Difficulty: Medium
Correct Answer: 200 kΩ or less
Explanation:
Introduction / Context:
“Stiffness” for a current source means the delivered load current stays close to the ideal source current regardless of reasonable load changes. A Norton current source has an internal resistance in parallel with the ideal current source. Current divides between the internal resistance and the external load; to keep the load current near the nominal value, we want most current to go through the load, not the internal shunt path.
Given Data / Assumptions:
Concept / Approach:
For a Norton source, the load current is I_L = I_N * (R_int / (R_int + R_L)). If R_L ≪ R_int, then R_int/(R_int + R_L) ≈ 1, making I_L ≈ I_N. A practical “stiff” guideline is that R_L should be at least one to two orders of magnitude smaller than R_int so that the current through R_int is negligible.
Step-by-Step Solution:
Compute order-of-magnitude: R_int = 20 MΩ; choose R_L ≤ 0.01 * R_int ≈ 200 kΩ for ≲1% error.Check current division: If R_L = 200 kΩ, factor = 20 MΩ / (20 MΩ + 200 kΩ) ≈ 20,000 kΩ / 20,200 kΩ ≈ 0.99 → I_L ≈ 9.9 mA.Thus, loads of 200 kΩ or less keep the source behavior close to ideal 10 mA.
Verification / Alternative check:
Try R_L = 20 kΩ: factor ≈ 20,000 kΩ / 20,020 kΩ ≈ 0.999 → even stiffer. As R_L decreases further, the approximation improves.
Why Other Options Are Wrong:
(a) “20 kΩ or more” includes large R_L values approaching MΩ, which reduce stiffness. (c) “200 kΩ or more” trends the wrong way for stiffness. (d) “20 km” is a units mistake (distance, not resistance).
Common Pitfalls:
Confusing Norton (parallel R_int) with Thevenin (series R_th); forgetting that for current sources, small R_L compared to R_int is desirable for “stiffness.”
Final Answer:
200 kΩ or less
Discussion & Comments