Difficulty: Easy
Correct Answer: source 400 µA of current to no more than 10 attached gates
Explanation:
Introduction / Context: TTL outputs have asymmetric drive: they can sink substantially more current when LOW than they can source when HIGH. Correctly estimating fan-out requires using the guaranteed IOH/IOL and the corresponding IIH/IIL of the receiving inputs.
Given Data / Assumptions:
Concept / Approach: Fan-out at HIGH is the number of inputs supported by total available sourcing current divided by the worst-case HIGH input current per gate. With 400 µA available and 40 µA per input, HIGH-state fan-out ≈ 10. This is a canonical figure in many TTL references.
Step-by-Step Solution:
Compute HIGH-state fan-out: 400 µA / 40 µA ≈ 10.Interpret IOH sign: IOH is specified negative because the output is sourcing.Thus, it can source about 400 µA total to roughly 10 standard TTL inputs.Verification / Alternative check:
Datasheets for the 74xx family commonly cite a nominal fan-out of 10.Why Other Options Are Wrong:
sink 16 mA: That is LOW-state sinking capability, not HIGH-state behavior.source 16 mA: TTL cannot source anywhere near 16 mA in the HIGH state.sink a maximum of 400 µA: Reverses the roles of source/sink.Common Pitfalls:
Forgetting that TTL HIGH drive is weak compared to LOW drive.Not using worst-case IIH and IOH to ensure reliability.Final Answer:
source 400 µA of current to no more than 10 attached gates
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