# Resonance

One of the most important phenomena in ac circuits, especially in RF engineering, is the property of resonance. This is a condition that occurs when capacitive and inductive reactance cancel each other out.

#### Series Resonance

**A series RLC circuit.**

Recall that capacitive reactance, XC, and inductive reactance, XL, can be equal in magnitude, although they are always opposite in effect. In any circuit containing an inductance and capacitance, there exists a frequency at which XL = −XC. This condition constitutes resonance. In a simple LC circuit, there is only one such frequency. But in some circuits involving transmission lines or antennas,there can be many such frequencies. The lowest frequency at which resonance occurs is called the resonant frequency, symbolized fo.

Refer to the schematic diagram of above figure. You should recognize this as a series RLC circuit. At some particular frequency, XL = −XC. This is inevitable if L and C are finite and nonzero. This frequency is fo for the circuit. At fo, the effects of capacitive reactance and inductive reactance cancel out. The result is that the circuit appears as a pure resistance, with a value that is theoretically equal to R. If R = 0, that is, if the resistor is a short circuit, then the circuit is called a series LC circuit, and the impedance at resonance will be theoretically 0 + j 0. The circuit will offer no opposition to the flow of alternating current at the frequency fo. This condition is series resonance. In a practical series LC circuit, there is always a little bit of loss in the coil and capacitor, so the real part of the complex impedance is not exactly equal to 0 (although it can be extremely small).

#### Parallel Resonance

**A parallel RLC circuit.**

Refer to the circuit diagram of above figure. This is a parallel RLC circuit. Remember that, in this sort of situation, the resistance R should be thought of as a conductance G, with G = 1/R. Then the circuit can be called a parallel GLC circuit. At some particular frequency fo, the inductive susceptance BL will exactly cancel the capacitive susceptance BC ; that is, BL = −BC. This is inevitable for some frequency fo, as long as the circuit contains finite, nonzero inductance and finite, nonzero capacitance. At the frequency fo, the susceptances cancel each other out, leaving theoretically zero susceptance. The admittance through the circuit is then very nearly equal to the conductance, G, of the resistor. If the circuit contains no resistor, but only a coil and capacitor, it is called a parallel LC circuit, and the admittance at resonance will be theoretically 0 + j 0. That means the circuit will offer great opposition to alternating current at fo, and the complex impedance will theoretically be infinite! This condition is parallel resonance. In a practical parallel LC circuit, there is always a little bit of loss in the coil and capacitor, so the real part of the complex impedance is not infinite (although it can be extremely large).

#### Calculating Resonant Frequency

The formula for calculating resonant frequency fo, in terms of the inductance L in henrys and the capacitance C in farads, is as follows:

**fo = 1/[2π(LC )1/2]**

Considering π = 3.14 to three significant figures, this formula can be simplified to: fo = 0.159/(LC )1/2 The 1⁄ 2 power of a quantity represents the positive square root of that quantity. The preceding formulas are valid for series-resonant and parallel-resonant RLC circuits. The formula will also work if you want to find fo in megahertz (MHz) when L is given in microhenrys (μH) and C is in microfarads (μF). These values are far more common than hertz, henrys, and farads in electronic circuits. Just remember that millions of hertz go with millionths of henrys, and with millionths of farads.

#### The Effects of R and G

Interestingly, the value of R or G does not affect the resonant frequency in either type of circuit. But these quantities are significant, nevertheless! The presence of nonzero resistance in a series-resonant circuit, or nonzero conductance in a parallel-resonant circuit, makes the resonant frequency less welldefined. Engineers say that the resonant frequency response becomes “more broad” or “less sharp.” In a series circuit, the resonant frequency response becomes more broad as the resistance increases. In a parallel circuit, the resonant frequency response becomes more broad as the conductance increases. The sharpest possible responses occur when R = 0 in a series circuit, and when G = 0 (that is, R = ) in a parallel circuit.

#### Problem 1

**Find the resonant frequency of a series circuit with an inductance of 100 μH and a capacitance of 100 pF.**

First, convert the capacitance to microfarads: 100 pF = 0.000100 μF. Then find the product LC = 100 × 0.000100 = 0.0100. Take the square root of this, getting 0.100. Finally, divide 0.159 by 0.100, getting fo = 1.59 MHz.

#### Problem 2

**Find the resonant frequency of a parallel circuit consisting of a 33-μH coil and a 47-pF capacitor.**

Again, convert the capacitance to microfarads: 47 pF = 0.000047 μF. Then find the product LC = 33 × 0.000047 = 0.00155. Take the square root of this, getting 0.0394. Finally, divide 0.159 by 0.0394, getting fo = 4.04 MHz.

#### Problem 3

**Suppose you want to design a circuit so that it has fo = 9.00 MHz. You have a 33-pF fixed capacitor available. What size coil will be needed to get the desired resonant frequency?**

Use the formula for the resonant frequency, and plug in the values. This will allow you to use simple arithmetic to solve for L. Convert the capacitance to microfarads: 33 pF = 0.000033 μF.

Then calculate as follows:

**fo = 0.159/(LC )1/2
9.00 = 0.159/(L × 0.000033)1/2
9.002 = 0.1592/(0.000033 × L)
81.0 = 0.0253/(0.000033 × L)
81.0 × 0.000033 × L = 0.0253
0.00267 × L = 0.0253
L = 0.0253/0.00267
= 9.48 μH **

#### Problem 3

**Padding capacitors (Cp ) allow limited adjustment of the resonant frequency in a series LC circuit (as shown at A), or in a parallel LC circuit (as shown at B).**

**Suppose a circuit must be designed to have fo = 455 kHz. A coil of 100 μH is available. What size capacitor is needed?**

Convert the frequency to megahertz: 455 kHz = 0.455 MHz. Then the calculation proceeds in the same way as with the preceding problem:

**fo = 0.159/(LC )1/2
0.455 = 0.159/(100 × C )1/2
0.4552 = 0.1592/(100 × C )
0.207 = 0.0253/(100 × C )
0.207 × 100 × C = 0.0253
20.7 × C = 0.0253
C = 0.0253/20.7
= 0.00122 μF
= 1220 pF **

In practical circuits, variable inductors and/or variable capacitors are often placed in tuned circuits, so that small errors in the frequency can be compensated for. The most common approach is to design the circuit for a frequency slightly higher than fo, and to use a padder capacitor in parallel with the main capacitor (above figure).