- 2072 Chaitra · 2 marks
Identify noise coupling mechanism by calculation
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.
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.
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).
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.
✓ Answer checked by Claude Sonnet 5.5
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