Memory Mapping MCQ Quiz in मल्याळम - Objective Question with Answer for Memory Mapping - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Binary  adder has 11 inputs i.e (5 + 5 + C in)

It has 6 outputs i.e (5 + Cout)

∴ Size of ROM = 2¹¹ × 6 = 2048 × 6

chip select for chip 1 ${\mathrm{A}}_8=1{\mathrm{A}}_9=0$

        $$\begin{gathered} \mathrm{C}\mathrm{S} \\ {\mathrm{A}}_9{\mathrm{A}}_8 \end{gathered}$$      ${\mathrm{A}}_7-{\mathrm{A}}_0$                 Address(in Hexadecimal)

        01      00000000                   0100

        01      11111111                     01FF

chip select for chip 2 ${\mathrm{A}}_8=0{\mathrm{A}}_9=1$

      $$\begin{gathered} \mathrm{C}\mathrm{S} \\ {\mathrm{A}}_9{\mathrm{A}}_8 \end{gathered}$$      ${\mathrm{A}}_7-{\mathrm{A}}_0$                  Address(in Hexadecimal)      

      10      00000000                    0200

      10      11111111                     02FF

Also see that ${\mathrm{A}}_{15}-{\mathrm{A}}_{10}$ are don't care.

Thus, in $\mathrm{F}800–\mathrm{F}9\mathrm{F}\mathrm{F}$, the range from $111110(0000000000)$ to $111110(0011111111)$ does not lie in the range of chip 1 and chip 2. The remaining address lies in the range. Thus, the range $\mathrm{F}800–\mathrm{F}9\mathrm{F}\mathrm{F}$ does not contain all the valid addresses.