Series and parallel resonance, Q-factor and bandwidth

Series and parallel resonance, Q-factor, bandwidth, half-power frequencies, magnification and the practical tank circuit.

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Why it matters

Resonant circuits select one frequency and reject the rest. They are the heart of tuned amplifiers, carrier-excited sensor demodulators, Q-meters, oscillators and notch filters that remove 50 Hz hum from biomedical signals. The Q-factor tells you how sharp the selection is and how much voltage can build up across L and C, which can be many times the supply.

Key ideas

Resonance. A circuit with L and C is at resonance when its input impedance (or admittance) is purely real, so the voltage and current are in phase and the pf is unity. At resonance the inductive and capacitive reactive powers are equal and cancel; energy swaps between L and C internally.

Series RLC. Z = R + j(ωL − 1/(ωC)).

  • Resonance at ω0 = 1/√(LC). Here Z = R is a minimum and the current I0 = V/R is a maximum.
  • Below ω0 the circuit is capacitive (current leads); above ω0 it is inductive (current lags).
  • The voltages across L and C are equal and opposite, each Q·V. This is voltage magnification: a 10 V supply with Q = 50 puts 500 V across the capacitor.

Quality factor.

  • General definition: Q = 2π × (peak energy stored)/(energy dissipated per cycle).
  • For series RLC: Q = ω0L/R = 1/(ω0CR) = (1/R)·√(L/C).
  • A high Q means a sharp, narrow response. Q is related to the damping ratio by Q = 1/(2ζ).

Bandwidth and half-power frequencies. At ω1 and ω2 the current falls to I0/√2 and the power to half its maximum. At these points |X| = R and the phase is ±45°.

  • BW = ω2 − ω1 = R/L (rad/s), or R/(2πL) in Hz.
  • BW = ω0/Q.
  • The resonant frequency is the geometric mean of the band edges: ω0 = √(ω1·ω2).
  • For high Q the response is nearly symmetric: ω1,2 ≈ ω0 ∓ BW/2.

Selectivity. Selectivity is the ability to reject neighbouring frequencies. It increases with Q. Adding series resistance (for example the source resistance) lowers Q and broadens the response.

Parallel (ideal) RLC. Y = G + j(ωC − 1/(ωL)).

  • Resonance at ω0 = 1/√(LC). Here the admittance is a minimum, the impedance Z = R is a maximum, and the line current is a minimum.
  • The branch currents in L and C are each Q times the line current. This is current magnification.
  • Q = R·√(C/L) = ω0RC = R/(ω0L), and BW = 1/(RC).
  • A larger R gives a higher Q, the opposite of the series case.

Practical parallel circuit (a coil with resistance r in parallel with C). This is the tank circuit used in practice.

  • Unity-pf resonance occurs at ω0 = √(1/(LC) − r²/L²), slightly below 1/√(LC), and exists only if r < √(L/C).
  • The impedance at this frequency is the dynamic resistance Rd = L/(C·r), which is large for a low-loss coil.

Duality. Series and parallel resonance are duals: series resonance is a current maximum (acceptor circuit); parallel resonance is a voltage or impedance maximum (rejector circuit).

Formulas

  • ω0 = 1/√(LC) (rad/s), f0 = 1/(2π√(LC)) (Hz), with L in H and C in F.
  • Series: Q = ω0L/R = (1/R)·√(L/C), BW = R/L (rad/s), VL = VC = Q·V at resonance.
  • Parallel (ideal): Q = R·√(C/L), BW = 1/(RC) (rad/s), IL = IC = Q·I at resonance.
  • BW = ω0/Q, ω0 = √(ω1·ω2).
  • Exact series half-power points: ω1,2 = ∓R/(2L) + √((R/(2L))² + 1/(LC)).
  • Practical tank: ω0 = √(1/(LC) − r²/L²), Rd = L/(C·r) (Ω).

Worked examples

Example 1 (standard): series resonance. Given: R = 10 Ω, L = 10 mH, C = 1 µF, supply 1 V RMS at variable frequency. Find f0, Q, the bandwidth, the current at resonance and VC at resonance.

  1. ω0 = 1/√(10×10⁻³ × 10⁻⁶) = 1/√(10⁻⁸) = 10⁴ rad/s, so f0 = 10⁴/(2π) = 1591.5 Hz.
  2. Q = ω0L/R = 10⁴ × 0.01/10 = 10.
  3. BW = R/L = 1000 rad/s, which is 159.2 Hz (equal to f0/Q).
  4. I0 = V/R = 1/10 = 0.1 A.
  5. VC = I0·XC = 0.1 × 1/(10⁴ × 10⁻⁶) = 0.1 × 100 = 10 V, which is Q·V.

Answer: f0 = 1.59 kHz, Q = 10, BW = 159 Hz, I0 = 0.1 A, VC = VL = 10 V.

Example 2 (GATE level): a practical tank circuit. Given: a coil with r = 10 Ω and L = 0.1 H in parallel with C = 10 µF, fed from 100 V RMS. Find the unity-pf resonant frequency, the dynamic resistance and the line current at resonance.

  1. 1/(LC) = 1/(0.1 × 10⁻⁵) = 10⁶ s⁻² and (r/L)² = (100)² = 10⁴ s⁻².
  2. ω0 = √(10⁶ − 10⁴) = 994.99 rad/s, so f0 = 158.4 Hz. Compare 159.2 Hz for the ideal case.
  3. Rd = L/(C·r) = 0.1/(10⁻⁵ × 10) = 1000 Ω.
  4. Line current: I = 100/1000 = 0.1 A, in phase with the voltage.
  5. Coil Q at resonance: ω0L/r = 99.5/10 = 9.95. The coil current is about 100/|10 + j99.5| = 1.0 A, roughly Q times the line current.

Answer: f0 = 158.4 Hz, Rd = 1 kΩ, line current 0.1 A.

Common mistakes

  • Using BW = R/L for a parallel RLC. For parallel it is 1/(RC).
  • Thinking a larger R sharpens a series circuit. It lowers Q and broadens the response.
  • Assuming a practical tank resonates exactly at 1/√(LC).
  • Treating the half-power frequencies as symmetric about f0 when Q is low.
  • Forgetting the voltage across C at series resonance can far exceed the supply. Component voltage ratings matter.
  • Mixing rad/s and Hz in BW and Q calculations.

For GATE IN

Expect: f0, Q and BW of series or parallel RLC (NAT); the voltage across L or C at resonance; half-power frequencies; the resonant frequency and dynamic impedance of a practical tank; the effect of changing R, L or C on Q and BW; Q-meter measurement principle. Practise keeping series and parallel formulas apart, and checking BW = ω0/Q.

Quick check

  1. What is f0 for L = 1 mH and C = 1 nF?
  2. A series RLC has Q = 20 and a supply of 5 V. What is the capacitor voltage at resonance?
  3. At parallel resonance, is the line current a maximum or a minimum?
  4. f0 = 1 MHz and Q = 100. What is the bandwidth?

Answers: 1. 1/(2π√(10⁻¹²)) = 159 kHz. 2. 100 V. 3. A minimum. 4. 10 kHz.

Interview questions

All Electrical Circuits interview questions

Try answering each one aloud before you open it.

  1. 1.What is resonance, and how do series and parallel resonance differ?Concept

    Resonance is the frequency at which an LC circuit's input impedance is purely resistive, so voltage and current are in phase, because the inductive and capacitive reactances cancel. In a series RLC, the impedance is then minimum (R) and the current maximum; it is an acceptor circuit with voltage magnification across L and C. In a parallel RLC, the admittance is minimum, so the impedance is maximum and the line current minimum; it is a rejector circuit with current magnification in the L and C branches. For ideal elements both occur at ω0 = 1/√(LC).

  2. 2.What is the Q-factor and how is it related to bandwidth?Concept

    Q is 2π times the peak energy stored divided by the energy dissipated per cycle. It measures how lightly damped and how selective the circuit is. For a series RLC, Q = ω0L/R = (1/R)·√(L/C); for a parallel RLC, Q = R·√(C/L). Bandwidth between the half-power points is BW = ω0/Q, so a high Q means a narrow, sharp response. Q also equals 1/(2ζ) in terms of the damping ratio.

  3. 3.Why can the capacitor voltage in a series resonant circuit exceed the supply voltage? Is that dangerous?Concept

    At resonance the current is V/R, and the capacitor voltage is I·XC = V·(XC/R) = Q·V. With a high Q, for example Q = 50 on a 10 V supply, the capacitor and inductor each see 500 V, in antiphase so that they cancel in the loop. Energy is not created: it builds up over many cycles and swaps between L and C. In practice the components must be rated for Q times the supply voltage, which is a real hazard in tuned and power circuits.

  4. 4.Why does a practical parallel tank circuit resonate below 1/√(LC), and what is its dynamic resistance?Concept

    A real coil has series resistance r. For the input admittance to be real, the capacitor susceptance must cancel the susceptance of the r–L branch, which gives ω0 = √(1/(LC) − r²/L²), slightly lower than the ideal value. Resonance exists only if r < √(L/C). At this frequency the circuit looks like a pure resistance Rd = L/(Cr), called the dynamic resistance. It is large for a low-loss coil, which is why tank circuits are used as high-impedance tuned loads.

  5. 5.A radio IF stage is tuned to 455 kHz with C = 100 pF. What inductance is needed, and what Q gives a 10 kHz bandwidth?Concept

    L = 1/(ω0²C) = 1/((2π × 455×10³)² × 100×10⁻¹²) = 1.22 mH. The required Q is f0/BW = 455/10 = 45.5. You then choose the circuit resistance to give that Q. In a series circuit, R = ω0L/Q = (2π × 455×10³ × 1.22×10⁻³)/45.5 ≈ 76.9 Ω.

  6. 6.Where is resonance used in instrumentation?Concept

    Examples include tuned and band-pass amplifiers in carrier-excited sensors such as LVDTs and capacitive transducers, and twin-T or LC notch filters that remove 50 Hz mains pickup from ECG and other low-level signals. The Q-meter finds the Q of a coil by measuring capacitor voltage magnification at series resonance (Q = VC/V). Resonance-based measurement also gives inductance or capacitance from f0 = 1/(2π√(LC)). Oscillators and crystal-controlled timebases depend on high-Q resonators.

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