Static and dynamic hazards and their elimination

A hazard is an unwanted short pulse (glitch) at the output of a combinational circuit. It happens when an input changes and different signal paths have different propagation delays, so the output briefly takes a wrong value even though the steady-state value is correct.

Static hazard

The output should stay at the same value, but it momentarily changes and comes back.

• Static-1 hazard: output should stay 1 but goes $1 \to 0 \to 1$. Occurs in SOP (AND-OR) circuits.
• Static-0 hazard: output should stay 0 but goes $0 \to 1 \to 0$. Occurs in POS (OR-AND) circuits.

Dynamic hazard

The output should change once ($0 \to 1$ or $1 \to 0$) but it changes three or more times, e.g. $0 \to 1 \to 0 \to 1$. It happens in multi-level circuits where one input reaches the output through three or more paths of different delay. A circuit that is free of static hazards at every level (a two-level hazard-free design) is also free of dynamic hazards.

Example of a static-1 hazard

$$F = AB + \overline{A}C$$

Let $B = C = 1$. Then $F = A + \overline{A} = 1$ for both values of A, so F should stay at 1.

A ──┬──────────[AND]── AB ──┐
    │     B ───┘            [OR]── F
    └─[NOT]──A'─[AND]─ A'C ─┘
          C ────┘

When A falls from 1 to 0, AB goes to 0 at once, but A'C rises only after the inverter delay. For that short time both AND outputs are 0, so F dips to 0: a static-1 hazard.

A    ‾‾‾‾‾|_________
AB   ‾‾‾‾‾|_________
A'C  ______|‾‾‾‾‾‾‾‾   (late by inverter delay)
F    ‾‾‾‾‾|_|‾‾‾‾‾‾‾   (glitch)

Eliminating the hazard

On the K-map, the two groups AB (cells 6, 7) and A'C (cells 1, 3) sit next to each other (cells 3 and 7) but no group covers both. A move between two adjacent 1-cells that are in different groups is a hazard.

The fix is to add a redundant group (the consensus term) that covers the adjacent cells:

$$F = AB + \overline{A}C + BC$$

When $B = C = 1$, the extra term $BC = 1$ holds F at 1 while A changes, so the glitch is removed. The logic function is unchanged because $BC$ is the consensus of $AB$ and $\overline{A}C$.

Rule: in an SOP circuit, make sure every pair of adjacent 1s on the K-map is covered by at least one common group. For POS circuits, do the same with the 0s to remove static-0 hazards.