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Question

In 8051 microcontroller, one machine cycle lasts in 12 oscillation period. Crystal oscillator's frequency is 11.0592 MHz. Find how long it takes to execute instruction MOV R4, # 25.

This question was previously asked in
UGC NET 2023 Paper 1 Question Paper (22-Jun-2023) (Shift 2)
The correct answer is

1.085 µs

The timing rule for the 8051. Instruction times are quoted in machine cycles, and in the standard 8051 one machine cycle occupies 12 oscillator periods (it comprises six states S1–S6, each of two oscillator phases).

Step 1 — find the machine-cycle time from the crystal frequency.

\(T_{mc} = \dfrac{12}{f_{osc}}\)

Step 2 — substitute fosc = 11.0592 MHz.

\(T_{mc} = \dfrac{12}{11.0592\times10^{6}} = 1.085\times10^{-6}\ \text{s} = 1.085\ \mu s\)

Equivalently, the machine-cycle frequency is \(11.0592\ \text{MHz}/12 = 921.6\ \text{kHz}\), whose period is 1.085 µs.

Step 3 — how many machine cycles does the instruction take? MOV R4, #25 loads an immediate constant into a register-bank register. It is a 2-byte instruction that executes in one machine cycle, so:

\(t = 1 \times 1.085\ \mu s = 1.085\ \mu s\)

Why 11.0592 MHz is the classic 8051 crystal. Dividing it by 12 and then by 32 gives exactly 28800, so the serial port can generate standard baud rates (9600, 19200, …) with zero error — which is why almost every 8051 timing question uses this value and the resulting 1.085 µs machine cycle.

Useful reference points. Most 8051 instructions take 1 or 2 machine cycles; MUL AB and DIV AB take 4. So a 2-cycle instruction here would take 2.17 µs, and MUL AB would take 4.34 µs.

Hence, the instruction MOV R4, #25 takes 1.085 µs to execute.

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