Series Resonant Circuits
A series resonant circuit is one of the most important building blocks in all of radio electronics. It is a resistor, inductor, and capacitor wired in series — one after another — and driven by an AC signal. At one specific frequency, this simple combination of three components behaves in an almost magical way: it passes that frequency with almost no opposition, while blocking frequencies above and below. Every radio receiver that lets you tune to one station while rejecting others relies on exactly this behavior.
In the previous lesson you learned that resonance occurs when XL equals XC and that the impedance of a series circuit drops to just its resistance at resonance. This lesson goes deeper: you will learn exactly how sharp that frequency selectivity is, what controls it, and how voltages inside the circuit can far exceed the supply voltage. These concepts — Q factor, bandwidth, and voltage magnification — are the working vocabulary of every antenna tuner, bandpass filter, and tuned RF amplifier in your shack.
- The Series Circuit at Resonance
- Minimum Impedance: Why Series Resonance Passes Maximum Current
- Q Factor: The Measure of Circuit Quality
- Bandwidth and the 3 dB Points
- Voltage Magnification at Resonance
- Series Resonant Circuit Calculator
- The Frequency Response Curve
- Series Resonance in Your Radio Station
A series RLC circuit. R, L, and C are wired end-to-end with a single current flowing through all three. At the resonant frequency, XL = XC, they cancel, and the circuit impedance drops to just R — current is maximum.
View LargerThe Series Circuit at Resonance
In a series RLC circuit, the same current flows through every component — the resistor, the inductor, and the capacitor all carry identical current at every moment. This is a crucial difference from parallel circuits and it explains much of what makes series resonance useful.
The total impedance of a series RLC circuit is:
Z = √(R² + (XL − XC)²)
where XL = 2πfL and XC = 1/(2πfC)
At frequencies below resonance, XC is large and XL is small, so XC − XL dominates and the circuit looks capacitive. At frequencies above resonance, XL is large and XC is small, so XL − XC dominates and the circuit looks inductive. At exactly the resonant frequency, XL = XC, the two reactive terms cancel completely, and the impedance formula reduces to:
Zres = R
This is the minimum impedance the circuit can have — equal to the series resistance alone.
This minimum-impedance condition at resonance is the defining characteristic of the series resonant circuit. Because current I = V/Z and Z is at its minimum, current is at its maximum at resonance. The circuit is said to be in resonance when it draws the most current from the source.
Minimum Impedance: Why Series Resonance Passes Maximum Current
It seems paradoxical that adding more components to a circuit — an inductor and a capacitor — can make it pass more current than just a resistor alone. But at resonance, this is exactly what happens: the series RLC circuit passes the same current as if only the resistor were present. The inductor and capacitor are "invisible" at the resonant frequency.
The reason is that XL and XC are equal and opposite in their effect on the circuit. Inductive reactance causes voltage to lead current; capacitive reactance causes voltage to lag current. At resonance these effects are equal and cancel, leaving the circuit's behavior determined entirely by R.
Think of it as a tug-of-war where both teams are equally matched. The net effect on the rope is zero — it doesn't move despite enormous forces being applied. In the RLC circuit, the inductor and capacitor exert large reactive forces, but they cancel each other, and the resistor is all that's left controlling current flow.
A series circuit has R = 10 Ω, L = 50 µH, C = 100 pF. Find the resonant frequency and compare impedance at f0, at 10% below f0, and at 10% above f0.
Step 1 — Resonant frequency:
f0 = 1 / (2π√(LC)) = 1 / (2π√(50×10⁻⁶ × 100×10⁻¹²))
= 1 / (2π√(5×10⁻¹⁵)) = 1 / (2π × 7.071×10⁻⁸) = 2.251 MHz
Step 2 — Impedance at resonance:
Z = R = 10 Ω
Step 3 — Impedance at 10% below resonance (f = 2.026 MHz):
XL = 2π × 2.026×10⁶ × 50×10⁻⁶ = 636.6 Ω
XC = 1 / (2π × 2.026×10⁶ × 100×10⁻¹²) = 785.6 Ω
Xnet = XL − XC = 636.6 − 785.6 = −149.0 Ω
Z = √(10² + 149²) = √(100 + 22201) = √22301 = 149.3 Ω
Step 4 — Impedance at 10% above resonance (f = 2.476 MHz):
XL = 2π × 2.476×10⁶ × 50×10⁻⁶ = 777.6 Ω
XC = 1 / (2π × 2.476×10⁶ × 100×10⁻¹²) = 642.8 Ω
Xnet = 777.6 − 642.8 = +134.8 Ω
Z = √(10² + 134.8²) = √(100 + 18171) = √18271 = 135.2 Ω
Result: At resonance Z = 10 Ω; 10% below: 149 Ω; 10% above: 135 Ω. Moving just 10% away from resonance increases impedance by a factor of 13–15, demonstrating the sharp selectivity of this circuit.
Q Factor: The Measure of Circuit Quality
The Q factor — Quality factor — is the single most important specification for a resonant circuit. It tells you how selective the circuit is: how sharply it responds to the resonant frequency while rejecting others. A high-Q circuit is very selective; a low-Q circuit is broad and accepts a wide range of frequencies.
For a series resonant circuit, Q is defined as the ratio of the reactive power stored in the circuit to the real power dissipated by the resistance. In practical terms, Q equals the reactance of either component at resonance divided by the series resistance:
Q = XL / R = XC / R = (2πf0L) / R = 1 / (2πf0CR)
where all values are evaluated at the resonant frequency f0
Because XL = XC at resonance, either reactance gives the same Q value. The formula using L is usually more convenient because inductors dominate loss in real circuits.
Q is a dimensionless number — no units. Typical values in amateur radio applications:
| Circuit Type | Typical Q Range | Application |
|---|---|---|
| Air-core coil, unshielded | 50–200 | Transmitter output networks, antenna tuners |
| Toroidal inductor (powdered iron) | 100–300 | Low-pass filters, impedance matching |
| Toroidal inductor (ferrite) | 20–80 | Audio and IF transformers |
| Crystal resonator | 10,000–100,000 | Oscillator frequency control, narrow filters |
| Ceramic resonator | 1,000–5,000 | Low-cost oscillators |
The resistance R in the Q formula includes all sources of loss in the circuit — the DC resistance of the wire, eddy-current losses in the core, skin-effect losses at high frequency, and any external resistance deliberately added. This total loss resistance is often called the equivalent series resistance (ESR) or the circuit resistance.
A 40-meter band transmitter uses a series resonant circuit with L = 2.5 µH, C = 36 pF, and coil ESR = 0.5 Ω. Find Q at the resonant frequency.
Step 1 — Resonant frequency:
f0 = 1 / (2π√(2.5×10⁻⁶ × 36×10⁻¹²))
= 1 / (2π√(9×10⁻¹⁷)) = 1 / (2π × 9.487×10⁻⁹) = 16.78 MHz
(This is in the 17-meter band — let's adjust to a true 40-meter example.)
For 7.1 MHz with L = 2.5 µH: C = 1/(4π² × f² × L) = 1/(39.478 × 50.41×10¹² × 2.5×10⁻⁶) = 201.6 pF
At 7.1 MHz:
XL = 2π × 7.1×10⁶ × 2.5×10⁻⁶ = 111.5 Ω
Q = XL / R = 111.5 / 0.5 = 223
This is an excellent Q for a transmitter tank circuit. With this Q, the circuit can generate significant voltages across the reactive components and efficiently transfer power to the antenna system.
Bandwidth and the 3 dB Points
Q factor and bandwidth are two sides of the same coin. A high-Q circuit is very selective — it responds strongly only to frequencies very close to f0. A low-Q circuit is broad — it passes a wide range of frequencies around f0. The bandwidth is the frequency range over which the circuit passes signals acceptably well.
The standard definition of bandwidth uses the half-power points — the frequencies at which the output power falls to half its maximum value. Since power is proportional to I², the half-power points occur where current falls to 1/√2 ≈ 0.707 of its peak value. A 3 dB drop in power corresponds to a 0.707 drop in current (because 10 log(0.5) = −3 dB). These frequencies are called the upper and lower −3 dB frequencies, or the half-power bandwidth points.
BW = f0 / Q
where:
BW = bandwidth in Hz
f0 = resonant frequency in Hz
Q = quality factor (dimensionless)
The −3 dB frequencies are: f1 = f0 − BW/2 and f2 = f0 + BW/2
This formula reveals a fundamental trade-off: higher Q gives better selectivity (narrower bandwidth) but also means a more sensitive circuit that requires more precise tuning. Lower Q gives a more tolerant circuit that accepts a broader range of frequencies but does a poorer job of rejecting adjacent signals.
A receiver uses a series resonant input filter centered at f0 = 7.150 MHz with Q = 80. Find the bandwidth and the −3 dB frequencies.
Bandwidth:
BW = f0 / Q = 7,150,000 / 80 = 89,375 Hz ≈ 89.4 kHz
Lower −3 dB frequency:
f1 = 7,150,000 − 89,375/2 = 7,150,000 − 44,688 = 7,105,313 Hz ≈ 7.105 MHz
Upper −3 dB frequency:
f2 = 7,150,000 + 44,688 = 7,194,688 Hz ≈ 7.195 MHz
Interpretation: The 40-meter amateur band runs from 7.0 to 7.3 MHz — a span of 300 kHz. This filter's 89 kHz bandwidth covers roughly 30% of the band, which is reasonable for a pre-selector filter but not nearly sharp enough for a final IF filter, which typically needs Q in the thousands (achieved with crystals or ceramic resonators).
Voltage Magnification at Resonance
One of the most striking properties of a series resonant circuit is that the voltage across the inductor or capacitor at resonance can be far larger than the supply voltage. This is not a measurement error — it is a real physical phenomenon called voltage magnification, and it is the reason that high-voltage capacitors are required in transmitter tank circuits even when operating at relatively modest supply voltages.
To understand why this happens, recall that at resonance the current is at its maximum: I = VS/R, where VS is the supply voltage. The voltage across the inductor is:
VL = I × XL = (VS/R) × XL = VS × (XL/R) = VS × Q
Similarly: VC = VS × Q
The voltage across both reactive components at resonance equals the supply voltage multiplied by Q. A circuit with Q = 100 and a supply voltage of 100 V will develop 10,000 V across both the inductor and the capacitor at resonance. This is why capacitors in transmitter tank circuits must be rated for voltages many times higher than the supply rail.
Note that VL and VC are equal in magnitude but exactly opposite in phase — they cancel in the KVL sum around the loop, so the net reactive voltage presented to the source is zero. Only VR = I × R = VS appears from the source's perspective.
A transmitter operates at 14.2 MHz (20 meters). The tank circuit has R = 2 Ω (coil loss) and L = 1.2 µH. The DC supply is 13.8 V and peak RF current is 5 A. Find the Q and the peak voltage across the tank capacitor.
Step 1 — XL at 14.2 MHz:
XL = 2π × 14.2×10⁶ × 1.2×10⁻⁶ = 107.0 Ω
Step 2 — Q:
Q = XL / R = 107.0 / 2 = 53.5
Step 3 — Peak voltage across capacitor:
VC(peak) = Ipeak × XC = 5 × 107.0 = 535 V peak
This means a capacitor rated for at least 600 V working voltage should be used — even though the supply is only 13.8 V DC. This is why QRP transmitter builders often use 500 V or 1000 V capacitors in their tank circuits.
Voltage magnification in a series resonant circuit. At resonance, the voltage across L and C each equal Q × VS — far exceeding the supply voltage. The voltage across R equals VS at resonance since all supply voltage appears across the resistance.
View LargerSeries Resonant Circuit Calculator
Series Resonant Circuit Calculator
Enter the circuit values to calculate resonant frequency, Q factor, bandwidth, impedance at resonance, and component voltages at resonance.
The Frequency Response Curve
The frequency response of a series resonant circuit — a plot of current or output voltage versus frequency — forms a bell-shaped curve centered on f0. The width of this bell is determined by Q: high Q produces a narrow, tall peak; low Q produces a wide, shallow peak.
The shape of the response curve follows this equation for normalized current (current as a fraction of peak current):
I/Imax = 1 / √(1 + Q²((f/f0) − (f0/f))²)
This shows that Q² multiplies the frequency deviation term — doubling Q makes the circuit four times more selective at the same frequency offset from resonance. This is why high-Q circuits are so desirable in receiver front ends: they suppress adjacent-channel interference by a factor proportional to Q².
The curve is not perfectly symmetrical on a linear frequency scale — it is slightly asymmetric, with the lower half wider than the upper half. However, on a logarithmic frequency scale (which is more natural for decade-spanning measurements) the curve is symmetric. For narrowband circuits (BW much less than f0), the asymmetry is negligible.
| Frequency (relative to f0) | Net Reactance | Relative Current (Q = 50) | Circuit Behavior |
|---|---|---|---|
| 0.90 × f0 | Strongly capacitive | ~10% | Highly attenuated |
| 0.97 × f0 | Moderately capacitive | ~50% | −6 dB below peak |
| 0.99 × f0 | Slightly capacitive | ~71% (−3 dB point) | Half-power bandwidth |
| f0 | Zero (cancelled) | 100% (maximum) | Full resonance |
| 1.01 × f0 | Slightly inductive | ~71% (−3 dB point) | Half-power bandwidth |
| 1.03 × f0 | Moderately inductive | ~50% | −6 dB below peak |
| 1.10 × f0 | Strongly inductive | ~10% | Highly attenuated |
Series Resonance in Your Radio Station
Series resonant circuits appear throughout amateur radio equipment in applications you interact with directly:
Antenna Tuner (ATU) Series Circuits
When you use an antenna tuner to match a non-resonant antenna to your transmitter's 50 Ω output, the tuner often contains series resonant elements. By adjusting L and C, the tuner creates a circuit that presents 50 Ω at the transmitter while delivering maximum power to the antenna. The series elements in the L-network and T-network configurations (covered in the Pi, L, and T Networks lesson) rely on the series resonant behavior you have just learned.
Transmitter Output Tank Circuit
The tank circuit in a tube or transistor transmitter is fundamentally a series resonant circuit (or parallel — depending on the topology) that determines the operating frequency and shapes the output waveform. The Q of this circuit determines the harmonic suppression: higher Q means greater suppression of unwanted harmonics that would violate FCC rules. Tank circuit design is a direct application of the voltage magnification and Q concepts from this lesson.
Crystal Filters in Receivers
The IF (intermediate frequency) filter in a superheterodyne receiver is the most critical selectivity element in the radio. Crystal filters use the extremely high Q of quartz crystals (often Q = 50,000–100,000) to achieve bandwidths as narrow as 250 Hz for CW reception or 2.7 kHz for SSB. These filters are arrays of series resonant crystal elements, each contributing to the overall filter response. The Q formula Q = XL/R still applies — the "L" is the acoustic inductance of the crystal and the "R" is the acoustic loss of the quartz.
Loading Coils on Short Antennas
A short vertical antenna is capacitive — it has a capacitive reactance at its base. Adding a series inductor (loading coil) cancels this capacitive reactance, creating a series resonant condition at the antenna's base. This is exactly series resonance applied to make the antenna appear as a pure resistance to the transmitter. The Q of the loading coil determines how much of the transmitter power is wasted in the coil versus radiated — high-Q coils give better antenna efficiency.
Experiment: Observing Series Resonance with an Audio Oscillator
You need: Audio frequency oscillator or signal generator (20 Hz–20 kHz), 100 mH inductor (available cheaply as an audio choke), 1 µF non-polarized capacitor, 100 Ω resistor, multimeter set to AC volts, connecting wires.
Theory: f0 = 1/(2π√(LC)) = 1/(2π√(0.1 × 10⁻⁶)) = 503 Hz. Q = XL/R = (2π × 503 × 0.1)/100 = 3.16 (low Q due to the resistor, but good for demonstration).
Procedure:
- Wire R, L, and C in series across the oscillator output.
- Connect the multimeter across R to measure current indirectly (VR = I × R).
- Sweep the oscillator from 100 Hz to 1500 Hz while watching VR.
- Find the frequency where VR is maximum — this is f0. Note the frequency; it should be close to 503 Hz.
- At resonance, move the multimeter across the capacitor and measure VC. Compare to the input voltage.
- Note how VC exceeds the input — this is voltage magnification.
- Record VR at f0/2 and 2f0 to see how much attenuation occurs away from resonance.
Expected result: VR peaks sharply at ~503 Hz. VC at resonance ≈ Q × Vin ≈ 3.16 × Vin. At half and double f0, VR is dramatically lower, demonstrating the frequency-selective behavior of the series resonant circuit.
Frequently Asked Questions
If VL and VC both equal Q × VS at resonance, does KVL still hold?
Yes, absolutely — KVL must always hold. VL and VC are equal in magnitude but exactly 180° out of phase with each other at resonance. They cancel when added as phasors, so the sum around the loop is VR + VL − VC = VS + 0 = VS. The large voltages across L and C are real and measurable, but they cancel in the series loop because they are opposite in sign. This is why measuring VC and VL separately with a voltmeter can give readings much higher than VS — the meter measures magnitudes, not the phasor sum.
Why does Q decrease as frequency increases even with the same coil?
Several effects increase coil losses at higher frequencies: skin effect concentrates current in a thin layer on the conductor surface, increasing effective resistance; eddy current losses in any core material increase; and distributed capacitance between coil turns becomes significant, changing the resonant behavior. These effects mean that the same coil has a lower effective Q at 14 MHz than at 7 MHz, which must be considered when designing RF circuits. Manufacturers typically specify Q at a specific test frequency for this reason.
Can series resonance be harmful to equipment?
Yes — the voltage magnification at resonance can destroy components not rated for the resulting voltages. Capacitors in transmitter circuits must be rated well above the supply voltage for exactly this reason. In power distribution systems, accidental series resonance between line inductance and power-factor correction capacitors can cause dangerously high voltages. In receivers, a series resonant condition in the wrong place can produce unexpectedly large signal voltages that overdrive front-end amplifiers. Understanding series resonance is essential for safe circuit design.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.