Static-0 hazard

Also called: static 0 hazard, static-zero hazard

A hazard where an output that should stay 0 during an input change can briefly pulse to 1, typical of product-of-sums circuits.

A static-0 hazard is the mirror image of a static 1 hazard. For some single input change, the output should stay at 0, but unequal delays can make it pulse to 1 for a moment.

It shows up in product of sums circuits. An AND of OR-factors is 0 as long as at least one factor is 0. During an input change, the job of holding the output at 0 can pass from one factor to another. If the old factor becomes 1 before the new one becomes 0, every factor is briefly 1 and the output pulses high.

The fix is the dual of the static-1 fix: add the consensus factor. For factors and , it's . It doesn't change the function, but it stays at 0 throughout the hand-over and holds the output down.

On a K-map, look at the 0s instead of the 1s: two adjacent 0s covered by different POS groups, with no group covering both, mean a static-0 hazard.

			F
0	0	0	0	0
0	0	1	0	0
0	1	0	0	0
0	1	1	1	1
1	0	0	1	1
1	0	1	0	0
1	1	0	1	1
1	1	1	1	1

Worked example

Example

Finding and fixing one

F = . With A = B = 0, the factors reduce to C and , so F = , which should always be 0.

1.
C reaches the first factor directly, but reaches the second through an inverter.
2.
When C rises, the first factor becomes 1 at once. The inverter hasn't fallen yet, so the second factor is still 1. For that moment F = 1·1 = 1: a static-0 glitch.
3.
The consensus factor of and is . With A = B = 0 it's 0 and doesn't depend on C, so it holds F at 0 during the change.
4.
The truth table above shows both columns are identical: the extra factor costs a gate but leaves the function unchanged.

Common mistakes

Mixing up the names. A static-0 hazard is where the output should stay 0.
Fixing a POS hazard with an extra product term. In a POS circuit you add a sum factor.
Covering adjacent 1s to fix it. A static-0 hazard is about adjacent 0s, so look at the 0s and their POS groups.