Difficulty: Easy
Correct Answer: There is a signal loss to all gates on the node, and the affected node will be stuck in the LOW state.
Explanation:
Introduction / Context:
In multi-fanout digital nets, several load (receiving) gate inputs share one driving source. If one load input becomes shorted to ground (or otherwise abnormally low impedance), the entire node can be clamped, preventing valid logic transitions and propagating failure across otherwise healthy loads. Recognizing this signature enables quick isolation of the faulty part.
Given Data / Assumptions:
Concept / Approach:
A shorted input presents a very low resistance to ground (or to Vcc, depending on the fault). The shared node voltage is pulled hard toward that rail. As a result, the driver cannot change the node level effectively; all other loads on the node lose their signal drive. With a ground short, the node appears permanently LOW, which then forces every downstream stage to interpret a constant 0 level.
Step-by-Step Solution:
Verification / Alternative check:
Use an ohmmeter (with power removed) to measure from node to ground/Vcc; an abnormally low resistance indicates a short. In-circuit, monitor current draw or use a thermal camera to spot the culprit IC.
Why Other Options Are Wrong:
Only the output of the defective gate is affected: false; the whole node is compromised.
Signal loss only, without stating stuck level: incomplete; the dominant symptom is also a stuck LOW for a ground short.
Stuck LOW only, without noting global signal loss: incomplete; it misses the network-wide consequence on all connected loads.
Common Pitfalls:
Assuming the driver is weak or faulty; forgetting that any single shorted load can overpower the driver and clamp the net; overlooking that a Vcc short would clamp HIGH instead.
Final Answer:
There is a signal loss to all gates on the node, and the affected node will be stuck in the LOW state.
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