A Boolean expression can be correct for every input and still produce a brief, unintended pulse while an input changes. This is a logic hazard: different gate-path delays let internal signals change at different times. A Karnaugh map can help spot a common static hazard, but ordinary minimization does not automatically remove hazards.
This guide works through a static-1 hazard, shows why the redundant consensus term removes it in a two-level SOP network, and explains when to consider the same issue for POS circuits.
What a static hazard means
A static-1 hazard occurs when the output is supposed to stay at 1 during a one-input transition, but a brief delay can make it dip to 0. The Boolean function has the right value before and after the transition; the glitch appears only in the physical implementation as signals pass through gates with unequal propagation delays.
A static-0 hazard is the dual case in a two-level POS implementation: the output should remain 0 but may briefly rise. These terms describe particular transition behavior. They do not mean every circuit glitch is covered by the same fix.
Worked example: F = AB + A′C
Consider the function F(A,B,C) = AB + A′C, which is 1 at minterms 1, 3, 6, and 7: F = Σm(1,3,6,7). The two product terms cover those 1s:
- A′C covers minterms 1 and 3, where A = 0 and C = 1.
- AB covers minterms 6 and 7, where A = 1 and B = 1.
Now hold B = 1 and C = 1 while A changes. At A = 0, term A′C is true; at A = 1, term AB is true. The output should remain 1 at both endpoints. But if A′C turns off before AB turns on, both product terms can momentarily be 0, creating a short low pulse.
The K-map view and the consensus term
On a three-variable K-map, minterms 3 (A = 0, B = 1, C = 1) and 7 (A = 1, B = 1, C = 1) are adjacent: only A changes. The minimized groups for A′C and AB cover these cells separately, so no one product term remains true across that transition.
Add the group covering minterms 3 and 7, which gives BC:
By the consensus theorem, BC is logically redundant in AB + A′C: it changes no truth-table output. Physically, it bridges the transition because B and C stay at 1 while A changes, so the BC path keeps the output asserted. The resulting expression is larger than the minimum SOP, but it is a hazard-aware implementation for this transition in a two-level SOP circuit.
In K-map terms, this adds an overlapping group so each adjacent pair of required 1s that must stay high during a single-input change shares at least one implicant. The group is redundant for Boolean equivalence but useful for transition coverage. This is a different goal from minimizing term count and literals; see why overlapping groups are valid and the consensus theorem.
When hazard-free logic matters
Analyze hazards when a brief pulse could be observed or acted on, especially in asynchronous control paths or combinational signals that can affect state without waiting for a clock. A synchronous design may tolerate some combinational glitches if they settle before the receiving register's setup time, but that must be established from the circuit's timing requirements rather than assumed.
- Identify transitions where exactly one input changes and the output is required to remain at 1 (for SOP) or 0 (for POS).
- Check whether each such adjacent pair of cells is covered together by a common implicant in the implementation.
- If a pair has no common term, add an appropriate redundant consensus group and recheck the truth table and timing assumptions.
Limits and next steps
This example addresses a static hazard in a two-level circuit under a single-input change. Multiple inputs changing together, reconvergent paths, multi-level factoring, and dynamic hazards require additional timing analysis. A Boolean minimizer returns an equivalent expression under its minimization objective; it is not a substitute for hazard analysis or circuit timing verification.
Review two-level and multi-level implementations, then compare the redundant term with the minimum-cover method. The combinational vs. sequential logic guide explains where K-maps fit in a larger circuit.