Electronic Devices & Circuits · Chapter 29

Oscillators: Barkhausen Criterion and Types

Part 6 · Feedback that is deliberately made unstable, and kept just unstable enough.

Dr. Mithun MondalEngineering DevotionDigital Textbook
i Learning Objectives

By the end of this chapter you should be able to:

  • Distinguish an oscillator from an amplifier in terms of loop gain, and state both parts of the Barkhausen criterion.
  • Explain why a practical design sets \(|A\beta|\) slightly above unity and what supplies the starting signal.
  • Compare thermistor, lamp, diode and FET amplitude stabilisation and say what each costs in distortion.
  • Derive \(\beta = 1/3\) at \(f = 1/2\pi RC\) for the Wien network and the resulting gain-of-three requirement.
  • Derive the \(1/29\) attenuation and \(f = 1/(2\pi RC\sqrt6)\) of the three-section RC phase-shift network.
  • Apply the general three-reactance condition to obtain the Hartley, Colpitts and Clapp oscillators.
  • Explain the crystal equivalent circuit, its series and parallel resonances, and why its \(Q\) and stability are so high.
  • Compute the period of a 555 or op-amp astable, and describe drift and phase noise qualitatively.

Every circuit in this course so far has answered to something. A rectifier answers to the mains, an amplifier answers to its input, and a filter answers to whatever is put into it. An oscillator answers to nothing: it is switched on and it produces a waveform, indefinitely, from a d.c. supply and nothing else. Since energy is conserved and nothing is being converted to a signal by magic, an oscillator is best understood as a device that converts d.c. power into a.c. power at a frequency the circuit itself decides.

The mechanism is one already familiar in its unwanted form. Chapter 26 wanted large loop gain and Chapter 27 wanted the feedback to be negative; when the phase shift round a high-gain loop reaches \(180^\circ\) at some frequency and the loop gain is still above unity there, the amplifier breaks into oscillation and the designer adds compensation to stop it. This chapter turns that failure into the specification. It builds loops that are deliberately unstable at exactly one frequency and stable everywhere else, then adds a mechanism to hold the instability at the boundary so that the amplitude neither dies away nor grows until the amplifier clips. Chapter 28 stopped one step short of this: the equal-component Sallen-Key section with \(K = 3\) is a Wien-bridge oscillator, and the whole of the present chapter is on the far side of that line.

The Barkhausen condition \(|A\beta| = 1\) cannot be satisfied by design, only by accident or by feedback. No resistor ratio holds to better than a per cent, and \(\beta\) drifts with temperature; a loop built for \(|A\beta| = 1.000\) will in practice have 0.98 or 1.02 and will either never start or will grow until it clips. Every real oscillator is therefore designed with the loop gain a few per cent above unity and then given a mechanism — a lamp, a thermistor, a pair of diodes, a FET used as a voltage-controlled resistance — whose job is to reduce \(A\) as the amplitude rises. The oscillation settles at whatever amplitude drives the loop gain down to exactly one, and the quality of an oscillator is very largely the quality of that amplitude-control loop rather than of the resonator.

1 From Amplifier to Oscillator: the Barkhausen Criterion

Take the standard feedback loop of Chapter 26: an amplifier of gain \(A\), a network that returns a fraction \(\beta\) of the output to the input, and a summing point. The closed-loop gain is

\[ A_f = \frac{A}{1 - A\beta} \]

with the sign convention that \(A\beta\) is the loop gain measured all the way round. For an amplifier we arrange \(A\beta\) to be negative and large, so \(A_f \approx -1/\beta\) and the gain is set by the passive network. Now ask what happens if \(A\beta\) is made equal to \(+1\). The denominator vanishes, and the equation says the circuit has a finite output for zero input. That is not a mathematical absurdity: it is the definition of an oscillator.

The Barkhausen criterion
\(|A\beta| = 1\) and \(\angle A\beta = 0^\circ\)

Sustained oscillation requires both conditions, at the same frequency. The phase condition selects the frequency — there is usually only one frequency at which the loop phase comes to zero — and the magnitude condition then determines whether the oscillation at that frequency grows, holds or dies. A loop that satisfies the phase condition at 1 kHz but has \(|A\beta| = 0.9\) there is an amplifier with a peak in its response, not an oscillator.

The loop, with nothing driving it0no inputAv_oβfeedback networksamples v_o, returns βv_oLoop gain AβBarkhausen: |Aβ| = 1and ∠Aβ = 0° (or 360°)The loop must return to thesumming point a signal equalin size and identical in phaseto the one it started from.What actually happens at switch-on|Aβ| < 1|Aβ| = 1|Aβ| > 1dies away — an amplifierthe wanted steady stategrows — how it startsA practical oscillator is designed with |Aβ| a few per cent above 1 and then pulled back to exactly 1 by the stabiliser of Section 2.
Figure 29.1 — The Barkhausen criterion and the three possible start-up behaviours

The commonest confusion is the belief that a real circuit is built with \(|A\beta|\) set to one. It is not, for two reasons.

Nothing would start it. With the loop gain at exactly unity the amplitude is whatever it was, and at switch-on it was zero. What actually starts an oscillator is noise: the thermal noise of the resistors, the shot noise of the transistor, and the step of the supply rail coming up. All of these contain a component at the oscillation frequency, of the order of microvolts, and the frequency-selective network passes it round the loop. If \(|A\beta| > 1\) that component is bigger on each circuit of the loop, and the envelope grows exponentially, as the right-hand panel of Figure 29.1 shows. If \(|A\beta| < 1\) it shrinks and the output stays at noise level for ever.

Nothing would hold it. Even if the loop gain could be set to 1.000 at switch-on, it would not stay there. A 1 per cent change in a feedback resistor, a 10°C change in temperature, or the ageing of a capacitor moves it, and the movement is one-directional in its consequences: a loop gain of 0.99 stops, a loop gain of 1.01 grows without limit until something saturates. There is no stable equilibrium at all, only a knife edge.

The resolution is to design for \(|A\beta| \approx 1.05\) so that the circuit starts reliably from noise, and to include an element whose effective gain falls as the amplitude rises. The oscillation then grows until the loop gain has been pulled down to exactly one, and stops growing there. The equilibrium is now stable: any excursion upward reduces the gain and pushes it back, any excursion downward increases it. Section 2 is about the choice of that element, and it is the part of oscillator design where most of the engineering effort actually goes.

If no such element is included, the amplifier limits by clipping on its supply rails. The oscillation is then stable in amplitude but badly distorted, the output closer to a square wave than a sine. For a signal generator that is unacceptable; for a digital clock it does not matter, which is why the relaxation oscillators of Section 7 run rail to rail.

2 Holding the Amplitude Steady

An amplitude stabiliser is a slow, non-linear negative-feedback loop wrapped round the fast, linear positive-feedback loop of the oscillator. Its input is the oscillation amplitude and its output is the loop gain, and it has one difficult requirement: it must respond slowly compared with one cycle of the oscillation. A stabiliser that responded within a cycle would be modulating the waveform, which is another way of saying that it would be distorting it.

  • The tungsten lamp. The oldest solution, and Hewlett’s original 1939 answer in the first Hewlett-Packard product. A small incandescent lamp is used as the lower resistor \(R_i\) of the non-inverting gain network. Its filament has a positive temperature coefficient, so as the amplitude rises the filament heats, its resistance rises, and the gain \(1 + R_f/R_i\) falls. The thermal time constant of a filament is tens of milliseconds, which is thousands of cycles at audio frequency, so the lamp sees only the r.m.s. amplitude and adds essentially no distortion — below 0.01 per cent is routine. Its weaknesses are that lamps are not specified as components, that the amplitude can "bounce" for a second after switch-on, and that at low frequencies the filament starts to follow the waveform.
  • The NTC thermistor. The same idea with the opposite temperature coefficient, so it goes in the \(R_f\) position instead: rising amplitude, rising temperature, falling resistance, falling gain. Cheaper and more repeatable than a lamp, with the same slow response and the same low distortion.
  • Diode limiting. A pair of back-to-back diodes across part of \(R_f\). Below about 0.5 V across them they are open circuits and the gain is the full designed value; above that they conduct, shunting \(R_f\) and reducing the gain. Cheap, immediate and certain, and what most laboratory Wien-bridge circuits use — but the worst of the four for distortion, because the limiting happens within each cycle rather than over many. The tips of the sine are compressed and 1 to 5 per cent distortion is typical; a resistor in series with each diode softens the knee considerably.
  • The FET as a variable resistance. The best modern solution. A JFET biased in its ohmic region, where Chapter 21 showed it behaves as a resistance controlled by \(V_{GS}\), forms part of the gain-setting network. The output is rectified and smoothed to a d.c. proportional to amplitude, and that d.c. drives the gate, so the loop holds the amplitude to a reference rather than to whatever a component happens to do. Distortion below 0.001 per cent is achievable, at the cost of a rectifier, a filter and a control amplifier.
i Why the stabiliser must be slow

The oscillation is a sinusoid of frequency \(f_0\); its amplitude envelope is a d.c. quantity. A stabiliser that responds in a time comparable with \(1/f_0\) cannot tell them apart, and its gain modulation appears as sidebands at \(f_0 \pm f_{\text{mod}}\) — distortion. A stabiliser whose time constant is 1000 cycles sees only the envelope. The penalty is a slow settling time after switch-on or after the frequency is retuned, and a well-known "thump" as the amplitude overshoots and comes back. Fast, clean and simple: choose two.

3 The Wien-Bridge Oscillator

The Wien-bridge oscillator is the standard audio-frequency sine source, and its selective network is a series \(RC\) arm and a parallel \(RC\) arm forming a divider from the output back to the amplifier input. Both arms use the same \(R\) and the same \(C\).

Let \(Z_s = R + 1/j\omega C\) and \(Z_p = R\,\|\,1/j\omega C = R/(1+j\omega RC)\). Then

\[ \beta = \frac{Z_p}{Z_s+Z_p} = \frac{j\omega RC}{(1+j\omega RC)^2 + j\omega RC} \]

Writing \(x = \omega RC\), the denominator is \(1 + 2jx - x^2 + jx = (1-x^2) + 3jx\), so

\[ \beta = \frac{jx}{(1-x^{2}) + 3jx} = \frac{1}{3 + j\left(x - \dfrac{1}{x}\right)} \]

A form worth memorising. The imaginary part vanishes when \(x = 1\), and only then; the phase shift of the network is zero at exactly one frequency and the magnitude is a maximum there too.

Setting \(x = \omega RC = 1\) gives the frequency, and substituting it back gives the attenuation:

\[ f_0 = \frac{1}{2\pi RC}, \qquad \beta(f_0) = \frac{1}{3}, \qquad \angle\beta(f_0) = 0^\circ \]

Since the network inverts nothing, the amplifier must not invert either — hence the non-inverting connection of Figure 29.2 — and to make \(|A\beta| = 1\) with \(\beta = 1/3\) the amplifier gain must be exactly 3. For a non-inverting op-amp stage \(A = 1 + R_f/R_i\), so \(R_f/R_i = 2\).

Wien-bridge oscillator, f_0 = 1591.5 Hznode PC 10 nFR 10 kΩseries armR10 kΩC10 nFparallel arm+v_outR_f 20 kΩlampR_i = 10 kΩ hotβ = 1/[3 + j(x − 1/x)], x = ωRCat x = 1: β = 1/3 exactly, ∠β = 0°so A must be 3: R_f/R_i = 2f_0 = 1/(2πRC) = 1591.5 HzGanging the two R values as a dualpotentiometer 1 kΩ–10 kΩ tunesthe oscillator over 1.59–15.92 kHzwith no change of amplitude.
Figure 29.2 — The Wien-bridge oscillator with lamp amplitude stabilisation
1 Worked Example 29.1 — A 1.59 kHz Wien-bridge oscillator

Take \(R = 10\ \text{k}\Omega\) and \(C = 10\) nF in both arms. Then

\[ f_0 = \frac{1}{2\pi(10^{4})(10^{-8})} = \frac{1}{6.2832\times10^{-4}} = 1591.5\ \text{Hz} \]

Checking the network at three frequencies confirms both parts of the criterion: at 795.8 Hz, \(|\beta| = 0.2981\) and \(\angle\beta = +26.6^\circ\); at 1591.5 Hz, \(|\beta| = 0.3333\) and \(\angle\beta = 0.0^\circ\); at 3183.1 Hz, \(|\beta| = 0.2981\) and \(\angle\beta = -26.6^\circ\). Only the middle frequency satisfies the phase condition, and it is also where \(|\beta|\) is largest, so the loop gain requirement is hardest to meet anywhere else — two independent reasons why the circuit oscillates at \(f_0\) and nowhere else.

Setting the gain. With \(R_i = 10\ \text{k}\Omega\), \(R_f = 20\ \text{k}\Omega\) gives \(A = 3.00\) and \(|A\beta| = 1.000\) — which, as Section 1 argued, will not start. Making \(R_f = 21\ \text{k}\Omega\) gives \(A = 3.10\) and \(|A\beta| = 1.033\), a 3.3 per cent excess: enough to start from noise in a few tens of milliseconds and small enough that a lamp can absorb it. The lamp is chosen so that its hot resistance is 10 kΩ at the wanted output amplitude; if the amplitude tries to rise, the filament heats, \(R_i\) rises, \(A\) falls towards 3, and the growth stops.

Tuning. Both \(R\) values must change together, which is why the classic instrument uses a two-gang potentiometer. A 1 kΩ–10 kΩ dual gang gives \(f_0\) from \(1/[2\pi(10^{4})(10^{-8})] = 1591.5\) Hz down at the top of its travel to \(1/[2\pi(10^{3})(10^{-8})] = 15\,915\) Hz at the bottom — a 10:1 range on one control, with \(\beta\) staying at exactly \(1/3\) throughout because it depends only on the two arms being equal, not on their value. Switched capacitors then move the whole decade. That property — wide tuning at constant amplitude — is the reason the Wien bridge became the standard audio oscillator.

4 The RC Phase-Shift Oscillator

The other classical \(RC\) oscillator gets \(180^\circ\) from its network and the other \(180^\circ\) from an inverting amplifier. One \(RC\) section approaches \(90^\circ\) but never reaches it, so three are needed, each contributing \(60^\circ\).

Analyse the three-section high-pass ladder — \(C\) in series, \(R\) to ground, three times over — driven from a voltage source and loaded by the (very high) input impedance of the amplifier. Writing \(\alpha = 1/\omega RC\), the mesh equations give

\[ \beta = \frac{V_f}{V_o} = \frac{1}{\left(1 - 5\alpha^{2}\right) + j\left(\alpha^{3} - 6\alpha\right)} \]

The phase condition demands that the loop phase be zero; the amplifier already supplies \(180^\circ\), so the network must supply the other \(180^\circ\), which means \(\beta\) must be real and negative. Setting the imaginary part to zero:

\[ \alpha^{3} - 6\alpha = 0 \;\Longrightarrow\; \alpha^{2} = 6 \;\Longrightarrow\; \frac{1}{\omega RC} = \sqrt6 \;\Longrightarrow\; f_0 = \frac{1}{2\pi RC\sqrt6} \]

Substituting \(\alpha^2 = 6\) back into the real part gives the attenuation:

\[ \beta(f_0) = \frac{1}{1 - 5(6)} = \frac{1}{-29} = -\frac{1}{29} = -0.03448 \]

The minus sign is the \(180^\circ\); the magnitude \(1/29\) is the price. The amplifier must therefore have a gain of at least 29, which for an inverting op-amp stage means \(R_f/R_1 = 29\) — with \(R_1 = 10\ \text{k}\Omega\), \(R_f = 290\ \text{k}\Omega\).

2 Worked Example 29.2 — Phase-shift oscillator, and why the phase condition is weak

With \(R = 10\ \text{k}\Omega\) and \(C = 10\) nF in each of the three sections, \(\sqrt6 = 2.4495\) and

\[ f_0 = \frac{1}{2\pi(10^{4})(10^{-8})(2.4495)} = \frac{1591.5}{2.4495} = 649.7\ \text{Hz} \]

Now look at how sharply the phase passes through \(180^\circ\). Evaluating the network 10 per cent below \(f_0\) gives \(\angle\beta = -173.9^\circ\) with \(|\beta| = 0.0276\); at \(f_0\) it is \(180.0^\circ\) with \(|\beta| = 0.0345\); 10 per cent above it is \(+174.4^\circ\) with \(|\beta| = 0.0418\). The phase moves by only about 12° for a 20 per cent change of frequency, which is a very shallow slope, and the frequency at which the loop actually settles is correspondingly sensitive to any additional phase shift anywhere — from the amplifier’s own roll-off, from stray capacitance, from a change in load. Compare the Wien network, whose phase passes through zero over a range of \(\pm 26.6^\circ\) within one octave, and the crystal of Section 6, whose phase swings by \(180^\circ\) within 768 parts per million.

The BJT version deserves a mention because it appears in every examination paper. With a common-emitter stage the network is loaded by the transistor’s input resistance, and the last resistor of the ladder is usually absorbed into it. Working through the loaded analysis gives a minimum current gain of

\[ h_{fe(\min)} = 4\frac{R_C}{R} + 23 + 29\frac{R}{R_C} \]

which is minimised by differentiating with respect to \(R_C/R\): the optimum is \(R_C/R = 2.693\) and the minimum required \(h_{fe}\) is 44.5. Any general-purpose transistor clears that comfortably, but the result explains why the ratio \(R_C/R\) is not free and why a phase-shift oscillator built with \(R_C = R\) needs \(h_{fe} \ge 56\).

The phase-shift oscillator is simpler than the Wien bridge and needs no amplitude stabilisation to give a usable output, because the network attenuates harmonics heavily — it is a three-pole high-pass filter, so the second harmonic is fed back much more strongly than the fundamental and the loop simply will not support it. What it cannot do is tune, since three ganged resistors are needed and \(\beta\) then depends on all three matching, and its frequency accuracy is poor for the reason just given. It is used for fixed low-frequency tones and as a teaching circuit; for anything requiring a settable frequency the Wien bridge wins.

5 LC Oscillators: Hartley, Colpitts and Clapp

Above about 100 kHz the \(R\) and \(C\) values an \(RC\) oscillator needs become inconveniently small, stray capacitance dominates, and op-amp gain has run out. The tuned circuit takes over, bringing a family of oscillators derivable from one condition.

Consider an inverting amplifier of voltage gain \(-A\) with three purely reactive impedances arranged as in the first panel of Figure 29.3: \(Z_1\) from input to common, \(Z_2\) from output to common, and \(Z_3\) bridging output back to input. The feedback fraction is the divider formed by \(Z_3\) and \(Z_1\) across the output, and setting the loop gain to unity with zero phase leads to two requirements:

\[ X_1 + X_2 + X_3 = 0 \qquad\text{and}\qquad |A| \ge \frac{X_2}{X_1} \]

Since the three reactances must sum to zero and an inductor and a capacitor have opposite signs, \(X_1\) and \(X_2\) must be of the same kind and \(X_3\) of the other. Two choices exist, and they are the two classical LC oscillators.

\(X_1\) (input–common)\(X_2\) (output–common)\(X_3\) (bridging)FrequencyFeedback fraction
Hartley\(L_2\)\(L_1\)\(C\)\(f = 1/2\pi\sqrt{L_TC}\), \(L_T = L_1+L_2+2M\)\(\beta = L_2/L_1\)
Colpitts\(C_2\)\(C_1\)\(L\)\(f = 1/2\pi\sqrt{LC_T}\), \(1/C_T = 1/C_1+1/C_2\)\(\beta = C_1/C_2\)
Clapp\(C_2\)\(C_1\)\(L + C_3\)\(1/C_T = 1/C_1+1/C_2+1/C_3\)\(\beta = C_1/C_2\)
General three-reactance form and the three tanks it generates−AinoutZ_1Z_2Z_3General formX_1 + X_2 + X_3 = 0|A| ≥ X_2/X_1collectorbaseL_1100 μHL_210 μHC1 nFHartleyf = 1/2π√(L_T C), L_T = L_1+L_2collectorbaseC_1 1 nFC_2 10 nFL 100 μHColpittsC_T = C_1C_2/(C_1+C_2) = 909 pFcoll.baseC_1C_2LC_3 100 pFClappC_3 dominates: 1/C_T = ∑1/C
Figure 29.3 — The general three-reactance oscillator and the Hartley, Colpitts and Clapp tanks

The Hartley uses a tapped inductor and a single capacitor. The tap is at signal common, the collector drives one end and the base is fed from the other, so the tank is an autotransformer and the feedback fraction is the turns ratio. With \(L_1 = 100\ \mu\)H, \(L_2 = 10\ \mu\)H and \(C = 1\) nF (taking the mutual inductance as negligible), \(L_T = 110\ \mu\)H and

\[ f = \frac{1}{2\pi\sqrt{(110\times10^{-6})(10^{-9})}} = 479.9\ \text{kHz} \]

with \(\beta = 10/100 = 0.1\), so the stage needs a voltage gain of at least 10. A tapped inductor is awkward to make and to buy, which is the Hartley’s main disadvantage.

The Colpitts exchanges the roles: two capacitors in series form the tapped element and a single plain inductor bridges them. This is much the more popular arrangement, because capacitors are far easier to obtain in matched pairs than a tapped coil is to wind. The two capacitors are in series as far as the tank is concerned, so with \(C_1 = 1\) nF at the collector and \(C_2 = 10\) nF at the base,

\[ C_T = \frac{C_1C_2}{C_1+C_2} = \frac{(1)(10)}{11}\ \text{nF} = 909.1\ \text{pF}, \qquad f = \frac{1}{2\pi\sqrt{(100\times10^{-6})(909.1\times10^{-12})}} = 527.9\ \text{kHz} \]

The feedback fraction is the reactance ratio \(\beta = X_{C1}/X_{C2} = C_1/C_2 = 0.1\), so again a gain of 10 is needed. Increasing \(C_2\) increases \(\beta\) and makes starting easier but lowers the frequency, so the two are not independent — which is the Colpitts’ own weakness.

The Clapp fixes the weakness that both share. In either circuit the tuning capacitors sit directly across the device terminals, so the transistor’s own capacitances — which vary with temperature and with bias — add to them and pull the frequency. The Clapp adds a third, much smaller capacitor \(C_3\) in series with the inductor. All three are then in series for the purpose of resonance, and because reciprocals add, the smallest one dominates:

\[ \frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
3 Worked Example 29.3 — What the Clapp’s third capacitor buys

Take the Colpitts above — \(L = 100\ \mu\)H, \(C_1 = 1\) nF, \(C_2 = 10\) nF, \(f = 527.9\) kHz — and add \(C_3 = 100\) pF in series with the inductor. Now \(1/C_T = 1/1000 + 1/10\,000 + 1/100\) pF\(^{-1}\), giving \(C_T = 90.09\) pF and \(f = 1.6768\) MHz. Note that \(C_T\) is 90.1 per cent of \(C_3\) alone: the small capacitor has taken over.

Now suppose 5 pF of transistor and stray capacitance appears across \(C_1\). In the Colpitts, \(C_1\) becomes 1005 pF, \(C_T\) becomes 913.6 pF and the frequency falls to 526.66 kHz — a shift of \(-2264\) parts per million. In the Clapp the same 5 pF changes \(C_T\) from 90.090 to 90.070 pF and the frequency from 1.676801 to 1.676426 MHz, a shift of \(-224\) ppm. The same stray causes ten times less frequency error, because \(C_3\) now sets the frequency and \(C_1\) and \(C_2\) have been demoted to setting only the feedback fraction. The price is a smaller circulating current in the tank and therefore a smaller loop gain, so a Clapp is harder to start and is not usually made tunable over more than about a 1.5:1 range.

6 Crystal Oscillators

A quartz crystal is a thin slice of quartz with metal electrodes plated on two faces. Quartz is piezoelectric: a voltage across it deforms it mechanically, and a mechanical deformation generates a voltage. Applying an alternating voltage therefore sets the slice vibrating, and because a mechanical resonator made of a stiff, low-loss, almost perfectly elastic material has extraordinarily small losses, the resonance is extraordinarily sharp. Electrically the crystal behaves exactly as the equivalent circuit of Figure 29.4, in which the mechanical properties appear as electrical ones: the mass of the slice as an inductance \(L_s\), its stiffness as a capacitance \(C_s\), and its mechanical losses as a resistance \(R_s\), all shunted by the ordinary electrostatic capacitance \(C_p\) of the two electrodes with quartz between them.

The numbers are what make the device remarkable. For a 1 MHz crystal a representative set is \(L_s = 3.3\) H, \(C_s = 0.00768\) pF, \(R_s = 200\ \Omega\) and \(C_p = 5\) pF. No wound inductor of 3.3 H could have anything like 200 Ω of loss, and no ordinary capacitor is made in units of thousandths of a picofarad; these are mechanical quantities wearing electrical clothes.

The motional arm alone is series-resonant at

\[ f_s = \frac{1}{2\pi\sqrt{L_sC_s}} = \frac{1}{2\pi\sqrt{(3.3)(7.68\times10^{-15})}} = 999.73\ \text{kHz} \]

where its reactance passes through zero and its impedance falls to \(R_s = 200\ \Omega\). Above that frequency the motional arm is net inductive, and at some slightly higher frequency it resonates with the holder capacitance \(C_p\). That is the parallel or anti-resonance, at which \(C_s\) and \(C_p\) appear in series:

\[ f_p = \frac{1}{2\pi\sqrt{L_s\dfrac{C_sC_p}{C_s+C_p}}} = f_s\sqrt{1+\frac{C_s}{C_p}} \approx f_s\left(1 + \frac{C_s}{2C_p}\right) = 1000.50\ \text{kHz} \]

The separation is \(f_p - f_s = 768\) Hz on 999.73 kHz, that is 768 parts per million or 0.077 per cent, and it is fixed entirely by the ratio \(C_p/C_s = 651\). Between those two frequencies, and only between them, the crystal looks inductive; everywhere else it is a small capacitance of about 5 pF.

The quality factor follows from the motional elements:

\[ Q = \frac{\omega_sL_s}{R_s} = \frac{2\pi(999\,730)(3.3)}{200} = 103\,645 \]
Why the stability is so good
A \(Q\) of \(10^{5}\) against an LC tank’s 150

Frequency stability in any oscillator comes from the steepness with which the resonator’s phase passes through zero, and that steepness is proportional to \(Q\). If some other part of the loop — the transistor, a stray capacitance, a temperature change — introduces an unwanted phase shift \(\Delta\phi\), the oscillation frequency must move until the resonator cancels it, and the movement is \(\Delta f/f = \Delta\phi/2Q\). A good LC tank has \(Q \approx 150\); the crystal above has 103 645, nearly 700 times better, so the same disturbance causes nearly 700 times less frequency error. That single ratio is the whole reason a crystal oscillator holds a few parts per million where an LC oscillator holds a few parts per thousand.

Equivalent circuit of a quartz crystalthe crystalC_p 5 pF (holder / electrode capacitance)L_s 3.3 HC_s 0.00768 pFR_s 200 ΩQ = ω_s L_s/R_s = 103 645 — three orders above any LC tankC_p/C_s = 651, and that ratio alone fixes how far f_p sits above f_sCrystal reactance+X−Xff_sf_pinductive only herecapacitivecapacitivef_s = 999.73 kHzf_p = 1000.50 kHzgap 768 Hz = 768 ppm(the frequency axis betweenf_s and f_p is expanded)
Figure 29.4 — The quartz crystal: equivalent circuit and reactance against frequency

Two ways of using the crystal follow from the reactance curve. In a series-mode oscillator the crystal is placed in the feedback path as a series element, where it is a low resistance at \(f_s\) and a high impedance everywhere else; the loop closes only at \(f_s\). In a parallel-mode oscillator — the Pierce circuit, and the one inside every microcontroller — the crystal replaces the inductor of a Colpitts, working in the narrow inductive band just below \(f_p\). The exact frequency there depends on the load capacitance the circuit presents, which is why parallel-mode crystals are specified with a load capacitance (typically 18 or 20 pF) and why using one with the wrong load capacitors puts the clock a few tens of parts per million out.

Two consequences follow. A crystal cannot be pulled by more than a few hundred parts per million, so its frequency is chosen when it is ordered; and it has so strong a preference for that frequency that crystal oscillators are reliable to build in a way LC oscillators are not. Temperature shifts it by an amount fixed by the cut — an AT-cut holds a few parts per million over the commercial range, and the same crystal in a small oven (an OCXO) reaches parts in \(10^{9}\).

7 Relaxation Oscillators, Drift and Phase Noise

Everything so far has produced a sinusoid by arranging for the loop to be marginally unstable at one frequency. The other family of oscillators abandons the sinusoid entirely. A relaxation oscillator has no resonator and no linear region: it charges a capacitor through a resistor towards a threshold, flips a switch when the threshold is reached, and charges the other way. The output is a square wave and the capacitor voltage an exponential sawtooth, and the frequency depends on an \(RC\) time constant and on two comparator thresholds rather than on any resonance.

The op-amp astable. Take a comparator with positive feedback — the Schmitt trigger of Chapter 27 — whose thresholds are \(\pm\beta V_{sat}\) with \(\beta = R_1/(R_1+R_2)\), and connect an \(RC\) network from the output to the inverting input. The capacitor charges towards \(+V_{sat}\) from \(-\beta V_{sat}\); when it reaches \(+\beta V_{sat}\) the output flips and it charges the other way. Solving the exponential for each half period and doubling gives

\[ T = 2RC\ln\!\left(\frac{1+\beta}{1-\beta}\right) \]

With \(R_1 = R_2\), \(\beta = 0.5\) and \(T = 2RC\ln3 = 2.1972RC\), a convenient near-round figure. For \(f = 1\) kHz with \(C = 10\) nF the requirement is \(R = 1/(2000\ln3\times10^{-8}) = 45.5\ \text{k}\Omega\).

The 555 astable. The 555 timer packages two comparators with thresholds at \(\tfrac13V_{CC}\) and \(\tfrac23V_{CC}\), a flip-flop and a discharge transistor. In the astable connection the capacitor charges through \(R_A + R_B\) and discharges through \(R_B\) alone, so the two halves of the cycle are unequal:

\[ t_{H} = 0.693(R_A+R_B)C,\qquad t_{L} = 0.693R_BC,\qquad T = 0.693(R_A+2R_B)C,\qquad D = \frac{R_A+R_B}{R_A+2R_B} \]

The 0.693 is \(\ln 2\), and it appears because the capacitor swings between one third and two thirds of the supply: the fraction of the remaining gap covered is the same either way and independent of \(V_{CC}\), which is why a 555’s frequency does not change when the supply does.

4 Worked Example 29.4 — A 1 kHz 555 astable

Take \(R_A = 10\ \text{k}\Omega\), \(R_B = 68\ \text{k}\Omega\) and \(C = 10\) nF. Then

\[ t_H = 0.6931(78\,000)(10^{-8}) = 0.5407\ \text{ms}, \qquad t_L = 0.6931(68\,000)(10^{-8}) = 0.4713\ \text{ms} \]

so \(T = 1.0120\) ms, \(f = 988.1\) Hz and the duty cycle is \(0.5407/1.0120 = 53.42\) per cent. The often-quoted shortcut \(f = 1.44/[(R_A+2R_B)C]\) gives 986.3 Hz, differing only because 1.44 is a rounding of \(1/\ln2 = 1.4427\).

The duty cycle cannot be brought below 50 per cent by choice of resistors alone, because \(t_H\) always contains \(R_A\) and \(t_L\) never does; the standard cure is a diode across \(R_B\) so that charging goes through \(R_A\) only, after which \(D = R_A/(R_A+R_B)\) and any value is reachable. Making \(R_B\) large compared with \(R_A\) approaches 50 per cent asymptotically: the values above already give 53.4 per cent with \(R_B = 6.8R_A\).

Frequency stability, drift and phase noise. Three separate things are meant when an oscillator is called stable, and it is worth keeping them apart.

  • Long-term drift is a slow, one-way change over months and years, caused by ageing of the resonator — stress relief in a quartz blank, loss of moisture from a capacitor dielectric, contamination migrating onto crystal electrodes. It is quoted in parts per million per year: a few ppm for an ordinary crystal, well under one for a good one, and nothing meaningful at all for an \(RC\) oscillator whose components drift much faster than that.
  • Short-term drift is the change over minutes to hours, dominated by temperature and by supply voltage acting through the transistor capacitances. An LC oscillator might hold \(10^{-3}\) to \(10^{-4}\); a plain crystal oscillator \(10^{-5}\); a temperature-compensated one (TCXO) \(10^{-6}\) to \(10^{-7}\); an oven-controlled one (OCXO) \(10^{-9}\). The Clapp’s extra capacitor is an attempt to reduce exactly this coupling from device into resonator.
  • Phase noise is the cycle-to-cycle randomness: the zero crossings do not occur at exactly equal intervals, because the loop is amplifying its own noise as well as its own signal. In the frequency domain it appears as skirts either side of the carrier, quoted in dBc/Hz at a stated offset — typically \(-120\) dBc/Hz at 10 kHz offset for a decent crystal oscillator. Because the resonator converts a phase perturbation into a frequency perturbation with the factor \(1/2Q\), phase noise falls as \(Q\) rises: the second great argument for the crystal. It also falls as the signal level in the resonator rises, so a hard-limited oscillator tends to be noisier than a linear one held by a slow AGC loop.

8 Summary and Key Results

Chapter 29 — oscillation conditions and the numbers from the worked circuits
OscillatorFrequencyFeedback fraction \(\beta\)Gain requiredWorked value
Barkhausen criterionset by \(\angle A\beta = 0^\circ\)\(|A\beta| = 1\)design for \(\approx 1.05\), stabiliser pulls it to 1
Wien bridge\(f_0 = 1/2\pi RC\)\(1/3\) at \(f_0\), \(\angle 0^\circ\)\(A = 3\), \(R_f/R_i = 2\)10 kΩ, 10 nF → 1591.5 Hz
RC phase-shift\(f_0 = 1/(2\pi RC\sqrt6)\)\(-1/29 = -0.03448\)\(A \ge 29\); BJT \(h_{fe} \ge 44.5\)10 kΩ, 10 nF → 649.7 Hz
General LC form\(X_1+X_2+X_3 = 0\)\(X_1/X_2\)\(|A| \ge X_2/X_1\)\(X_1,X_2\) alike; \(X_3\) opposite
Hartley\(1/2\pi\sqrt{L_TC}\), \(L_T=L_1+L_2+2M\)\(L_2/L_1\)\(\ge L_1/L_2\)110 µH, 1 nF → 479.9 kHz, \(\beta=0.1\)
Colpitts\(1/2\pi\sqrt{LC_T}\), \(C_T = C_1C_2/(C_1{+}C_2)\)\(C_1/C_2\)\(\ge C_2/C_1\)100 µH, 909.1 pF → 527.9 kHz
Clapp\(1/C_T = 1/C_1+1/C_2+1/C_3\)\(C_1/C_2\)\(\ge C_2/C_1\)\(C_3=100\) pF → 1.6768 MHz; 5 pF stray shifts 224 ppm, not 2264
Crystal, series resonance\(f_s = 1/2\pi\sqrt{L_sC_s}\)impedance \(= R_s\) there3.3 H, 0.00768 pF → 999.73 kHz
Crystal, parallel resonance\(f_p = f_s\sqrt{1+C_s/C_p}\)1000.50 kHz; gap 768 ppm
Crystal \(Q\)\(Q = \omega_sL_s/R_s\)103 645, against \(\approx 150\) for an LC tank
Op-amp astable\(T = 2RC\ln[(1+\beta)/(1-\beta)]\)\(R_1/(R_1+R_2)\)comparator\(\beta = 0.5\): \(T = 2.1972RC\)
555 astable\(T = 0.693(R_A+2R_B)C\)thresholds \(\tfrac13,\tfrac23 V_{CC}\)10 kΩ, 68 kΩ, 10 nF → 988.1 Hz, 53.42 % duty

9 Common Mistakes

! Designing for \(|A\beta| = 1\) exactly, and building an oscillator that never starts

The Barkhausen criterion states the condition for a sustained oscillation, not the condition a circuit should be built to. A Wien bridge with \(R_f = 20\) kΩ and \(R_i = 10\) kΩ has \(A = 3.000\) and \(|A\beta| = 1.000\) on paper; on the bench the resistors are 1 per cent parts, so the loop gain is somewhere between 0.98 and 1.02, and half the circuits built will sit silent. Design for a 3 to 5 per cent excess — \(R_f = 21\) kΩ gives \(|A\beta| = 1.033\) — and let the lamp, thermistor or diode pair remove the excess once the amplitude has built up. The excess also fixes the start-up time: too little and the circuit takes seconds to reach amplitude, too much and it overshoots and clips before the stabiliser catches it.

! Confusing the crystal&rsquo;s series and parallel resonances, or ignoring the load capacitance

\(f_s\) and \(f_p\) differ by only 768 ppm in the worked example, which sounds negligible until one remembers that the whole point of using a crystal is to hold a few ppm. A crystal specified for parallel (load-capacitance) operation and used in a series-mode circuit runs 768 ppm low — 768 Hz on 1 MHz, or 27 seconds a month on a clock. Worse, a parallel crystal specified for an 18 pF load and given 33 pF of load by over-large oscillator capacitors runs measurably slow, because the operating point slides down the steep inductive region towards \(f_s\). Always read which mode the crystal is specified for, and in a Pierce circuit make the two load capacitors give the specified \(C_L\) after the few picofarads of board and pin capacitance are counted.

! Expecting a phase-shift network to fix the frequency as precisely as a tuned circuit

The frequency an oscillator settles at is where the total loop phase is zero, so any stray phase shift moves it, and how far it moves depends on how steeply the network’s phase varies with frequency. In the three-section RC ladder the phase moves only about 12° for a 20 per cent change of frequency, so a 5° phase error from the amplifier’s own roll-off shifts the frequency by roughly 8 per cent. A tuned circuit of \(Q = 150\) shifts by \(\Delta\phi/2Q\), which for the same 5° is 290 ppm, and a crystal of \(Q = 103\,645\) shifts by 0.42 ppm. If a specification quotes frequency accuracy in per cent, an RC oscillator may do; in parts per million, only a crystal will.

10 Chapter Review

  1. 1. A Wien-bridge oscillator is to run at 2 kHz using 15 nF capacitors. Find \(R\), choose \(R_f\) and \(R_i\) for reliable starting, and state what the lamp must do once the circuit is running.

    The frequency condition is \(f_0 = 1/2\pi RC\), so \(R = 1/[2\pi(2000)(15\times10^{-9})] = 5305\ \Omega\); the nearest preferred value is 5.1 kΩ, giving \(f_0 = 1/[2\pi(5100)(1.5\times10^{-8})] = 2080\) Hz, 4.0 per cent high, or 5.6 kΩ giving 1895 Hz, 5.3 per cent low. If 2.00 kHz matters, use a dual gang trimmer; if not, 5.1 kΩ is the closer. Both arms must use the same value. The amplitude condition needs \(A = 3\) at equilibrium, so \(R_f/R_i = 2\); building it at exactly 2 will not start, so take \(R_i = 10\ \text{k}\Omega\) (the lamp’s hot resistance) and \(R_f = 22\ \text{k}\Omega\), giving a cold-start gain of \(A = 3.20\) and \(|A\beta| = 3.20/3 = 1.067\). At switch-on the lamp is cold and its resistance is well below 10 kΩ, so the gain is higher still and the oscillation builds quickly. As the amplitude rises, the filament heats and \(R_i\) rises until \(1 + R_f/R_i = 3.000\), which requires \(R_i = R_f/2 = 11\ \text{k}\Omega\) hot. The equilibrium amplitude is therefore whatever value drives the lamp to 11 kΩ, and the loop settles there: any increase heats the lamp further, drops the gain below 3, and the amplitude falls back.

  2. 2. Derive the frequency and attenuation of the three-section RC phase-shift network, and explain why two sections will not do.

    For the three-section ladder with \(\alpha = 1/\omega RC\), analysis gives \(\beta = 1/[(1-5\alpha^2) + j(\alpha^3-6\alpha)]\). The amplifier is inverting and supplies \(180^\circ\), so the network must supply the other \(180^\circ\), which requires \(\beta\) to be real and negative — hence \(\alpha^3-6\alpha = 0\), so \(\alpha = \sqrt6\), \(\omega RC = 1/\sqrt6\) and \(f_0 = 1/(2\pi RC\sqrt6)\). Substituting \(\alpha^2 = 6\) into the real part gives \(\beta = 1/(1-30) = -1/29\), so the amplifier needs a gain of at least 29. Two sections cannot work for a reason of principle rather than of arithmetic: a single \(RC\) section has phase \(\phi = \tan^{-1}(1/\omega RC)\), which tends to \(90^\circ\) only as \(\omega \to 0\), and at that limit the attenuation tends to zero as well. Two sections can therefore approach \(180^\circ\) only at a frequency where \(|\beta| \to 0\), so no finite amplifier gain can close the loop. Three sections reach \(180^\circ\) at a finite frequency with a finite attenuation of \(1/29\); four sections reach it with an attenuation of only \(1/18.4\) and a steeper phase slope, which is why four-section versions are sometimes used when a lower gain or a better-defined frequency is wanted.

  3. 3. A Colpitts oscillator uses \(L = 220\ \mu\)H, \(C_1 = 470\) pF at the collector and \(C_2 = 4.7\) nF at the base. Find the frequency, the feedback fraction and the minimum gain. Then convert it to a Clapp with \(C_3 = 68\) pF and recompute.

    The two capacitors are in series across the inductor, so \(C_T = C_1C_2/(C_1+C_2) = (470)(4700)/5170 = 427.3\) pF and \(f = 1/[2\pi\sqrt{(220\times10^{-6})(427.3\times10^{-12})}] = 519.1\) kHz. The feedback fraction is the reactance ratio across the tapped pair, \(\beta = X_{C1}/X_{C2} = C_1/C_2 = 470/4700 = 0.100\), so the stage needs a voltage gain of at least \(1/\beta = 10\). Adding \(C_3 = 68\) pF in series with the inductor gives \(1/C_T = 1/470 + 1/4700 + 1/68\ \text{pF}^{-1} = 0.002128+0.000213+0.014706 = 0.017047\), so \(C_T = 58.66\) pF and \(f = 1/[2\pi\sqrt{(220\times10^{-6})(58.66\times10^{-12})}] = 1.4010\) MHz. The feedback fraction is unchanged at 0.100, because \(C_3\) sits in the inductive branch and takes no part in the divider — which is exactly the point of the Clapp. \(C_T\) is now 86.3 per cent of \(C_3\), so the frequency is set overwhelmingly by one small, stable capacitor rather than by two large ones that the transistor is shunting.

  4. 4. A crystal has \(L_s = 0.52\) H, \(C_s = 0.0122\) pF, \(R_s = 82\ \Omega\) and \(C_p = 4.0\) pF. Compute \(f_s\), \(f_p\) and \(Q\), and comment on the result.

    The series resonance is \(f_s = 1/[2\pi\sqrt{(0.52)(1.22\times10^{-14})}] = 1/[2\pi\sqrt{6.344\times10^{-15}}] = 1/[2\pi(7.965\times10^{-8})] = 1.9982\) MHz — a nominal 2 MHz crystal. For the parallel resonance, \(C_s\) and \(C_p\) appear in series: \(C_sC_p/(C_s+C_p) = (0.0122)(4.0)/4.0122 = 0.012163\) pF, so \(f_p = 1/[2\pi\sqrt{(0.52)(1.2163\times10^{-14})}] = 2.0012\) MHz. The separation is 3.045 kHz, or 1524 ppm, and it is set entirely by the ratio \(C_p/C_s = 4.0/0.0122 = 327.9\); the shortcut \(f_p \approx f_s(1+C_s/2C_p)\) gives \(1.9982(1.001525) = 2.0012\) MHz, agreeing to five figures. The quality factor is \(Q = \omega_sL_s/R_s = 2\pi(1.9982\times10^{6})(0.52)/82 = 79\,617\). The comment worth making is on the size of these numbers: 0.52 H with only 82 Ω of series loss is an inductor no coil could imitate, and it is what a \(Q\) of eighty thousand means physically. The pulling range, the whole band over which any circuit can persuade this crystal to move, is the 1524 ppm between \(f_s\) and \(f_p\) — 0.15 per cent, and in practice rather less than half of it is usable.

  5. 5. Design a 555 astable for 5 kHz with a duty cycle as close to 50 per cent as the plain circuit allows, using a 10 nF capacitor. Then say what changes if a diode is placed across \(R_B\).

    The plain circuit has \(D = (R_A+R_B)/(R_A+2R_B)\), which approaches 50 per cent only as \(R_B/R_A \to \infty\), so make \(R_A\) as small as the 555 allows — about 1 kΩ, below which the discharge transistor is asked for too much current. Take \(R_A = 1\ \text{k}\Omega\). Then \(T = 1/5000 = 200\ \mu\)s and \(T = 0.6931(R_A+2R_B)C\) gives \(R_A + 2R_B = 200\times10^{-6}/(0.6931\times10^{-8}) = 28\,854\ \Omega\), so \(R_B = (28\,854-1000)/2 = 13\,927\ \Omega\); take 14 kΩ (E96) or 15 kΩ (E24). With \(R_A = 1\) kΩ and \(R_B = 14\) kΩ, \(t_H = 0.6931(15\,000)(10^{-8}) = 103.97\ \mu\)s, \(t_L = 0.6931(14\,000)(10^{-8}) = 97.04\ \mu\)s, so \(T = 201.0\ \mu\)s, \(f = 4975\) Hz and \(D = 51.72\) per cent. That is as close to square as the plain circuit gets at this frequency. Placing a diode across \(R_B\), anode at the discharge pin, lets the charging current bypass \(R_B\) so that \(t_H = 0.693R_AC\) and \(t_L = 0.693R_BC\), giving \(D = R_A/(R_A+R_B)\). Now equal resistors give exactly 50 per cent: \(R_A = R_B = R\) with \(T = 2(0.6931)R(10^{-8}) = 200\ \mu\)s requires \(R = 14.43\ \text{k}\Omega\), so two 14.3 kΩ resistors give \(f = 5044\) Hz at 50.0 per cent. The diode’s forward drop makes the charging threshold slightly different from the discharging one, so in practice the duty cycle comes out one or two per cent off square and a trimmer in the \(R_A\) leg is usual.