8085 has 16-bit address bus which makes _____bytes of total memory accessible.

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Examoogle · Accepted Answer

Both

Explanation: The Intel 8085 microprocessor utilizes a 16-bit address bus, which determines its maximum addressable memory space. Since each bit can represent two states (0 or 1), 16 bits provide $2^{16}$ unique memory addresses, totaling 65,536 bytes of memory.

Working Principle: The number of address lines ($n$) directly determines the addressable memory capacity according to the formula $2^n$ bytes. For $n=16$, the calculation is $2^{16} = 65,536$. Since 1024 bytes are defined as 1 kilobyte (KB), the capacity is $\frac{65536}{1024} = 64$ KB.

Key Formulas: $Memory Capacity = 2^{n}$ bytes where $n$ is the number of address lines; $1 KB = 2^{10} bytes = 1024 bytes$

Key Points: The 8085 address bus width is 16 bits.; The address space ranges from $0000H$ to $FFFFH$.; Total addressable memory is exactly 65,536 bytes.; 65,536 bytes is equivalent to 64 KB.

Advantages: Direct addressing of up to 64KB simplifies memory management.; Compatibility with byte-addressable memory architecture.

Disadvantages: Limited address space compared to modern 32-bit or 64-bit processors.; Requires bank switching if memory needs exceed 64KB.

Applications: Embedded controller designs.; Educational hardware for microprocessor assembly language programming.; Legacy industrial control systems.

Additional Info: Option A is mathematically correct as $2^{16}$ defines the address space.. Option B is the standard shorthand notation for 65,536 bytes in binary computer architecture.. Both A and B are equivalent representations, making C the most accurate choice.

Answer: C is correct — Both $2^{16}$ and 64KB are equivalent representations of the total memory addressable by a 16-bit address bus.

Exam Tip: Remember that for any microprocessor, if the address bus has $n$ lines, the total memory capacity is always $2^n$ bytes. This rule is fundamental for all CPU architecture questions.

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