Using a latch for switch debouncing — contact-bounce eliminator role Evaluate: “A latch can act as a contact-bounce eliminator.”

Introduction / Context:
Mechanical switches bounce: a single press causes rapid on-off transitions before settling. Digital systems must convert this chattering waveform into a single, clean transition. Latches and simple RS configurations are classic debouncing techniques, often with RC networks or Schmitt triggers to shape edges.

Given Data / Assumptions:

• Switch contacts exhibit multiple transitions within a few milliseconds.
• A latch has two stable states and can be driven by appropriately conditioned signals.
• Power rails and reference levels are solid; no excessive noise beyond contact bounce.

Concept / Approach:
An SR latch with cross-coupled NOR or NAND gates can be driven by two switch positions so that only one valid assertion is seen at a time. Once set or reset, the latch holds its state despite brief chatter on the mechanical contacts, effectively eliminating multiple transitions at its output. Additional RC filtering or Schmitt-trigger buffering can improve noise immunity.

Step-by-Step Solution:
Map the switch poles to produce clean SET and RESET signals (often using a DPDT or steering diodes).Feed these into an SR latch (NOR-based for active-high or NAND-based for active-low logic).On a press, the first valid transition asserts SET/RESET; the latch changes state.Subsequent bounces do not change the state because the opposite input is inactive.

Verification / Alternative check:
Oscilloscope traces show a chattering switch input while the latch output presents one clean edge. Many vendor app notes present latch-based debouncers as standard solutions.

Why Other Options Are Wrong:
“Incorrect” ignores established practice. Requiring Schmitt inputs or edge-triggered clocks is unnecessary; they can help but are not mandatory for latch-based debouncing.

Common Pitfalls:
Driving both S and R simultaneously; neglecting proper pull-ups/pull-downs; forgetting that SR latches can become invalid if both inputs assert.

Final Answer:
Correct