Parallel Resonance

Parallel Resonance and Q-factor

Chapter 29: Parallel Resonance and Q-factor

29.1 Introduction to Parallel Resonance

A parallel resonant circuit is usually used to establish a condition of stable frequency or current designed to produce oscillation.[cite: 181, 182, 183] In this circuit, the capacitor and inductor are connected directly together, exchanging energy with each other.[cite: 183, 184] They work in much the same way the electrical frequency of a pendulum stabilizes an oscillator circuit.[cite: 185, 186, 188, 189]

Another use is where greatly reduced or increased impedance at a certain frequency is desired.[cite: 187, 190, 191] They can be used to block or tune a certain frequency range.[cite: 191, 192] These circuits are called filters.[cite: 195]

29.2 LR-C Parallel Network

This is a practical network where the coil has inductance L and resistance R, and next to it (in parallel) is pure capacitance C.[cite: 9, 11, 12, 199, 200]

Again we sum the admittances and make the imaginary parts sum to zero in order to start to derive an equation for resonance.[cite: 13, 14, 15, 16, 17]

Ycoil = 1 / (R + jXL) = (R – jXL) / (R2 + XL2)[cite: 4]
YC = 1 / (-jXC) = 1 / XC = jωC[cite: 7]

The equation to find the resonant frequency is given by:[cite: 18]

fr = (1 / 2π) × √(1 / LC – R2 / L2)[cite: 19, 108]

When R2 is very small, then fr = 1 / (2π√LC).[cite: 21, 22, 23]

29.3 Dynamic Resistance

Since the current at resonance is in phase with the voltage, the impedance of the network acts as a pure resistance.[cite: 34, 35, 38] The dynamic resistance RD is given by:[cite: 38]

RD = V / Ir[cite: 38, 39]

Where Ir is the current at resonance.[cite: 40] Substituting Yr = R / (R2 + ωr2L2) and following a few steps leads to:[cite: 41, 42]

RD = L / CR[cite: 43]

29.4 LR-CR Parallel Network

RC and RL are the resistances of the capacitor and coil respectively.[cite: 45, 46, 48, 49] The total network admittance is YT = YC + YL.[cite: 52]

The resonant frequency is given by:[cite: 53]

fr = (1 / (2π√LC)) × √((RL2 – L/C) / (RC2 – L/C))[cite: 53]

29.5 Q-factor in a Parallel Network

The Q-factor in a series R-L-C circuit is a measure of the voltage magnification.[cite: 55] In a parallel network, it is a measure of current magnification.[cite: 56, 58, 59] Energy flows between the inductor and capacitor. It leaves the capacitor and establishes a magnetic field in the inductor; this field then collapses and charges the capacitor, and so on.[cite: 57]

Circulating currents (i.e., IC or IL) may be several times greater than the supply current at resonance for the parallel network.[cite: 61, 62, 63]

Qr = Circulating Current / Current at Resonance = IC / Ir[cite: 66, 67, 68, 69, 70]

Given the capacitor current IC = V / XC = VωrC and current at resonance Ir = V / RD = V / (L/CR) = VCR / L:[cite: 71, 72]

Qr = (VωrC) / (VCR/L) = ωrL / R[cite: 74]

Which is the same equation for series resonance and can also be expressed as:[cite: 75, 78]

Qr = (1 / R) × √(L / C)[cite: 78]

Natural Frequency vs. Forced Resonant Frequency

fn is the natural frequency at which energy would naturally resonate in a closed loop of a capacitor and inductor.[cite: 93, 100, 101]

fn = 1 / (2π√LC)[cite: 96, 112]

fr is called the forced resonant frequency seen at the terminals. For a series circuit, the forced and natural frequencies coincide.[cite: 102, 105, 106]

fr = fn √(1 – 1 / Q2)[cite: 115]

Since the smallest Q we may come across is 10, we can see that even with a small Q, the difference between fn and fr tends to be small.[cite: 118, 120, 121]

Worked Problems

Admittance and Resonance Calculation

Problems re-started.[cite: 149] First, find the admittance then sum.[cite: 163] The total admittance is YT = YC + YL.[cite: 150, 158]

YC = 1 / ZC = 1 / (4 – j10)[cite: 151, 158]
YL = 1 / ZL = 1 / (3 + jXL)[cite: 151, 158]

Rationalizing the complex denominators:[cite: 165, 166, 167, 168]

YC = (4 + j10) / (42 + 102) = (4 + j10) / 116[cite: 166, 167]
YL = (3 – jXL) / (32 + XL2)[cite: 168]

At resonance, the complex parts of the admittances sum to zero:[cite: 169, 170, 171]

j10 / 116 = jXL / (32 + XL2)[cite: 173, 174, 175]
10(9 + XL2) = 116XL[cite: 176]
10XL2 – 116XL + 90 = 0[cite: 176, 177]

This forms a quadratic equation which can be solved for XL using:[cite: 178]

XL = (-b ± √(b2 – 4ac)) / 2a[cite: 178]