Driving capability (fan-out) Approximately how many 74LS-TTL logic gate inputs can be reliably driven from the output of a single standard 74TTL gate?
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A10
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B20
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C200
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D400
Answer
Correct Answer: 20
Explanation
Introduction / Context:Fan-out is the number of standard inputs a gate output can drive while meeting guaranteed logic levels. Because different TTL subfamilies have different input currents, the fan-out depends on both the driver family and the load family.
Given Data / Assumptions:
- Driver: standard 74TTL output.
- Loads: 74LS-TTL inputs, which draw less current than standard TTL inputs.
- Use guaranteed VOL/VOH and IOL/IOH vs IIL/IIH numbers.
Concept / Approach:Fan-out (LOW) ≈ IOL(max of driver) / |IIL(max per load)|; Fan-out (HIGH) ≈ IOH(max of driver) / IIH(max per load). 74LS inputs typically require much less current than standard TTL, so a standard TTL output can drive more LS inputs than standard TTL inputs. Typical textbook/handbook values round to about 20 LS loads per standard TTL output.
Step-by-Step Solution:
Identify current capabilities of the driver (IOL/IOH).Identify input currents of 74LS loads (IIL/IIH are small).Compute fan-out for HIGH and LOW; pick the limiting value.Result aligns with ≈ 20 LS inputs from a single 74TTL output.Verification / Alternative check:Reference logic charts in textbooks: standard TTL → fan-out 10 for standard TTL loads, and about 20 when driving LS inputs due to reduced current demand.
Why Other Options Are Wrong:
- 10: Applies to standard-to-standard TTL loading, not LS loads.
- 200 / 400: Far exceed current limits; would violate VOL/VOH specs.
Common Pitfalls:Ignoring both HIGH-state and LOW-state limits; true fan-out is the minimum of the two calculations. Also, temperature and supply variations can reduce margins.
Final Answer:20