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Question

The period of machine cycle of an 8051 system with crystal frequency 16 MHz is

The correct answer is

0.75 μs

8051 Machine Cycle Basics

The 8051 system is a popular 8-bit microcontroller widely used in embedded applications. A fundamental concept in understanding the operation and speed of the 8051 is its machine cycle. A machine cycle represents the time taken by the microcontroller to complete one basic operation or step of an instruction. For the standard 8051 microcontroller architecture, one machine cycle is defined as 12 oscillator periods. This characteristic is critical for timing calculations.

Crystal Frequency and Oscillator Period Relationship

The crystal frequency is the primary clock source for the 8051 system. An external crystal oscillator provides a precise clock signal, which dictates the internal timing of the microcontroller. The frequency of this crystal directly influences the speed at which the 8051 operates.

The relationship between frequency and period is inverse, as described by the formula:

$$ \text{Period } (T) = \frac{1}{\text{Frequency } (f)} $$

In this problem, the given crystal frequency is 16 MHz. This frequency allows us to determine the duration of a single oscillator period.

  • Given: Crystal Frequency (\(f_{\text{crystal}}\)) = 16 MHz = \(16 \times 10^6\) Hz

Calculating the 8051 Machine Cycle Period

To find the period of the machine cycle for the 8051 system with a 16 MHz crystal frequency, we follow these two steps:

  1. Calculate the Oscillator Period (\(T_{\text{oscillator}}\)):

    The oscillator period is the time taken for one complete cycle of the crystal oscillation.

    $$ T_{\text{oscillator}} = \frac{1}{f_{\text{crystal}}} $$ $$ T_{\text{oscillator}} = \frac{1}{16 \times 10^6 \text{ Hz}} $$ $$ T_{\text{oscillator}} = 0.0625 \times 10^{-6} \text{ seconds} $$ $$ T_{\text{oscillator}} = 0.0625 \text{ } \mu\text{s} $$
  2. Calculate the Machine Cycle Period (\(T_{\text{machine cycle}}\)):

    As stated earlier, one machine cycle in an 8051 takes 12 oscillator periods.

    $$ T_{\text{machine cycle}} = 12 \times T_{\text{oscillator}} $$ $$ T_{\text{machine cycle}} = 12 \times 0.0625 \text{ } \mu\text{s} $$ $$ T_{\text{machine cycle}} = 0.75 \text{ } \mu\text{s} $$

Therefore, the period of the machine cycle for an 8051 system operating with a 16 MHz crystal frequency is 0.75 µs. This calculation is vital for accurately timing instruction execution and peripheral operations within the 8051 microcontroller.

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Important Questions from Microprocessors and Microcontrollers

  1. Cycle stealing mode of DMA operation involves

  2. Which of the following instructions will move the contents of register 3 to the accumulator ?

  3. In 8086, which instruction at the end of a sub-routine takes the execution back to the main program?

  4. Which one of the following is not the feature of 8051?

  5. What is the operation performed by the following assembly language program of 8051?

               CLR        A

                 MOV       R1, # 100H

                 MOV       R7, # 21H

    AGAIN: MOV      @ R1, A

                 INC         R1

                 DJNC      R7, AGAIN

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