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Bandwidth of series resonant circuit with resonance curve

Short answer

Bandwidth is the frequency range where current is at least I0/√2. From |X|=R, ω1,2 = ∓R/2L + √[(R/2L)²+1/LC], giving BW = ω2−ω1 = R/L rad/s = R/(2πL) Hz = ω0/Q. A smaller R (higher Q) gives a sharper, narrower resonance curve.

What do you understand by the bandwidth of a series resonant circuit? Explain with the help of the resonance curve and also derive its expression both in terms of ω and f.

Bandwidth of a series resonant circuit is the range of frequencies, f2 − f1, over which the circuit current is at least 1/√2 (70.7 %) of its maximum value at resonance, i.e. the power delivered is at least half of the maximum power. f1 and f2 are the lower and upper half power (cut-off) frequencies.

Resonance curve

 I
 I0 |          ***
    |        **   **
0.707 I0 |- -*- - - - *-
    |     **|       |**
    |   **  |       |  **
    |__*____|___|___|____*__ f
            f1  f0  f2
            |<-- BW -->|
  • The current is maximum (I0 = V/R) at f0 and falls on both sides.
  • A small R gives a tall, narrow (sharp) curve, so a high Q and a narrow bandwidth. A large R gives a flat, wide curve.
  • A narrow bandwidth means a highly selective circuit, as used in radio tuners.

Derivation in terms of ω

At the half power points |Z| = √2 R, which requires |ωL − 1/ωC| = R.

ω²LC ∓ ωRC − 1 = 0
ω2 = R/2L + √[(R/2L)² + 1/LC]
ω1 = −R/2L + √[(R/2L)² + 1/LC]

BW = ω2 − ω1 = R/L  rad/s

In terms of f

BW = f2 − f1 = (ω2 − ω1)/2π = R/(2πL)  Hz

In terms of Q

Q = ω0L/R, so R/L = ω0/Q:

BW (rad/s) = ω0/Q        BW (Hz) = f0/Q
f0 = √(f1 f2)  (geometric mean)

For example, R = 10 Ω, L = 0.1 H gives BW = 100 rad/s (15.9 Hz); with R = 1 Ω the bandwidth drops to 10 rad/s.

Answer: BW = R/L rad/s = R/(2πL) Hz = ω0/Q. The bandwidth decreases when R is reduced (Q increases).

✓ Answer checked by Claude Opus 5.5 (high effort)

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