Parallel Resonant Circuits
The parallel resonant circuit — often called a tank circuit — is the opposite of the series resonant circuit in almost every respect. Where the series circuit reaches minimum impedance at resonance, the parallel circuit reaches maximum impedance. Where the series circuit draws maximum current at resonance, the parallel circuit draws minimum current. Where the series circuit is used to pass a specific frequency with minimum loss, the parallel circuit is used to present a high impedance at one frequency — and this is exactly what is needed in transistor and tube amplifier output stages.
Every HF and VHF transmitter output stage relies on a parallel resonant tank circuit to accomplish three tasks simultaneously: selecting the operating frequency, suppressing harmonics, and transforming the amplifier's output impedance to the 50 Ω needed by the transmission line. Understanding the parallel resonant circuit is therefore central to understanding how transmitters work.
A parallel resonant (tank) circuit. L and C are connected in parallel, with resistance R representing inductor losses. At resonance the circulating current in the LC loop is Q times the current supplied by the source — even though the source current drops to a minimum.
View LargerParallel vs. Series: The Key Differences
The fundamental difference between series and parallel resonant circuits comes from how current flows. In a series circuit, all current must travel through every component — one path only. In a parallel circuit, each branch carries its own current, and currents in different branches can be very different from the total current entering the circuit.
At resonance in a parallel LC circuit, the inductive current and capacitive current are equal in magnitude but exactly opposite in phase. They cancel in the external circuit — very little current is drawn from the source — but these currents still flow inside the LC loop. The energy circulates continuously between the inductor's magnetic field and the capacitor's electric field, around and around the loop, with minimal interaction with the source. This circulating energy is what gives the parallel resonant circuit its unique properties.
Think of two people on a see-saw. From outside the playground, you see very little energy flowing in or out — the two people exchange energy back and forth internally. You supply only enough energy to overcome the friction. The see-saw is in "resonance." A parallel LC circuit at resonance does the same thing: the inductor and capacitor swap energy back and forth internally, and the source supplies only the small amount lost to resistance.
Maximum Impedance at Resonance
In a parallel circuit, the total admittance (Y = 1/Z, the ease of current flow) is the sum of the branch admittances:
Y = G + j(BC − BL)
where G = 1/R (conductance), BC = ωC (capacitive susceptance), BL = 1/(ωL) (inductive susceptance)
At resonance, the capacitive susceptance equals the inductive susceptance (BC = BL), so the imaginary part of Y cancels to zero. Total admittance equals just the conductance G = 1/R, and the impedance is:
Zres = 1/G = R
This is the maximum impedance the parallel circuit can have — the circuit is purely resistive at resonance and presents maximum impedance to the source.
Away from resonance, the reactive susceptance no longer cancels, and the admittance is larger (impedance is smaller). Below resonance, inductive susceptance dominates — the circuit looks inductive and lower-impedance. Above resonance, capacitive susceptance dominates — the circuit looks capacitive and lower-impedance. The parallel circuit always presents its highest impedance at the resonant frequency.
This is the opposite of the series circuit, which presents its lowest impedance at resonance. Two sides of the same coin — one designed to pass maximum current, the other to present maximum impedance.
Dynamic Resistance: The Impedance Peak
In a real parallel resonant circuit, the resistance R is usually the series resistance of the inductor — the coil's wire resistance and core losses — not a separate component. This makes the analysis slightly more complex because R appears in series with L (inside the inductor branch) rather than as a separate parallel element.
For a practical parallel circuit where R is in series with L and C is ideal (lossless), the impedance at resonance is called the dynamic resistance RD:
RD = L / (C × RS)
where RS is the series resistance of the inductor
Equivalently: RD = Q² × RS = Q × XL
Dynamic resistance is the effective parallel resistance that the tank circuit presents to the external circuit at resonance. Several important facts about RD:
- RD can be very large — thousands of ohms for a high-Q tank circuit
- RD increases with Q² — doubling Q quadruples the dynamic resistance
- RD is purely resistive at resonance — no reactive component
- RD is always much larger than RS for any reasonable Q
A 20-meter transmitter tank circuit has L = 1.5 µH, C = 63 pF, and inductor series resistance RS = 1.2 Ω. Find the resonant frequency, Q, and dynamic resistance.
Step 1 — Resonant frequency:
f0 = 1 / (2π√(1.5×10⁻⁶ × 63×10⁻¹²))
= 1 / (2π√(94.5×10⁻¹⁸)) = 1 / (2π × 9.721×10⁻⁹) = 16.37 MHz
(Close to 14 MHz — adjusting C to 90 pF gives 14.1 MHz.)
At 14.1 MHz with L = 1.5 µH:
XL = 2π × 14.1×10⁶ × 1.5×10⁻⁶ = 132.9 Ω
Step 2 — Q:
Q = XL / RS = 132.9 / 1.2 = 110.7
Step 3 — Dynamic resistance:
RD = L / (C × RS) = 1.5×10⁻⁶ / (90×10⁻¹² × 1.2) = 1.5×10⁻⁶ / 108×10⁻¹² = 13,889 Ω ≈ 13.9 kΩ
Verification: RD = Q² × RS = 110.7² × 1.2 = 12,254 × 1.2 = 14,705 Ω (small discrepancy due to rounding)
Meaning: At 14.1 MHz, this tank circuit presents ~14 kΩ to the transistor collector. To match this to 50 Ω (the antenna system), a tap point on the coil or a coupling capacitor divider is used — reducing the impedance presented to the load by the square of the turns ratio.
Current Magnification in the Tank Loop
While a parallel resonant circuit draws minimum current from the source at resonance, the currents inside the LC loop are far larger than the source current. This is current magnification — the parallel circuit equivalent of voltage magnification in the series circuit.
The circulating current in the tank loop at resonance equals Q times the source current:
Itank = Q × Isource
where Isource is the current drawn from the driving source
A transistor amplifier supplying 100 mA to a tank circuit with Q = 80 circulates 8 A of RF current in the LC loop — even though only 100 mA flows in or out of the tank from the source. This circulating current is not wasted; it represents stored energy sloshing between L and C each cycle. The only real power dissipated is I²RS where I is the circulating current — this is why coil losses matter so much in tank circuits.
Inductor leads in high-power transmitter tank circuits must be heavy-gauge wire or copper strap capable of carrying these large circulating currents, even when the DC supply current is modest. This is a practical consequence of the current magnification in parallel resonant circuits.
Impedance versus frequency for a parallel resonant circuit. At resonance the impedance peaks at RD (dynamic resistance). Above and below resonance the reactive branches dominate and impedance falls. A higher Q gives a taller, narrower peak.
View LargerQ Factor and Bandwidth
The Q factor and bandwidth formulas for parallel resonant circuits use exactly the same relationships as for series circuits — which is one of the elegant symmetries of resonant circuit theory:
Q = RD / XL = RD / XC = 2πf0CRD = RD / (2πf0L)
BW = f0 / Q
The −3 dB frequencies are at f0 ± BW/2 (where impedance falls to RD/√2)
For a parallel circuit, the Q formula uses RD (the parallel dynamic resistance) in the numerator divided by the reactance — opposite to the series formula where Q = XL/RS. Both give the same numerical result for the same inductor because RD = Q² × RS, so Q = RD/XL = Q²RS/(QRS) = Q. The algebra is circular but the physics is consistent.
In practice, the Q of a parallel resonant circuit can be reduced deliberately by connecting an external resistance in parallel with the tank. This is called "damping" the circuit and is used to control bandwidth in IF amplifiers and filter designs. Connecting a resistive load (like an antenna) to a tank circuit always reduces Q and broadens the bandwidth.
A tank circuit has f0 = 7.2 MHz, L = 3.5 µH, RS = 1.0 Ω, and a 100 kΩ load resistor is connected in parallel. Find Q and bandwidth with the load.
Step 1 — XL at resonance:
XL = 2π × 7.2×10⁶ × 3.5×10⁻⁶ = 158.3 Ω
Step 2 — Unloaded Q:
Qunloaded = XL / RS = 158.3 / 1.0 = 158.3
Step 3 — Unloaded dynamic resistance:
RD = Q² × RS = 158.3² × 1.0 = 25,059 Ω ≈ 25.1 kΩ
Step 4 — Loaded dynamic resistance (parallel combination):
RD(loaded) = (25,059 × 100,000) / (25,059 + 100,000) = 2,505,900,000 / 125,059 = 20,038 Ω ≈ 20 kΩ
Step 5 — Loaded Q:
Qloaded = RD(loaded) / XL = 20,038 / 158.3 = 126.6
Step 6 — Bandwidth with load:
BW = f0 / Qloaded = 7,200,000 / 126.6 = 56,872 Hz ≈ 56.9 kHz
Result: The 100 kΩ load reduced Q from 158 to 127 and widened the bandwidth from 45.5 kHz to 56.9 kHz. Any resistive load connected to a tank circuit broadens its response — a fundamental consideration in receiver and transmitter design.
Parallel Resonant Circuit Calculator
Parallel Resonant Circuit Calculator
Enter the inductor series resistance, L, and C to calculate resonant frequency, Q, dynamic resistance, bandwidth, and tank loop current magnification.
The Transmitter Tank Circuit
The most important practical application of the parallel resonant circuit in amateur radio is the transmitter output tank circuit. Almost every HF transmitter — from classic tube rigs to modern solid-state transceivers — uses a parallel LC tank in the output stage, and understanding how it works lets you understand why transmitters are designed the way they are.
Function of the Tank Circuit
The transistor or tube driving the tank circuit generates a rich harmonic spectrum — it is acting as a switch, conducting current in pulses rather than smoothly. These current pulses contain the fundamental frequency plus harmonics at 2f0, 3f0, 4f0, and so on. The tank circuit must accomplish three things:
- Frequency selection: Present high impedance at f0 so maximum voltage swing develops at the fundamental
- Harmonic suppression: Present low impedance at harmonic frequencies so little voltage develops there
- Impedance transformation: Transform the high impedance of the tank (typically 1–10 kΩ) down to the 50 Ω required by the antenna system
The tank accomplishes all three simultaneously through its parallel resonant behavior. At f0, impedance is RD (high — typically 1–20 kΩ). At 2f0 and higher harmonics, the tank is detuned and presents low impedance, so harmonic voltages are naturally small. The impedance transformation is achieved by taking the output from a tap partway down the coil (for tube rigs) or through a coupling capacitor to a second coil (for solid-state rigs).
Why the Tank Q Matters for Harmonic Suppression
FCC regulations require that amateur transmitters suppress harmonics by at least 40 dB below the fundamental on HF bands. The tank circuit provides the first stage of this suppression. Higher Q gives greater harmonic suppression: a tank with Q = 12 provides about 20 dB of second-harmonic suppression, while Q = 30 provides about 30 dB. Additional low-pass filters after the tank circuit complete the required attenuation.
The minimum Q recommended for transmitter tank circuits by most authorities is around 10–15 to ensure adequate harmonic suppression without excessive circuit losses. The practical maximum is limited by the Q of available inductors — once you have picked the best inductor you can afford, the maximum Q is fixed by coil resistance.
Experiment: Observing Parallel Resonance Impedance Peak
You need: Signal generator (audio or RF), 10 mH inductor, 2.5 µF non-polarized capacitor (gives f0 ≈ 1 kHz), 10 kΩ resistor in parallel (simulates tank load), oscilloscope or AC voltmeter, 1 kΩ series resistor (current-sensing resistor).
Setup: Connect signal generator → 1 kΩ series resistor → parallel LC tank (inductor + capacitor in parallel) → ground. The output voltage across the tank represents the tank impedance (high voltage = high impedance).
Procedure:
- Sweep the generator from 200 Hz to 5 kHz.
- Watch the voltage across the tank. It should peak near 1 kHz.
- Find the exact frequency of maximum voltage — this is f0.
- Note the voltage at f0 (call it Vmax).
- Find the two frequencies where voltage drops to Vmax/√2 ≈ 0.707 × Vmax. These are the −3 dB points.
- Calculate BW = f2 − f1 and Q = f0/BW.
- Compare measured Q to the theoretical value XL/RS where RS is the inductor's DC resistance (measure it with the ohmmeter).
Expected result: The tank voltage peaks sharply near 1 kHz. Measured Q will typically be slightly lower than theoretical Q because of inductor losses and any loading from the measurement instrument.
Series vs. Parallel: Side-by-Side Comparison
| Property | Series Resonant | Parallel Resonant |
|---|---|---|
| Impedance at resonance | Minimum = RS (ohms to tens of ohms) | Maximum = RD (kilohms to megaohms) |
| Current from source at resonance | Maximum | Minimum |
| Phase angle at resonance | 0° (purely resistive) | 0° (purely resistive) |
| Magnification effect | Voltage magnification: VL = VC = Q × VS | Current magnification: Itank = Q × Isource |
| Q formula | Q = XL / RS | Q = RD / XL |
| Bandwidth | BW = f0 / Q | BW = f0 / Q |
| Below resonance behavior | Capacitive, impedance rises | Inductive, impedance falls |
| Above resonance behavior | Inductive, impedance rises | Capacitive, impedance falls |
| Typical application | Series bandpass filter, antenna loading coil, crystal filter | Transmitter tank circuit, RF amplifier output, tuned IF stage |
| Driving source type | Low-impedance voltage source (series) | High-impedance current source (parallel) |
Frequently Asked Questions
Why is it called a "tank" circuit?
The name comes from the analogy with a water tank. A water tank stores a large volume of water and can supply it on demand without draining the pump line — it acts as a reservoir that smooths out fluctuations. The LC tank circuit stores a large amount of reactive energy (circulating between L and C) and smooths out the pulsed current from the amplifier, presenting a nearly sinusoidal voltage to the antenna. The tank "stores" RF energy just as a water tank stores water.
If the parallel circuit draws minimum current at resonance, how does it deliver power to a load?
The minimum current refers to the reactive (circulating) current drawn from the driving source — the current that would flow with no resistive load. When you connect a real load (like an antenna through an impedance transformer), a real component of current flows to deliver power. This real current is in phase with the tank voltage and represents the useful power transfer. The tank draws minimum reactive current while delivering maximum output voltage — which is exactly what you want in a transmitter output stage.
What happens to a parallel resonant circuit if its Q is very low?
A low-Q parallel resonant circuit has a broad, flat impedance peak — it no longer provides sharp frequency selectivity. The dynamic resistance RD is low (RD = Q² × RS), so the impedance advantage at resonance is modest. For Q below about 5, the resonant behavior becomes so broad that the circuit barely functions as a tuned element at all. This is why coil losses (the dominant source of low Q) are so critical in resonant circuit design — every fraction of an ohm of series resistance directly degrades circuit performance.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.