Fan-out calculation: A TTL NAND gate specifies IIH(max) = 40 µA per input at logic HIGH. If it drives ten standard TTL inputs, how much source current must its output provide in the HIGH state?

Introduction / Context:Fan-out quantifies how many inputs a logic output can drive while still meeting guaranteed logic levels. Calculations rely on datasheet currents: IIH for inputs at HIGH and IIL for inputs at LOW. This problem asks you to compute the required source current for a driver at logic HIGH.

Given Data / Assumptions:

• IIH(max) per input = 40 µA.
• Number of loads N = 10.
• All loads are at logic HIGH simultaneously.

Concept / Approach:Total current the driver must source at HIGH equals the sum of the IIH currents of all connected inputs. Ensuring the driver’s IOH(max) exceeds this sum preserves VOH ≥ VOH(min). If the total exceeds capability, VOH droops and logic margins are violated.

Step-by-Step Solution:Compute total IIH: Itotal = IIH(max) * N.Substitute values: Itotal = 40 µA * 10.Calculate: Itotal = 400 µA = 0.4 mA.Compare to driver IOH(max) from datasheet to confirm compliance.

Verification / Alternative check:If fan-out rating is given, it equals IOH(max) / IIH(max). Rearranging confirms the same current requirement.

Why Other Options Are Wrong:40 µA or 200 µA underestimate the sum; 800 µA and 4 mA overestimate for the stated loads.

Common Pitfalls:Forgetting that fan-out must be evaluated for both HIGH and LOW states; IIL at LOW can be significantly larger and may become the limiting case in TTL.

Final Answer:400 µA