Difficulty: Easy
Correct Answer: fanout
Explanation:
Introduction / Context:When interconnecting digital devices, a single output often drives multiple inputs. Each input draws some current (or presents capacitance), and the driving gate has limited sourcing/sinking capability. The term that quantifies “how many loads” can be connected is fundamental to reliable system design.
Given Data / Assumptions:
Concept / Approach:Fanout is defined as the number of identical inputs that a single output can drive while guaranteeing valid logic levels. Engineers compare the driver’s guaranteed output current to the sum of the receivers’ required input currents to determine maximum loading for both logic states.
Step-by-Step Solution:
Check datasheet: obtain I_OH and I_OL for the driver, I_IH and I_IL for the receivers.Compute HIGH-state fanout: Fanout_H ≈ |I_OH| / |I_IH|.Compute LOW-state fanout: Fanout_L ≈ |I_OL| / |I_IL|.Choose the minimum of the two as the allowable fanout to ensure both states meet specs.Verification / Alternative check:Classic TTL example: I_OH ≈ −0.4 mA and I_IH ≈ 40 µA → ≈10 loads; I_OL ≈ 16 mA and I_IL ≈ 1.6 mA → ≈10 loads, yielding a fanout of 10.
Why Other Options Are Wrong:
Propagation delay: time for a signal to traverse a gate; unrelated to count of loads.Noise margin: allowable noise amplitude before errors, not a load-count metric.Speed-power product: a figure of merit balancing delay and power, not loading capacity.Common Pitfalls:Ignoring capacitive loading and rise/fall-time degradation—fanout is a DC specification; dynamic performance may limit practical loading further.
Final Answer:fanout
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