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The filter that punctured its capacitor

EASA Module 3 · Put your knowledge to work in context.

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The filter that punctured its capacitor

A series LC bypass leg is designed to present minimum impedance at 400 Hz on a 115 V, 400 Hz supply. It uses L = 100 mH with a coil resistance of 15 Ω, and C = 1.583 µF in a plastic-film case rated 400 V DC. On its first full-power test the unit trips on overcurrent after a few seconds and the capacitor is found punctured. The only spare inductor in stores is 120 mH, which a technician fits. He then measures 2.2 A in the circuit and declares that the capacitor is 'still faulty'. The correct 100 mH inductor is later found in another bin. The supply is a stiff 115 V, 400 Hz source.

1. Show that the original circuit was resonant at 400 Hz and explain the capacitor failure.

Read the worked answer

Reactances at 400 Hz:

  • X_L = 2πfL = 2π × 400 × 0.100 = 251.3 Ω
  • X_C = 1/(2πfC) = 1/(2π × 400 × 1.583 µF) = 251.3 Ω

X_L and X_C are equal, so they cancel; Z = R = 15 Ω and I = 115/15 = 7.67 A.

The voltage across the capacitor is not the supply voltage - it is I × X_C = 7.67 × 251.3 = 1928 V.

That is the classic series-resonance voltage magnification, with Q = X_L/R = 251.3/15 = 16.8, so V_C ≈ Q × V_supply ≈ 1.93 kV. A 400 V DC film capacitor is stressed to nearly five times its rating and punctures. Even the coil's 15 Ω did not limit the current to a safe value (7.67 A in a filter leg), which is why the unit tripped.

Judgement point: at resonance, component voltage ratings must be selected from Q × supply voltage, not from the supply voltage. Resonance is a voltage amplifier across the reactive elements even though the source only sees R.

2. Using the 2.2 A measurement, show that the circuit is merely detuned and not faulty.

Read the worked answer

With L = 120 mH:

  • X_L = 2π × 400 × 0.120 = 301.6 Ω
  • X_C = 251.3 Ω (unchanged)
  • Net reactance X = 301.6 − 251.3 = 50.3 Ω (inductive)
  • Z = √(15² + 50.3²) = √(225 + 2530) = 52.5 Ω
  • I = 115/52.5 = 2.19 A - the measured 2.2 A, exactly.

So the 2.2 A is the correct current for a 120 mH inductor; nothing is faulty. The new resonant frequency is f₀ = 1/(2π√(LC)) = 1/(2π√(0.120 × 1.583 µF)) = 365 Hz, i.e. 35 Hz below the 400 Hz operating point.

Why such a small inductance change matters: Q = X_L/R = 301.6/15 = 20, so the half-power bandwidth is f₀/Q ≈ 365/20 ≈ 18 Hz. At 400 Hz the circuit is about two bandwidths away from resonance, so the current falls from 7.67 A to 2.19 A - a 71 % drop for a 20 % inductance change. Series resonance is sharp; this sensitivity is a design feature, not a fault.

Note also that even at 2.19 A the capacitor sees 2.19 × 251.3 = 551 V, still above a 400 V rating. The capacitor specification was wrong regardless of tuning.

3. State how this filter should have been tested, and what you would specify for the replacement capacitor.

Read the worked answer

Test method:

  1. Do not apply full 115 V to an untried resonant circuit. Feed it from a variable autotransformer or a low-voltage 400 Hz source and wind up slowly, watching current.
  2. Current-limit the first application (series resistor or bench supply current limit) so that a mistuned or shorted component cannot draw destructive current.
  3. Predict before you measure: at resonance with 15 Ω, expect ~7.7 A and ~1.9 kV across C. If the measured current is far from prediction, stop and re-check L and C values rather than raising the voltage.
  4. Fit a discharge/bleed path and prove the capacitor dead with a meter before handling - it will hold a dangerous charge.

Capacitor specification:

  • Voltage rating ≥ 2 × the calculated V_C for margin, i.e. ≥ 4 kV DC (or a properly rated AC/pulse part), not 400 V.
  • Adequate RMS current rating for 7.67 A at 400 Hz - film capacitors have a current limit set by ESR and self-heating, often the real constraint in a resonant leg.
  • Verify the capacitance tolerance: a ±10 % part moves f₀ by about ±5 % and, with Q ≈ 17, that alone changes the current substantially.
  • Physically secure the part and check creepage/clearance for the resonant voltage, not the supply voltage.

Judgement point: specify resonant components from the voltages and currents that actually appear in service, and prove the design with a current-limited low-voltage sweep before full power.