The program counter in 8085 micro-processor is a 16-bit register, because_____.

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There are 16 address lines

Explanation: The Program Counter (PC) in the 8085 microprocessor is a 16-bit register because it must hold the memory address of the next instruction to be executed. Since the 8085 has a 16-bit address bus, it can address $2^{16} = 65,536$ unique memory locations (64 KB).

Working Principle: The microprocessor uses the address bus to communicate with external memory. Because the 8085 is a 16-bit addressable machine, the PC must be 16 bits wide to generate the full physical address required to select any location within the 64 KB memory map. When the processor executes an instruction, the PC provides the address on the address bus ($A_{15}-A_0$).

Key Formulas: $Memory_{capacity} = 2^{n}$ where $n$ is the number of address lines; $Addressable Range = 2^{16} = 64KB$

Key Points: The 8085 has 16 address pins ($A_{15}$ to $A_{0}$), enabling it to access 64KB of memory.; The Program Counter always points to the next instruction to be fetched.; The PC is automatically incremented by the microprocessor after every opcode fetch.

Advantages: Allows direct addressing of up to 64KB memory space.; Enables structured program flow control.

Disadvantages: Cannot address more than 64KB without bank switching or external logic.; Consumes two internal 8-bit registers ($PCH$ and $PCL$) to store a single address.

Applications: Instruction sequencing; Program flow management; Subroutine handling via stack operations

Additional Info: Option A is incorrect because the PC is not a counter for data bits, but a pointer for addresses.. Option C is incorrect as the stack pointer or general-purpose register pairs (BC, DE, HL) are used for data, not the PC.. Option D is incorrect; fetching is determined by the control unit and timing logic, not the bit-width of the PC itself.

Answer: B is correct — The 8085 has a 16-bit address bus capable of addressing 64KB of memory, necessitating a 16-bit Program Counter.

Exam Tip: Always remember: Number of addressable locations = $2^{	ext{number of address lines}}$. For any microprocessor, the PC width matches the Address Bus width.

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