TTL input sink current calculation A TTL NAND gate specifies IIL(max) = –1.6 mA per input. If its output in the LOW state drives eight standard TTL inputs, how much current must the driving output sink?

Introduction / Context:
Fan-out in TTL logic is limited by the amount of current a driving output can sink (LOW) or source (HIGH) while maintaining valid logic levels. Understanding the IIL(max) specification per input allows you to compute the total sink requirement placed on a driver's LOW output level.

Given Data / Assumptions:

• IIL(max) per TTL input = –1.6 mA (current into the input when it is receiving a LOW).
• Number of inputs driven = 8.
• Driver output is LOW (thus it must sink the sum of input currents).

Concept / Approach:

When a TTL output is LOW, current flows from each receiving gate into the driver. The total sink current is the sum of all input currents. Multiply the per-input IIL(max) by the number of inputs to get the worst-case sink demand on the driver.

Step-by-Step Solution:

Verification / Alternative check:

Compare to standard TTL driver capability (IOL(max) for 74xx is typically around 16 mA). The computed 12.8 mA is within this limit, consistent with a fan-out rating near 10 for many families.

Why Other Options Are Wrong:

• –8 mA: assumes only five loads or a smaller IIL.
• –1.6 mA: counts a single load, not eight.
• –25.6 mA: doubles the correct value; would exceed many standard TTL driver limits.

Common Pitfalls:

• Forgetting that the worst-case calculation uses IIL(max), not a typical value.
• Mixing source and sink specs; LOW requires sinking, HIGH requires sourcing (often much smaller current in TTL).

Final Answer:

–12.8 mA