8051 reset behavior: “The RST pin of the 8051 microcontroller requires a HIGH level to generate a reset.” Decide if this statement is accurate.

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Reset sequencing is critical when bringing a microcontroller into a known state. For the 8051 family, understanding the polarity and duration of the reset signal prevents intermittent startup faults and undefined behavior.


Given Data / Assumptions:

  • Device: classic 8051 family.
  • RST pin polarity and timing: active-high, pulse width requirements apply.
  • Oscillator must be running for a proper reset sequence.


Concept / Approach:
The 8051’s RST input is active-high. A logic high applied to RST for at least two machine cycles (with the clock oscillating) forces the CPU to reset registers, set PC to reset vector, and place I/O in default states. This behavior is uniform across common 8051 derivatives.


Step-by-Step Solution:

Confirm polarity: RST is active-high.Apply timing rule: keep RST high for ≥ 2 machine cycles.Ensure oscillator is active so timing is honored.On deassertion, code execution begins at the reset vector (0000h).


Verification / Alternative check:
Datasheets specify RST timing diagrams showing minimum high duration and that a low level releases reset. External RC networks or supervisor ICs are commonly used to satisfy these conditions during power-up.


Why Other Options Are Wrong:

  • Incorrect / conditional options: Suggest other polarities or dependencies (e.g., EA) that do not change RST’s active-high nature.


Common Pitfalls:
Using too-short reset pulses; relying solely on slow power ramps without a POR circuit; ignoring oscillator startup time before releasing reset.


Final Answer:
Correct

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