Chapter 11: Tuned Circuits

Digital techniques

Until late last century tuned circuits of L and C did the selecting in practically every receiver; modern radios mostly tune digitally. A crystal filter passes one fixed band and cannot be retuned, so it does not tune.

Inductive ones are added and capacitive ones subtracted

Inductive reactance counts as positive and capacitive as negative, so XTotal = XL1 + XL2 − XC1 − XC2. Both kinds of part have reactance. Swapping the signs gives the right size but calls an inductive total capacitive.

Its phase shift is the opposite of an inductor's

Current leads voltage in a capacitor but lags it in an inductor, so the two reactances work against each other and carry opposite signs. Size has nothing to do with it: XC is often the larger one.

200 Ω

XTotal = XL − XC = 340 − 140 = 200 Ω, positive and therefore inductive; adding the two as if both were positive gives 480 Ω.

450 Ω

Reactances of the same kind in series simply add: 175 + 275 = 450 Ω, as resistances would. 107 Ω, less than either, is what the reciprocal rule for parallel reactances gives, the wrong rule here.

One of half the value of either capacitor

Equal capacitors in series act as one of half the value, whose reactance is twice that of each: the same as adding the two reactances. Twice the value is what the pair gives in parallel.

207 Ω

Each XC = 1 ÷ (2π × 7 × 10⁶ × 220 × 10⁻¹²) = 103,3 Ω, and in series they add: 207 Ω, capacitive. Reading 220 pF as 22 pF gives ten times too much, 2,07 kΩ.

545 Ω

In series the inductances add to 6,2 µH, so XL = 2π × 14 × 10⁶ × 6,2 × 10⁻⁶ = 545 Ω, the two reactances added. Reading µH as nH gives a thousand times too little, 545 mΩ.

320 Ω

In series they act as 23,5 pF: XC = 1 ÷ (2π × 21 × 10⁶ × 23,5 × 10⁻¹²) = 322,5 Ω, nearest 320 Ω, twice one capacitor's 161 Ω. Stopping at one capacitor gives about 160 Ω.

1/XTotal = 1/X1 + 1/X2 + …

Parallel reactances combine by adding reciprocals, 1/XTotal = 1/X1 + 1/X2 + …, as parallel resistors do. Simply adding them is the rule for reactances in series.

220 Ω

1/XTotal = 1/330 + 1/660 = 1/220, so XTotal = 220 Ω, less than either. Adding them, 990 Ω, is the rule for reactances in series; 4,55 mΩ forgets to turn the sum of reciprocals over.

Half as large

Two equal capacitors in parallel make twice the capacitance, and doubling C halves XC; adding reciprocals, 1/X + 1/X = 2/X, agrees. Twice as large is what the same pair gives in series.

48,4 Ω

In parallel they act as 940 pF: XC = 1 ÷ (2π × 3,5 × 10⁶ × 940 × 10⁻¹²) = 48,4 Ω, half that of one. Reading 470 pF as 470 nF gives a thousand times too little, 48,4 mΩ.

176 Ω

In parallel they act as 1 ÷ (1/12 + 1/6) = 4 µH, so XL = 2π × 7 × 10⁶ × 4 × 10⁻⁶ = 176 Ω. Reading µH as mH gives a thousand times too much, 176 kΩ.

110 Ω

In parallel they make 3,2 nF: XC = 1 ÷ (2π × 455 × 10³ × 3,2 × 10⁻⁹) = 109 Ω, nearest 110 Ω. Taking 455 kHz as 455 MHz gives a thousand times too little, about 0,11 Ω.

Zero, as the two reactances cancel

At resonance XL and XC are the same size with opposite signs, so XTotal = XL − XC = 0; the frequency where this happens is the resonant frequency. An unlimited reactance belongs to an ideal parallel circuit.

Low near resonance and high elsewhere

Near resonance XL and XC nearly cancel, so the series total XL − XC is small; away from resonance it grows, which is why the circuit passes signals near resonance. High near resonance and low elsewhere describes a parallel circuit.

Like a capacitor

Below resonance XC is larger than XL, so the series total XL − XC is negative: the circuit acts as a capacitor. It turns inductive only above resonance, where XL has grown past XC.

Inductive, as XL is now the larger

XL rises with frequency while XC falls, so the two are equal at only one frequency, here 7 MHz. Above it XL is the larger and the series total is inductive; XC is larger only below resonance.

XTotal = XL − XC

The coil's reactance counts as positive and the capacitor's as negative, so XTotal = XL − XC. L and C cannot be added directly: work out each reactance at the frequency first. Adding the reactances ignores their opposite signs.

The coil's wire and the capacitor's leakage current

A little resistance comes from the wire the coil is wound with and from the capacitor's leakage current; good components keep it very low. A capacitor's losses are small, usually far below the coil's, but not nil.

212 Ω

XL = 2π × 7 × 10⁶ × 10 × 10⁻⁶ = 439,8 Ω and XC = 227,4 Ω, so the total is 439,8 − 227,4 = 212 Ω, inductive. Reading 7 MHz as 70 MHz gives XL 4398 Ω, XC 22,7 Ω: 4,38 kΩ.

162 Ω

XL = 2π × 5 × 10⁶ × 3,3 × 10⁻⁶ = 103,7 Ω and XC = 265,3 Ω, so the total is 265,3 − 103,7 = 162 Ω, capacitive. Reading 3,3 µH as 33 µH makes XL ten times larger and gives 771 Ω.

518 Ω

XL = 2π × 14 × 10⁶ × 8,2 × 10⁻⁶ = 721 Ω and XC = 203 Ω, so the total is 721 − 203 = 518 Ω, inductive. 721 Ω, the coil's reactance alone, leaves out the capacitor.

Inductive reactance is positive and capacitive reactance negative

Inductive reactance is taken as positive and capacitive reactance as negative, the signs standing for their opposite phase shifts. With the signs swapped, every total would come out with the wrong kind.

85 Ω

XTotal = 250 + 90 − 70 − 185 = 85 Ω, positive and so inductive: inductive reactances add and capacitive ones subtract. Adding all four, 595 Ω, ignores the capacitors' opposite sign; any resistance is added separately.

Its voltage divided by its current

Impedance, in ohms, is the ratio of voltage to current, Z = V ÷ I; with DC, which meets no reactance, it equals the resistance. Voltage multiplied by current gives power, in watts, not ohms.

150 Ω

Z = V ÷ I = 12 ÷ 0,08 = 150 Ω; dividing by 80 instead of 0,08 A gives 0,15 Ω.

6 mA

I = V ÷ Z = 9 ÷ 1500 = 0,006 A = 6 mA: Ohm's Law with the impedance in place of the resistance. Taking 1,5 kΩ as 1,5 Ω gives a thousand times too much, 6 A.

130 Ω

Resistance and reactance are 90° apart, so they combine as Z = √(R² + X²) = √(120² + 50²) = √16 900 = 130 Ω. Simply adding them, 170 Ω, ignores the 90° between them.

190 Ω

XL = 2π × 21 × 10⁶ × 1,2 × 10⁻⁶ = 158 Ω, so Z = √(100² + 158²) = 187 Ω, nearest 190 Ω. Reading 21 MHz as 21 kHz makes XL almost nothing, leaving about 100 Ω.

Hardly any, close to 0°

The phase shift depends on the ratio of resistance to reactance: mostly resistance gives a small shift, and mostly reactance gives nearly 90°.

Nearly 90°, with the voltage leading the current

Mostly reactance means a shift close to 90°, and because the reactance is inductive, positive, the voltage leads the current, as CIVIL reminds us. Nearly 0° is what mostly resistance gives.

Its ratio of resistance to reactance

An impedance combines resistance and reactance. Lots of resistance with little reactance gives little shift; lots of reactance gives nearly 90°, the shift of a pure reactance. So the ratio decides, not the resistance alone.

It is undefined: it grows without limit

At resonance the reciprocals 1/XL and 1/XC cancel, so 1/XTotal = 0 and XTotal = 1 ÷ 0: undefined, growing without limit. Zero is the total of the series circuit, not of the parallel one.

Blocks signals close to its resonant frequency, passing the rest

Its impedance is very high near resonance and small elsewhere, so in the path it holds back signals near resonance and lets others through. Passing signals near resonance is what a series tuned circuit in the path does.

A parallel LC circuit, resonant at that frequency

In the signal path, a parallel tuned circuit presents a very high reactance at its resonant frequency and blocks it, whereas a series one would pass it.

From inductive to capacitive

Below resonance the coil's smaller reactance dominates the pair, so the circuit is inductive; above resonance the capacitor's does, so it is capacitive. Capacitive first, then inductive, is the order of a series circuit.

720 Ω

1/XTotal = 1/240 + 1/(−360) = 1/720, so XTotal = 720 Ω, inductive and larger than either; ignoring the minus sign gives 144 Ω.

2,1 kΩ

1/XTotal = 1/1050 − 1/700 = −1/2100, so XTotal = −2100 Ω, 2,1 kΩ in size, capacitive as the capacitor has the smaller reactance; ignoring the signs gives 420 Ω.

310 Ω

XL = 2π × 5 × 10⁶ × 5 × 10⁻⁶ = 157 Ω and XC = 318 Ω; 1/X = 1/157 − 1/318, so X = 310 Ω, inductive, as the coil's smaller reactance dominates. Reading 5 MHz as 50 MHz gives 32,5 Ω.

390 Ω

XL = 2π × 3 × 10⁶ × 12 × 10⁻⁶ = 226 Ω and XC = 531 Ω; 1/X = 1/226 − 1/531, so X = 394 Ω, nearest 390 Ω. Reading 12 µH as 1,2 µH gives about 24 Ω.

It passes back and forth between the capacitor and inductor

The capacitor holds energy in its electric field and the coil in its magnetic field, and at resonance the energy keeps passing from one to the other and back. It never settles in the inductor.

Keeps the current going, recharging the capacitor with reversed polarity

An inductor resists any change in its current, so the collapsing field keeps the current moving the same way and charges the capacitor with the opposite polarity.

Both keep swapping energy between two forms

A pendulum trades potential energy at the top of its swing for kinetic energy at the bottom, as the tuned circuit trades electric and magnetic energy. Neither swings faster when pushed harder: its own parts set the rate.

The current flowing round the L and C loop

It is the current flowing round the closed loop of L and C, through both, as the energy passes between them. The current drawn from the source is a separate one, and far smaller.

It is often around 100 times as large

The circulating current flows round the loop of L and C, not in from the source, and can be far larger: 100 times the input is common, so 0,3 mA fed in may circulate 30 mA.

A very high impedance, so it draws little current

Near resonance the parallel pair's opposition is very high, so little current enters from outside even though a large current circulates inside. A very low impedance that draws a large current is how a series circuit behaves.

Their reactances are equal in size

At resonance XL = XC in size, one positive and one negative, so they cancel. The combined impedance is zero only for ideal parts in series; in parallel it is very high. Neither reactance is itself zero.

f = 1 ÷ (2π√LC)

The resonant frequency is f = 1 ÷ (2π√LC); leaving out the square root, as in 1 ÷ (2πLC), is a common slip.

7,07 MHz

f varies as 1 ÷ √C, so doubling C divides f by √2: 10 ÷ 1,414 = 7,07 MHz; forgetting the square root gives 5 MHz.

7,2 MHz

f varies as 1 ÷ √L, so a quarter of the inductance doubles f: 2 × 3,6 = 7,2 MHz; forgetting the square root gives 14,4 MHz.

Reducing the capacitance

In f = 1 ÷ (2π√LC) a smaller L or C gives a higher frequency, so reducing C raises f; increasing L, or both, lowers it. Resistance does not appear in the formula: it changes the Q, not f.

Made four times larger

f varies as 1 ÷ √C, so halving f needs √C to double, which takes four times the capacitance; merely doubling C divides f by only 1,41.

They are the same frequency

One formula, f = 1 ÷ (2π√LC), serves series and parallel circuits alike, so the same L and C resonate at the same frequency, not a higher one in parallel: XL equals XC at one frequency however they are joined.

Henry and farad

f = 1 ÷ (2π√LC) needs L in henry and C in farad to give hertz: 10 µH goes in as 10 × 10⁻⁶ H. Microhenry with picofarad, entered as they stand, gives the answer in gigahertz.

7,34 MHz

f = 1 ÷ (2π√(4,7 × 10⁻⁶ × 100 × 10⁻¹²)) = 1 ÷ (2π × 2,168 × 10⁻⁸) = 7,34 MHz. Reading 4,7 µH as 4,7 mH makes LC a thousand times bigger and f 31,6 times lower: 232 kHz.

159 MHz

LC = 10⁻⁷ × 10⁻¹¹ = 10⁻¹⁸, so √LC = 10⁻⁹ and f = 1 ÷ (2π × 10⁻⁹) = 159 MHz. Reading 10 pF as 100 pF makes LC ten times bigger and f √10 times lower: 50,3 MHz.

15,8 MHz

f = 1 ÷ (2π√(1,5 × 10⁻⁶ × 68 × 10⁻¹²)) = 1 ÷ (2π × 1,01 × 10⁻⁸) = 15,8 MHz. Reading 1,5 µH as 15 µH makes f √10 times lower, 4,98 MHz.

1 MHz

f = 1 ÷ (2π√(25 × 10⁻⁶ × 1 × 10⁻⁹)) = 1 ÷ (2π × 1,58 × 10⁻⁷) = 1,01 MHz, nearest 1 MHz. Reading 25 µH as 2,5 µH gives √10 times too much, about 3,2 MHz.

52 pF

C = 1 ÷ (4π²f²L) = 1 ÷ (39,5 × (7 × 10⁶)² × 10 × 10⁻⁶) = 51,7 × 10⁻¹² F = 51,7 pF, nearest 52 pF. Calling 10⁻¹² F a nanofarad gives a thousand times too much, 52 nF.

1,6 µH

L = 1 ÷ (4π²f²C) = 1 ÷ (39,5 × (14 × 10⁶)² × 82 × 10⁻¹²) = 1,58 × 10⁻⁶ H, nearest 1,6 µH. Reading 14 MHz as 1,4 MHz gives a hundred times too much, about 160 µH.

1,22 nF

C = 1 ÷ (4π²f²L) = 1 ÷ (39,5 × (455 × 10³)² × 100 × 10⁻⁶) = 1,22 × 10⁻⁹ F = 1,22 nF. Calling 10⁻⁹ F a picofarad gives a thousand times too little, 1,22 pF.

The resistance of the inductor's wire

Capacitors lose a little energy too, but normally far less than the coil, whose wire resistance causes most of the loss.

Low, as only a small resistance remains

The reactances cancel at resonance, but the circuit still contains a little resistance, so its impedance is low rather than exactly zero.

High, but held to a finite value by the losses

Ideal parts would give an undefined, infinite value, but real losses limit the impedance of a parallel circuit at resonance to a high, finite figure.

0,5 A

At resonance the coil's 400 Ω is cancelled by the capacitor's equal reactance, leaving only the 4 Ω: I = V ÷ R = 2 ÷ 4 = 0,5 A. Dividing by the coil's 400 Ω, 5 mA, forgets the capacitor.

Its losses: the higher the Q, the lower the losses

Q measures the losses: a high Q means low losses, a low Q high losses. More resistance with the same L and C lowers the Q, and with it the selectivity, though the resonant frequency stays put.

Q = XL ÷ R, at resonance

Q is the reactance of the coil at resonance, or of the capacitor, the same size there, divided by the circuit resistance: Q = XL ÷ R. R ÷ XL is upside down; more resistance must lower the Q.

95

Q = XL ÷ R = 380 ÷ 4 = 95. R ÷ XL = 0,0105 is upside down; adding the capacitor's equal reactance as well would wrongly double it to 190.

5 Ω

Q = XL ÷ R rearranges to R = XL ÷ Q = 625 ÷ 125 = 5 Ω; dividing the other way round, 125 ÷ 625, gives 0,2 Ω.

It is Q times the current drawn

The circulating current divided by the current fed in equals Q, so with a Q of 80 the circulating current is 80 times the input. Dividing by Q would make it smaller than the input.

110 mA

XL = 2π × 3,5 × 10⁶ × 20 × 10⁻⁶ = 440 Ω, so Q = 440 ÷ 4 = 110 and the circulating current is 110 × 1 mA = 110 mA. 1 mA is the current drawn, not the circulating current.

90

Q equals the circulating current divided by the input current: 180 ÷ 2 = 90. Reading 2 mA as 20 mA gives 9; dividing the other way round gives 0,011.

Between 50 and 200

Practical tuned circuits typically have a Q between 50 and 200; a Q below 1 would mean the losses outweigh the reactance.

The one with a Q of 160 is more selective

Selectivity increases with Q, so the circuit with a Q of 160 separates wanted from unwanted signals better than the one with a Q of 40.

How well it passes wanted signals and rejects unwanted ones

Selectivity is a tuned circuit's ability to pass the signals you want and block the rest, and it improves as the Q rises. Holding its frequency steady is stability, a different property.

84,8

XL = 2π × 3,6 × 10⁶ × 15 × 10⁻⁶ = 339 Ω, so Q = XL ÷ R = 339 ÷ 4 = 84,8. Leaving out the 2π gives XL = 54 Ω and a Q of 13,5.

100

XC = 1 ÷ (2π × 3,5 × 10⁶ × 150 × 10⁻¹²) = 303 Ω, so Q = XC ÷ R = 303 ÷ 3 = 101, nearest 100. Leaving out the 2π makes the Q 2π times too high, about 630.

50 kHz

Q = f ÷ bandwidth (the exam's formula sheet, p318), so bandwidth = f ÷ Q = 3 500 000 ÷ 70 = 50 000 Hz = 50 kHz. Reading 3,5 MHz as 350 kHz gives 5 kHz; multiplying, 245 MHz, is upside down.