--- title: "Identify noise coupling mechanism by calculation" subject: "Instrumentation II" chapter: 6 key: "semester_5/instrumentation_ii:switching-noise-calc" source: https://www.bctengineer.com/q/semester_5/instrumentation_ii/switching-noise-calc site: BCT Engineering --- # Identify noise coupling mechanism by calculation Instrumentation II · Chapter 6: Grounding And Shielding - Asked in: 2072 Chaitra, 2 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/instrumentation_ii/switching-noise-calc - Whole chapter: https://www.bctengineer.com/md/semester_5/instrumentation_ii/chapter-6 ## Question An electronic circuit receives noise from a switching element. If voltage on the switching device swings from 4V to 6V within 100 μs, during which current makes a transition from 10 μA to 25 μA in 10 ns, what might be the noise coupling mechanism? Identify it using suitable calculations. ## Short answer The source impedance (240–400 kΩ) is much larger than 377 Ω, indicating capacitive (electric-field) coupling. The high dV/dt and negligible dI/dt confirm negligible inductive coupling. ## Answer Capacitive coupling depends on a high-voltage swing with high source impedance (electric field), and inductive coupling depends on a large current change (magnetic field). To identify the mechanism, compute the **source (pseudo) impedance** Z = V/I. If Z is much greater than the free-space wave impedance of 377 ohm, the field is mainly electric (capacitive coupling); if much less, magnetic (inductive). Given: V swings 4 V to 6 V (ΔV = 2 V in 100 μs); I swings 10 μA to 25 μA (ΔI = 15 μA in 10 ns). ```text Z at 4 V, 10 uA = 4 / 10e-6 = 4.0 x 10^5 ohm = 400 kohm Z at 6 V, 25 uA = 6 / 25e-6 = 2.4 x 10^5 ohm = 240 kohm dV/dt = 2 V / 100 us = 2.0 x 10^4 V/s dI/dt = 15 uA / 10 ns = 1.5 x 10^3 A/s ``` The impedance of the source (240-400 kohm) is much greater than 377 ohm, so the source is a **high-voltage, low-current** one, which produces a predominantly **electric field**. The current is tiny (microamps), so the magnetic field and the induced voltage M·dI/dt (with dI/dt only 1.5 × 10³ A/s, M of the order of nH gives about 1.5 μV) are negligible. **Answer: capacitive (electric-field) coupling** is the likely mechanism, since Z ≈ 240-400 kΩ ≫ 377 Ω. Remedy: electrostatic shielding, spacing, lower receiver impedance. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the noise coupling mechanism identified in this problem? A. Compute only the frequency of the source B. Compare the temperature rise with 377 K C. Compute source impedance Z = V/I and compare with 377 ohm D. Count the number of switching events Answer: C. High Z (much above 377 ohm) means electric field/capacitive; low Z means magnetic/inductive. (Key point: Identification method) 2. What range of source impedance results from 4 V/10 uA and 6 V/25 uA? A. About 0.24 ohm to 0.4 ohm B. About 240 kohm to 400 kohm C. About 24 ohm to 40 ohm D. About 377 ohm Answer: B. 4/10e-6 = 400 kohm and 6/25e-6 = 240 kohm. (Key point: Computed impedance) 3. Since Z is much greater than 377 ohm, what is the likely mechanism? A. Capacitive (electric-field) coupling B. Inductive (magnetic) coupling C. Conductive coupling via common ground D. Pure far-field radiation Answer: A. High voltage with tiny current means the field is mainly electric. (Key point: Conclusion) 4. Why is inductive coupling negligible here? A. The current changes very slowly in picoseconds B. The voltage swing is zero C. Mutual inductance is infinite D. Current change is only microamps, so M dI/dt is about microvolts Answer: D. dI/dt is only 1.5 x 10^3 A/s, giving around 1.5 uV for M of nH. 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