Static Hazards

A static hazard in a combinational circuit drives a glitch into a latch enable pin or a multiplexer select line at exactly the wrong moment, causing a wrong bit to be stored. Every K-map simplification exercise in a GATE Digital Design question is really a hazard elimination exercise in disguise.

Core Concept

A static hazard is a momentary wrong output that occurs in a combinational circuit when a single input variable changes, even though the output should remain constant. A Static-1 hazard causes the output to glitch to 0 briefly during a 1-to-1 transition. A Static-0 hazard causes a glitch to 1 during a 0-to-0 transition.

Static hazards appear at the boundary between adjacent K-map groups that are not bridged by an overlapping group. When the input transitions from a cell covered by one group to a cell covered only by the other, there is a brief moment where neither gate has a 1 output, causing the OR gate to output 0 momentarily.

The standard fix is to add a consensus term — a redundant K-map group that spans the boundary between the two original groups. In hardware this is an extra AND gate feeding the sum-of-products OR gate. The 74LS08 AND gate (propagation delay 14 ns) and 74LS32 OR gate (14 ns) are the usual SOP implementation components. The consensus gate ensures at least one gate is always outputting 1 during the transition.

Boolean Expression

For the function F = A'C + AB (which has a static-1 hazard when A transitions from 1→0 with B=1, C=1), the consensus theorem gives the hazard-free expression: F = A'C + AB + BC. The added term BC is the consensus of A'C and AB with respect to A. It covers the transition cells and eliminates the hazard. In SOP, static-0 hazards are eliminated similarly using consensus in the POS form.

Given:
F(A,B,C) = Σm(0,1,2,3,5,6) in 3-variable K-map
Simplified SOP (two groups): F = A'  + BC
Check for static-1 hazard at transition:
  From m(2): A=0,B=1,C=0 → F=1 (covered by A')
  To   m(6): A=1,B=1,C=0 → F=1 (covered by BC)

Hazard check:
During A: 0→1 transition with B=1,C=0:
  A' goes 1→0 (A' gate output falls)
  BC = 1*0 = 0 (BC gate output stays 0)
  OR output: 0 OR 0 = 0  (GLITCH!)

Formula / Rule:
Add consensus term: term1=A', term2=BC
Consensus w.r.t A → eliminate A' variable:
Consensus = remaining literals = B * C... wait:
Consensus of A' and BC: complement A from each:
  From A': drop A' → remaining = 1
  From BC: keep BC as-is
  Consensus = BC  — already in expression
Actually check: consensus of (A')(BC) = B*C... 
Re-examine: A' covers m(0,1,2,3); BC covers m(3,5,6,7)
Boundary at m(2,6): A=0,B=1,C=0 to A=1,B=1,C=0
Add consensus group B (covers m(2,3,6,7)):
F_hazard_free = A' + BC + B

Final Answer:
F = A' + B  (BC is redundant once B is added)
No static-1 hazard remains.

Exam Tip: Static hazards only occur at input transitions where exactly one variable changes and the output stays constant. They are found at K-map group boundaries not covered by a shared overlapping group. In SOP form, adding the consensus term eliminates the static-1 hazard. In POS form, adding a consensus term eliminates the static-0 hazard. Anna University questions often ask you to identify the hazardous input transition and write the hazard-free expression — always state which literal the consensus is taken with respect to.

Key Properties

• Static-1 hazard: output glitches to 0 during a 1→1 output transition
• Static-0 hazard: output glitches to 1 during a 0→0 output transition
• Occurs at K-map boundaries between non-overlapping adjacent groups
• Fix: add redundant consensus group bridging the two groups in the K-map
• Gate ICs involved: 74LS08 AND (14 ns), 74LS32 OR (14 ns) for SOP implementation
• Hazard pulses last for the duration of one gate's propagation delay difference
• Glitches are harmless in purely combinational outputs but destructive if feeding a latch or edge-triggered register

Quick Revision

• Static hazards cause momentary wrong output during a constant-output input transition
• Static-1 hazard in SOP; Static-0 hazard in POS form
• Found where two K-map groups share no overlapping cells at the boundary
• Fix: add the consensus term (redundant group) to the expression
• Consensus of terms P and Q: drop the complementary literal and AND the rest
• Only one input variable changes during a hazardous transition
• Static hazards are distinct from dynamic hazards which involve multiple transitions
• Exam trap: adding the consensus group to the K-map but forgetting to include the term in the final Boolean expression