Learn › Topic 9

Resonance in Series and Parallel RLC Circuits

7 min read · bitGateLab Team

When you put an inductor and a capacitor in the same circuit, a very useful phenomenon appears: resonance. At a certain frequency the reactances of the two cancel out, and the circuit responds very differently than it does across the rest of the spectrum. Thanks to that, a radio receiver picks one station out of hundreds, a filter selects a tone and an oscillator produces a stable frequency. Here you'll learn to calculate the resonant frequency, the quality factor Q and the bandwidth of series and parallel RLC circuits.

The series RLC circuit

With R, L and C in series, the total impedance is the sum of the three:

Z = R + j(X_L − X_C) = R + j(ωL − 1/(ωC))

The inductor's reactance grows with frequency and the capacitor's decreases. That defines three regions:

Below resonance, X_C > X_L: the circuit is capacitive and the current leads the voltage.
At resonance, X_L = X_C: the reactances cancel, Z = R, the impedance is at its minimum, and the current is at its maximum and in phase with the voltage.
Above resonance, X_L > X_C: the circuit is inductive and the current lags.

Setting ωL = 1/(ωC) gives the resonant frequency:

f₀ = 1 / (2π√(LC))

Notice that f₀ depends only on the product LC. The resistance does not change where resonance occurs, but it does change how pronounced it is.

Quality factor Q

The Q factor measures how "sharp" the resonance is. For the series circuit it is defined as the reactance at resonance divided by the resistance:

Q = X_L0 / R = (1/R) · √(L/C)

A high Q has a consequence that is surprising the first time you see it: at resonance, the voltage across the inductor and across the capacitor is Q times the source voltage. The two voltages are equal and opposite, so they cancel each other and you do not see them in the sum, but each component still has to withstand them.

In a series RLC circuit with Q = 30 driven with 1 V, the capacitor can see 30 V. Always check the voltage rating of the components in a resonant circuit.

Bandwidth

The bandwidth (BW) is the range of frequencies over which the current stays above 0.707 times its maximum value, that is, where the power delivered to R is at least half the maximum. Its limits, f₁ and f₂, are the half-power or −3 dB frequencies:

BW = f₂ − f₁ = f₀ / Q       (series circuit: BW = R / (2πL))

The resonant frequency is the geometric mean of the band edges, f₀ = √(f₁·f₂). When Q is greater than 10, the curve is nearly symmetric and you can approximate f₁ ≈ f₀ − BW/2 and f₂ ≈ f₀ + BW/2.

Normalized current of a series RLC circuit as a function of frequency. With Q = 2 (solid line) the bandwidth runs from 0.78·f₀ to 1.28·f₀, that is, f₀/2. With Q = 5 (dashed line) the curve is narrower.

Example 1: series RLC. L = 10 mH, C = 100 nF and R = 10 Ω, driven with 1 V RMS.

LC = 0.01 · 100×10⁻⁹ = 1×10⁻⁹; √(LC) = 3.162×10⁻⁵ s.
f₀ = 1/(2π · 3.162×10⁻⁵) = 5032.9 Hz.
Reactance at resonance: X_L0 = 2π · 5032.9 · 0.01 = 316.2 Ω (and X_C0 is also 316.2 Ω).
Q = 316.2/10 = 31.6.
BW = f₀/Q = 5032.9/31.6 = 159.2 Hz. Check: R/(2πL) = 10/(0.0628) = 159.2 Hz.
Band edges: f₁ ≈ 4954 Hz and f₂ ≈ 5113 Hz.
Current at resonance: I = 1 V/10 Ω = 0.1 A RMS. Power in R: 0.1 W.
Voltage across the inductor and across the capacitor: 0.1 A · 316.2 Ω = 31.6 V RMS each, Q times the source voltage.

The parallel RLC circuit

With R, L and C in parallel, it is easier to work with admittances (Y = 1/Z), which add:

Y = 1/R + j(ωC − 1/(ωL))

Resonance occurs at the same frequency, f₀ = 1/(2π√(LC)), but the behavior is the reverse of the series circuit: the admittance is at its minimum, the impedance is at its maximum and equal to R, and the current delivered by the source is at its minimum. Inside, however, energy swings back and forth between L and C: the current circulating through each of them is Q times the source current. For the parallel circuit:

Q = R / X₀ = R · √(C/L)        BW = f₀/Q = 1/(2πRC)

Notice that in the parallel circuit a larger R gives a higher Q, the opposite of the series circuit. This circuit, called a tank circuit, is the one used in oscillators and tuned stages.

Example 2: parallel RLC. The same L = 10 mH and C = 100 nF, now in parallel with R = 10 kΩ and driven with 1 V RMS.

f₀ = 5032.9 Hz, the same as before, because LC has not changed.
Q = 10 000 · √(100×10⁻⁹/0.01) = 10 000 · 3.162×10⁻³ = 31.6.
BW = 1/(2π · 10 000 · 100×10⁻⁹) = 159.2 Hz.
Source current at resonance: only the current through R, 1 V/10 kΩ = 0.1 mA.
Current in the inductor: 1 V/316.2 Ω = 3.16 mA, and the same in the capacitor: 31.6 times the source current.

In a real circuit the inductor has the resistance of its wire in series, so the ideal parallel circuit is an approximation. If that resistance is small compared with X_L0, the resonant frequency barely changes.

Common mistakes

Forgetting the square root in f₀ = 1/(2π√(LC)), or the 2π, which gives the angular frequency ω₀ instead of f₀.
Using the series formula for Q in a parallel circuit. In the series circuit Q = X₀/R and in the parallel circuit Q = R/X₀.
Believing that at resonance the voltage across L or C cannot exceed the source voltage. It can be Q times larger.
Thinking that the resistance shifts the resonant frequency in the ideal circuit. It only changes Q and the bandwidth.
Sweeping the frequency in steps that are too large and "jumping over" the peak when Q is high.

Try it in the simulator

With the AC analysis you can see the resonance peak and measure its bandwidth directly on the plot.

Place a 1 V sine-wave source and connect a 10 Ω resistor, a 10 mH inductor and a 100 nF capacitor in series.
Run an AC sweep from 1 kHz to 20 kHz with enough points per decade and look at the current or the voltage across the resistor.
Find the peak near 5 kHz and measure the two frequencies where the curve is down 3 dB.
Look at the capacitor voltage at resonance: it should be around 31.6 V.
Change R to 100 Ω and watch the peak get wider and lower while f₀ stays put.
Build the parallel version with R = 10 kΩ and check that the impedance is now at its maximum at f₀.

Exercises

Calculate f₀ for L = 1 mH and C = 10 nF.

Show answer
LC = 1×10⁻¹¹, √(LC) = 3.162×10⁻⁶ s. f₀ = 1/(2π · 3.162×10⁻⁶) = 50.33 kHz.
If that inductor and capacitor form a series RLC circuit with R = 5 Ω, what are Q and the bandwidth?

Show answer
√(L/C) = √(10⁵) = 316.2 Ω, so Q = 316.2/5 = 63.2. BW = f₀/Q = 50 329/63.2 = 796 Hz (also R/(2πL) = 5/0.006283 = 796 Hz).
What capacitor do you need for a 100 mH inductor to resonate at 1 kHz?

Show answer
C = 1/((2πf₀)² · L) = 1/((6283)² · 0.1) = 1/(3.948×10⁶) = 253 nF.
In a series RLC circuit operating above f₀, is the circuit inductive or capacitive? Does the current lead or lag?

Show answer
Above f₀, X_L > X_C, so the circuit is inductive and the current lags the source voltage.