6.1 Reactance, Impedance, Resonance & Quality Factor (Q)

6.1 Reactance, Impedance, Resonance & Quality Factor (Q)

In direct current (DC) circuits, electrical resistance ($R$) is the sole property that opposes the flow of electric charge, governed strictly by Ohm's Law ($V = I \cdot R$). In alternating current (AC) and radio frequency (RF) circuits, however, current and voltage fluctuate continuously in magnitude and reverse direction periodically. This continuous oscillation introduces time-dependent electromagnetic phenomena where energy is cyclically stored in and released from magnetic and electric fields. The opposition to AC current flow arising from this stored energy is called reactance ($X$), and the total opposition resulting from the combination of resistance and reactance is known as impedance ($Z$).

Mastering reactance, impedance, resonance, and the Quality Factor ($Q$) is fundamental to amateur radio. These principles govern the operation of antenna matching networks, intermediate frequency (IF) filters, transmitter output tank circuits, transmission line stubs, and RF oscillators.

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1. Inductive & Capacitive Reactance

Reactance represents the opposition that an inductive or capacitive circuit component presents to alternating current. Unlike pure resistance—which permanently dissipates electrical energy as heat—pure reactance stores energy temporarily during one quarter of the AC cycle and returns it to the source during the subsequent quarter cycle, consuming zero net average power.

+-----------------------------------------------------------------------------------------+
|                        REACTANCE & PHASE RELATIONSHIP SUMMARY                           |
|                                                                                         |
| Component    Symbol   Formula                  Frequency Trend   Phase Relationship     |
| --------------------------------------------------------------------------------------- |
| Inductor     X_L      X_L = 2 * π * f * L      Increases (f ↑)   Voltage LEADS Current  |
|                                                                  by 90° (+90° phase)    |
| Capacitor    X_C      X_C = 1 / (2 * π * f * C) Decreases (f ↑)  Current LEADS Voltage  |
|                                                                  by 90° (-90° phase)    |
+-----------------------------------------------------------------------------------------+

Inductive Reactance ($X_L$)

When an alternating current flows through an inductor (a coil of wire), the changing magnetic flux induces a counter-electromotive force (back-EMF) that opposes changes in current, in accordance with Lenz's Law. This opposition is inductive reactance ($X_L$).

$$X_L = 2\pi f L$$

Where:

• $X_L$ is inductive reactance in ohms ($\Omega$)
• $f$ is frequency in Hertz (Hz)
• $L$ is inductance in Henrys (H)
• $\pi \approx 3.14159$

Key Characteristics of Inductive Reactance:

1. Direct Proportionality: $X_L$ is directly proportional to both frequency and inductance. Doubling the operating frequency or doubling the inductance doubles the inductive reactance.
2. Phase Relationship (Voltage Leads Current): Because the back-EMF opposes current buildup, current cannot change instantaneously. In an ideal inductor, the voltage wave leads the current wave by 90 degrees (a phase angle of $+90^\circ$).
3. DC Behavior: At DC ($f = 0\text{ Hz}$), $X_L = 0,\Omega$. An ideal inductor acts as a dead short circuit to direct current.

Capacitive Reactance ($X_C$)

A capacitor consists of two conductive plates separated by an insulating dielectric. As alternating voltage is applied, charge accumulates on the plates until the electrostatic field voltage opposes the applied voltage. This opposition is capacitive reactance ($X_C$).

$$X_C = \frac{1}{2\pi f C}$$

Where:

• $X_C$ is capacitive reactance in ohms ($\Omega$)
• $f$ is frequency in Hertz (Hz)
• $C$ is capacitance in Farads (F)

Key Characteristics of Capacitive Reactance:

1. Inverse Proportionality: $X_C$ is inversely proportional to both frequency and capacitance. Doubling the frequency or doubling the capacitance cuts the capacitive reactance in half.
2. Phase Relationship (Current Leads Voltage): Current flows at its maximum when the capacitor plates are uncharged (zero voltage). As charge accumulates and voltage reaches its peak, current drops to zero. Thus, in an ideal capacitor, the current wave leads the voltage wave by 90 degrees (voltage lags current by 90 degrees, a phase angle of $-90^\circ$).
3. DC Behavior: At DC ($f = 0\text{ Hz}$), $X_C = \infty,\Omega$. An ideal capacitor completely blocks direct current, acting as an open circuit.

[!TIP] The Classic Mnemonic: "ELI the ICE man"

• ELI: In an Inductor (L), Voltage (E, electromotive force) leads Current (I).
• ICE: In a Capacitor (C), Current (I) leads Voltage (E).

Worked Example: Reactance Calculations at HF

Problem 1: Calculate the inductive reactance of a $10,\mu\text{H}$ ($10 \times 10^{-6}\text{ H}$) inductor at $14.15\text{ MHz}$ ($14.15 \times 10^6\text{ Hz}$).

$$X_L = 2\pi f L = 2 \times 3.14159 \times (14.15 \times 10^6) \times (10 \times 10^{-6}) = 2 \times 3.14159 \times 141.5 \approx 889.1\,\Omega$$

Problem 2: Calculate the capacitive reactance of a $100\text{ pF}$ ($100 \times 10^{-12}\text{ F}$) capacitor at $14.15\text{ MHz}$.

$$X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14159 \times (14.15 \times 10^6) \times (100 \times 10^{-12})} = \frac{1}{0.008891} \approx 112.5\,\Omega$$

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2. Complex Impedance ($Z$) in Series AC Circuits

When resistance, inductance, and capacitance coexist in a single series circuit, the total opposition to alternating current is called impedance ($Z$). Because the voltages across resistors, inductors, and capacitors are out of phase with each other, their oppositions cannot simply be added algebraically. Instead, they must be combined as orthogonal vectors (phasors) in the complex impedance plane.

                    +j (Inductive Reactance: +X_L)
                               ^
                               |         * Impedance Vector (Z)
                               |       / |
                               |      /  |
                               |     /   | Net Reactance
                               |    /    | (X = X_L - X_C)
                               |   / θ   |
  (Purely Resistive) 0 --------+--/------+--------> +R (Resistance)
                               |  \      |
                               |   \     |
                               |    \    |
                               |     \   |
                               |      \  |
                               |       v |
                               v
                    -j (Capacitive Reactance: -X_C)

The Mathematical Formulas for Series Impedance

In a series RLC circuit, net reactance ($X$) is the algebraic difference between inductive reactance and capacitive reactance:

$$X = X_L - X_C$$

Total impedance magnitude ($|Z|$) is calculated using the Pythagorean theorem:

$$Z = \sqrt{R^2 + X^2} = \sqrt{R^2 + (X_L - X_C)^2}$$

The phase angle ($\theta$) between the applied total voltage and circuit current is:

$$\theta = \arctan\left(\frac{X_L - X_C}{R}\right) = \arctan\left(\frac{X}{R}\right)$$

Circuit Behavioral States

• Inductive Circuit ($X_L > X_C$): The net reactance is positive ($+jX$). Total voltage leads current by phase angle $\theta$ ($0^\circ < \theta < 90^\circ$).
• Capacitive Circuit ($X_C > X_L$): The net reactance is negative ($-jX$). Total current leads voltage by phase angle $\theta$ ($-90^\circ < \theta < 0^\circ$).
• Resistive / Resonant Circuit ($X_L = X_C$): Net reactance is zero ($X = 0$). Impedance equals pure resistance ($Z = R$), and phase angle is exactly $0^\circ$.

Step-by-Step Worked Example: Series RLC Circuit

Problem: A series circuit contains a $30,\Omega$ non-inductive resistor, an inductor with $X_L = 80,\Omega$, and a capacitor with $X_C = 40,\Omega$. What is the total impedance and phase angle?

1. Calculate net reactance: $X = X_L - X_C = 80,\Omega - 40,\Omega = +40,\Omega$ (inductive).
2. Calculate total impedance magnitude: $$Z = \sqrt{R^2 + X^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\,\Omega$$
3. Calculate phase angle: $$\theta = \arctan\left(\frac{40}{30}\right) = \arctan(1.333) \approx +53.13^\circ$$ The circuit presents an impedance of $50,\Omega$ with voltage leading current by $53.13^\circ$.

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3. Electrical Resonance: Series vs. Parallel Resonant Circuits

Resonance occurs in an AC circuit containing both inductance and capacitance at the exact frequency where inductive reactance equals capacitive reactance ($X_L = X_C$). Setting the two reactance equations equal reveals the fundamental resonant frequency formula ($f_0$):

$$2\pi f_0 L = \frac{1}{2\pi f_0 C} \implies f_0^2 = \frac{1}{4\pi^2 L C} \implies f_0 = \frac{1}{2\pi\sqrt{LC}}$$

Where $f_0$ is resonant frequency in Hertz, $L$ is inductance in Henrys, and $C$ is capacitance in Farads.

+-----------------------------------------------------------------------------------------+
|                   SERIES RESONANCE VS. PARALLEL RESONANCE COMPARISON                   |
|                                                                                         |
| Characteristic         Series Resonant Circuit        Parallel Resonant Circuit (Tank)  |
| --------------------------------------------------------------------------------------- |
| Reactance Balance      X_L = X_C                      X_L = X_C                         |
| Net Reactance (X)      Zero (0 Ω)                     Zero (0 Ω)                        |
| Terminal Impedance     MINIMUM (Equal to series R)    MAXIMUM (Purely resistive, high)  |
| Line (Source) Current  MAXIMUM (I = V / R)            MINIMUM (Near zero line current)  |
| Internal Circulating I Equal to line current          EXTREMELY HIGH (Circulates in LC) |
| Phase Angle at f₀      0° (Purely resistive)          0° (Purely resistive)             |
| Below Resonance (f<f₀) Capacitive (X_C > X_L)         Inductive (Draws inductive current|
| Above Resonance (f>f₀) Inductive (X_L > X_C)          Capacitive (Draws capacitive I)   |
| Typical Application    Band-pass filters, traps       Amplifier plate/collector loads,  |
|                        (series notch), antenna feeds  antenna traps, oscillators        |
+-----------------------------------------------------------------------------------------+

Series Resonant Circuit Dynamics

In a series RLC circuit at resonance:

• The $+90^\circ$ voltage across the inductor and the $-90^\circ$ voltage across the capacitor are equal in magnitude and $180^\circ$ out of phase, completely canceling each other out ($V_L + V_C = 0$).
• Total impedance collapses to its absolute minimum value, which is simply the series internal resistance $R$.
• The source delivers maximum current ($I = V_{\text{source}} / R$).

Parallel Resonant Circuit (Tank Circuit) Dynamics

In a parallel LC network (commonly called an LC tank circuit):

• The branch currents through the inductor and capacitor are equal in magnitude and $180^\circ$ out of phase. They circulate back and forth between the electric field of the capacitor and the magnetic field of the inductor.
• Because these circulating branch currents cancel at the external nodes, the net current drawn from the external source (the line current) drops to a minimum.
• Consequently, the terminal impedance across the parallel tank reaches a maximum ($Z_{\text{tank}} = L / (R C) \approx Q \cdot X_L$).
• Parallel tank circuits are universally employed as resonant load impedances in RF power amplifiers to extract maximum output power at the fundamental frequency while suppressing harmonics.

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4. Quality Factor ($Q$) & Resonant Bandwidth

The Quality Factor ($Q$) is a dimensionless figure of merit that quantifies the energy storage efficiency and frequency selectivity of a resonant circuit. Mathematically, $Q$ is defined as $2\pi$ times the ratio of energy stored to energy dissipated per cycle:

$$Q = \frac{\text{Energy Stored per Cycle}}{\text{Energy Dissipated per Cycle}}$$

Formulas for Calculating Quality Factor

For a series resonant circuit with series loss resistance $R$:

$$Q = \frac{X_L}{R} = \frac{X_C}{R} = \frac{2\pi f_0 L}{R}$$

For a resonant circuit where center frequency ($f_0$) and half-power (-3 dB) bandwidth ($\text{BW}$) are known:

$$Q = \frac{f_0}{\text{BW}} \iff \text{BW} = \frac{f_0}{Q}$$

      Signal Amplitude
             ^
             |                 * Center Frequency (f₀)
       1.0 --+                / \
             |               /   \
     0.707 --+--------------*-----*--------- Half-Power (-3 dB) Level
   (1/√2)    |             /|     |\
             |            / |     | \
             |           /  |     |  \
             |          /   |     |   \
       0.0 --+---------+----+-----+----+-----> Frequency
                      f₁   f₀    f₂
                      |<--- BW --->|
                      (BW = f₂ - f₁)

The Half-Power (-3 dB) Bandwidth

The bandwidth ($\text{BW}$) of a tuned resonant circuit is formally defined as the frequency span between the lower half-power cutoff frequency ($f_1$) and upper half-power cutoff frequency ($f_2$):

$$\text{BW} = f_2 - f_1$$

At frequencies $f_1$ and $f_2$:

• Output voltage drops to $70.7%$ ($1/\sqrt{2} \approx 0.7071$) of the peak resonant voltage.
• Power delivered to the load drops to exactly $50%$ (half-power, or $-3\text{ dB}$) of maximum resonant power.

High $Q$ vs. Low $Q$ Engineering Trade-offs

• High $Q$ ($Q > 50$): Produces a very sharp, steep resonance curve with a narrow bandwidth. High $Q$ provides excellent adjacent-channel selectivity in receiver IF filters and tight harmonic suppression in transmitters. However, excessive $Q$ can cause ringing, clip audio sidebands, and require frequent retuning across the band.
• Low $Q$ ($Q < 10$): Produces a broad, flat frequency response with wide bandwidth. Low $Q$ is desirable in broadband matching networks and wideband antennas, but provides poor rejection of out-of-band signals.

Worked Example: Resonant Bandwidth Calculation

Problem: A tuned RF band-pass filter centered at $7.15\text{ MHz}$ ($7,150\text{ kHz}$) has a Quality Factor ($Q$) of $50$. What is the half-power ($-3\text{ dB}$) bandwidth of the filter?

$$\text{BW} = \frac{f_0}{Q} = \frac{7,150\text{ kHz}}{50} = 143\text{ kHz}$$

The filter passes frequencies within a $143\text{ kHz}$ window (from approximately $7.0785\text{ MHz}$ to $7.2215\text{ MHz}$) before signal power drops by more than $3\text{ dB}$.