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Hazard types and hazard cover in K-maps

Short answer

A hazard is a momentary glitch caused by unequal propagation delays. Static-1 hazard: output should stay 1 but briefly goes to 0 in SOP circuits; static-0: stays 0 but glitches to 1 in POS; dynamic: multiple transitions. Hazard cover: add a redundant prime implicant covering adjacent 1s not in the same group (e.g., add BC to F=AB+A'C) to eliminate static hazards.

What is hazard? Explain types of hazards with hazard cover techniques used in K-map simplification.

A hazard is an unwanted momentary pulse (glitch) at the output of a combinational circuit, caused by unequal propagation delays along different paths when an input changes. The steady-state output is correct, but the transient is wrong; in sequential systems such glitches can trigger flip-flops falsely.

Types of hazards

TypeExpected outputActual outputUsually found in
Static-1stays 11→0→11 \to 0 \to 1SOP (AND-OR) circuits
Static-0stays 00→1→00 \to 1 \to 0POS (OR-AND) circuits
Dynamicone change, 0→10 \to 10→1→0→10 \to 1 \to 0 \to 1multi-level circuits
  • Static hazard: the output should not change, but it briefly changes and returns.
  • Dynamic hazard: the output should change once but changes several times, because the changing input reaches the output by three or more paths of different delay.

Hazard cover in K-maps (example)

F=AB+A‾CF = AB + \overline{A}C
        BC
A      00  01  11  10
     +---+---+---+---+
 0   | 0 | 1 | 1 | 0 |   A'C  = cells 1, 3
     +---+---+---+---+
 1   | 0 | 0 | 1 | 1 |   AB   = cells 7, 6
     +---+---+---+---+

Cells 3 (011) and 7 (111) are adjacent 1s but lie in different groups. With B=C=1B = C = 1, F should stay 1 while A changes. When A goes 1→01 \to 0, the term ABAB falls at once but A‾C\overline{A}C rises only after the inverter delay, so F momentarily becomes 0: a static-1 hazard.

Hazard cover technique: add a redundant group that covers every pair of adjacent 1s not already in a common group. Here the extra group is cells 3 and 7, i.e. the consensus term BCBC:

F=AB+A‾C+BCF = AB + \overline{A}C + BC

While A changes with B=C=1B = C = 1, BC=1BC = 1 keeps F at 1, so the glitch disappears. The function is unchanged (consensus theorem: AB+A‾C+BC=AB+A‾CAB + \overline{A}C + BC = AB + \overline{A}C), only the circuit gains one gate.

Rules

  1. For SOP circuits: every two adjacent 1-cells must be covered by at least one common prime implicant (removes static-1 hazards).
  2. For POS circuits: every two adjacent 0-cells must be covered by a common sum term (removes static-0 hazards).
  3. A two-level circuit built with full hazard cover is also free of dynamic hazards; dynamic hazards are avoided by not using multi-level paths for the same variable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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