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Standard TTL output drive (HIGH-state sourcing ability): When a standard TTL gate output is HIGH, what current can it source and to how many typical TTL inputs can it reliably fan out?

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:

  • Standard TTL IOH(min) ≈ -400 µA (negative sign denotes sourcing).
  • Standard TTL IIH(max) ≈ 40 µA per input at HIGH.
  • Standard TTL IOL(max) ≈ 16 mA; IIL(max) ≈ -1.6 mA per input at LOW (for reference).


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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