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A microprocessor with a 16-bit address bus is used in a linear memory selection configuration (i.e., Address bus lines are directly used as chip selects of memory chips) with 4 memory chips. The maximum addressable memory space is
  • a)
    64 K
  • b)
    16 K
  • c)
    8 K
  • d)
    4 K
Correct answer is option 'A'. Can you explain this answer?
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A microprocessor with a 16-bit address bus is used in a linear memory ...
Explanation:

Given, a microprocessor with a 16-bit address bus is used in a linear memory selection configuration with 4 memory chips.

Linear Memory Selection Configuration:

In the linear memory selection configuration, the address bus lines are directly used as chip selects of memory chips. Here, each memory chip is selected with its own set of address lines.

In this configuration, the memory chips are connected in a linear fashion, i.e., the output of one chip is connected to the input of the next chip. Hence, the configuration is also called a daisy-chain configuration.

Addressable Memory Space:

The addressable memory space is the maximum amount of memory that can be accessed by the microprocessor. It is determined by the number of address lines available in the microprocessor.

In this case, the microprocessor has a 16-bit address bus, which means it has 2^16 = 65,536 possible addresses.

However, in the linear memory selection configuration with 4 memory chips, each chip is selected with 4 address lines. Hence, only the first 4 address lines are used for chip selection, and the remaining 12 address lines are used for accessing the memory locations within the selected chip.

Therefore, each memory chip can access a maximum of 2^12 = 4,096 memory locations.

Since there are 4 memory chips, the total addressable memory space is:

4 x 4,096 = 16,384 bytes = 16 Kb

However, the question asks for the maximum addressable memory space, which is given by the total number of possible addresses, i.e., 65,536 bytes = 64 Kb.

Hence, the correct answer is option 'A', 64 Kb.
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