TTL fan-out and output HIGH level: In a TTL circuit, if too many load gate inputs are connected, what happens to the driver’s HIGH-level output?

Difficulty: Medium

Correct Answer: VOH drops below VOH(min)

Explanation:


Introduction / Context:
Fan-out is the number of standard inputs a logic output can reliably drive. Exceeding fan-out loads the driver beyond its guaranteed current sourcing/sinking capability, risking violation of specified logic levels. This question targets the effect on the output HIGH level of a TTL driver overloaded with inputs.


Given Data / Assumptions:

  • Each TTL input requires a defined input current at HIGH (IIH) and at LOW (IIL).
  • The driving output specifies VOH(min) at a guaranteed IOH(max).
  • Connecting excessive inputs increases current demand above IOH(max).


Concept / Approach:
When IOH demand exceeds the driver’s rating, the output transistor and pull-up network cannot maintain the guaranteed HIGH voltage. Therefore, the actual VOH falls below its guaranteed minimum (VOH(min)), potentially entering the undefined region between VIH(min) and VIL(max) and causing logic errors down the chain.


Step-by-Step Solution:
Compute total IIH = IIH(per input) * number of inputs.Compare to IOH(max) of the driving gate.If total IIH > IOH(max), the driver VOH sags below VOH(min).


Verification / Alternative check:
Datasheet fan-out = IOH(max) / IIH(max). Observe VOH on a scope under overload; it droops, confirming the violation.


Why Other Options Are Wrong:
Option A reverses the relationship; VOH(min) is a spec limit, not a measured value that “drops.”Options C and D contradict the overload effect.Option E is an input threshold parameter, not directly affected by the driver’s overload.


Common Pitfalls:
Ignoring IIH at HIGH and considering only LOW-state fan-out; both states must be checked for reliable design margins.


Final Answer:
VOH drops below VOH(min)

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