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A 16-bit microprocessor has twenty address lines (A0 to A19) and 16 data lines. The higher eight significant lines of the data bus of the processor are tied to the 8-data lines of a 16 Kbyte memory that can store one byte in each of its 16K address locations. The memory chip should map onto contiguous memory locations and occupy only 16 Kbyte of memory space. Which of the following statement(s) is/are correct with respect to the above design?
1. If the 16 Kbyte of memory chip is mapped with a starting address of 80000H, then the ending address will be 83FFFH.
2. The active high chip-select needed to map the 16 Kbyte memory with a starting address at F0000H is given by the logic expression (A19 · A18 · A17 · A16).
3. The 16 Kbyte memory cannot be mapped with contiguous address locations with a starting address as 0F000H using only A19 to A14 for generating chip select.
4. The above chip cannot be interfaced as the width of the data bus of the processor and the memory chip differs.

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Correct Answer - Option :

Concept:

The type of microprocessor → 16 bit

Data bus length → 16 bit

Number of address lines→ 20 (A19 – A0)

Memory chip connected → 16 kB

Given condition is memory chip of 16 kB should be mapped onto contiguous memory

Locations. For a 16 kB memory chip, number of address lines required,

2n = N = 16 kB = 214

n = 14 i.e. A13 – A0

The remaining 6 lines can be used for chip select i.e. A19 – A14.

As per option, if the starting address is 0F000H i.e.

A19 A18 A17 A16 A15 A14 A 13A12 A11................A0..           .

0     0     0     0     1    1     1    1     0   0   0   0  0

Chip select lines:

The ending for starting address of 0F000H is 12FFFH but the chip select lines get modified.

If starting addressing 80000H then the ending address would be 83FFFH which does not

Change A19 to B14 lines i.e. 100000.

For 16 kB to find the ending address 3FFF H should be added to starting address

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