}\bar A_{12})$

Explanation

The selected high nibbles are binary 0001 and 0010. Thus $A_{15}=A_{14}=0$, and $A_{13},A_{12}$ must differ. Lower 12 bits range freely within each 4 Kbyte block.

Active-low select must be zero exactly for those conditions. The OR of the two upper bits with the XNOR of the next pair has that property:
$$\overline{CS}=A_{15}+A_{14}+A_{13}A_{12}+\bar A_{13}\bar A_{12}.$$
This is **A**.

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