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Question

Calculate the time required to execute the entire instruction cycle if two machine codes, 0011 1110 and 0011 0010, are stored in memory locations 2000H and 2001H, respectively.

If the clock frequency is 2 MHz, the first machine code represents opcode to load data byte in the accumulator and the second code represents data to be loaded in the accumulator.

The correct answer is

3.5 μs

Instruction Cycle Execution Time Calculation

To accurately determine the time required to execute an entire instruction cycle, we need to understand the components of the instruction, the number of machine cycles and T-states involved, and the system's clock frequency.

Clock Period Determination

The clock frequency is a fundamental parameter that dictates the duration of each clock cycle (T-state).

  • The given clock frequency (\(f\)) is \(2\) MHz.
  • We know that \(1\) MHz is equal to \(10^6\) Hz.
  • Therefore, \(f = 2 \times 10^6\) Hz.

The clock period (\(T\)), which is the time duration of one T-state, is calculated as the reciprocal of the clock frequency:

\[T = \frac{1}{\text{frequency}}\]

Substituting the given frequency value:

\[T = \frac{1}{2 \times 10^6 \text{ Hz}}\]

\[T = 0.5 \times 10^{-6} \text{ seconds}\]

\[T = 0.5 \text{ μs}\]

So, each T-state lasts for \(0.5\) microseconds.

Instruction Analysis and T-States

The problem describes two machine codes stored sequentially in memory:

  • Memory location 2000H contains 0011 1110. This is identified as the opcode to load a data byte into the accumulator. In microprocessors like the 8085, this corresponds to an MVI A, data (Move Immediate to Accumulator) instruction.
  • Memory location 2001H contains 0011 0010. This is the actual data byte that will be loaded into the accumulator.

An MVI A, data instruction is a 2-byte instruction. Its execution typically involves two machine cycles in a microprocessor architecture like the 8085:

  1. Opcode Fetch (OF) Cycle: This cycle fetches the first byte, which is the opcode (0011 1110), from memory address 2000H. An Opcode Fetch cycle is crucial for the processor to identify what operation needs to be performed. This cycle typically consumes 4 T-states.
  2. Memory Read (MR) Cycle: This cycle fetches the second byte, which is the data (0011 0010), from memory address 2001H. This data will be loaded into the accumulator. A Memory Read cycle typically consumes 3 T-states.

To find the total number of T-states required for the entire instruction cycle, we sum the T-states of these two machine cycles:

Total T-states = T-states (Opcode Fetch) + T-states (Memory Read)

Total T-states = \(4 \text{ T-states} + 3 \text{ T-states}\)

Total T-states = \(7 \text{ T-states}\)

Execution Time Calculation for Instruction Cycle

With the total number of T-states and the clock period determined, we can now calculate the total time required to execute the entire instruction cycle:

Execution Time = Total T-states \(\times\) Clock Period

Execution Time = \(7 \times 0.5 \text{ μs}\)

Execution Time = \(3.5 \text{ μs}\)

Final Conclusion

Based on the analysis of the instruction (MVI A, data) and the given clock frequency, the time required to execute the entire instruction cycle is \(3.5 \text{ μs}\).

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Important Questions from Instruction Set of 8085

  1. The technique of assigning a memory address to each I/O device in the computer system is called -

  2. In an 8085 microprocessor, the instruction CMP B has been executed while the contents of accumulator is less than that of register B. As a result, carry flag and zero flag will be respectively -

  3. A fetch cycle is the

  4. Determine the contents of accumulator if the instruction RAL is executed twice. Assume the contents of accumulator is AAH and CY = 0.

  5. Following instruction is executed in 8085, LDB 4000H

    Identify the correct statement.

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