RC, LC and Crystal Oscillators

Sinusoidal oscillators analysed through the Barkhausen criterion: the RC phase-shift and Wien-bridge audio oscillators with their 29 and 3 gain conditions, the Colpitts and Hartley LC oscillators with their tapped-divider feedback fractions, and the quartz crystal, whose series and parallel resonances and enormous Q are read off its equivalent RLC-plus-mounting-capacitance model.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

An oscillator is a feedback amplifier with no input. The loop consists of a non-inverting op-amp stage of gain \(A = 1 + R_f/R_1\) driving a frequency-selective network of transfer ratio \(\beta\), whose output returns to the amplifier input. Here the network is a Wien bridge, for which \(\beta = 1/3\) with zero phase shift at one frequency \(f_0\) and less than \(1/3\) with non-zero phase everywhere else. The amplifier uses \(R_f = 22~\text{k}\Omega\) and \(R_1 = 10~\text{k}\Omega\).

1. State the Barkhausen criterion in its two parts and explain what each part fixes.

2. Evaluate the loop gain \(A\beta\) at \(f_0\) for the resistors given, and say what the output waveform actually does.

3. Find the value of \(R_f\) for marginal oscillation, and explain why a practical design deliberately misses it.

• \(R_f = 22~\text{k}\Omega\), \(R_1 = 10~\text{k}\Omega\)

• Wien network: \(\beta = 1/3\) and \(\angle\beta = 0^\circ\) at \(f = f_0\)

• Ideal op-amp; positive feedback through the network, negative through \(R_f\)-\(R_1\).

Part (a) — the criterion

Sustained oscillation requires the signal that goes once round the loop to come back identical to itself. Writing the loop gain as \(T(j\omega) = A\beta\):

• Magnitude: \(|A\beta| = 1\). This fixes the amplitude — below unity any disturbance decays, above unity it grows.

• Phase: \(\angle A\beta = 0^\circ\) (or a multiple of \(360^\circ\)). This fixes the frequency — oscillation occurs at the one frequency where the loop phase comes out to zero.

Note the division of labour: the selective network chooses \(f_0\), the amplifier gain chooses whether the oscillation survives.

\(|A\beta| = 1\) sets the amplitude; \(\angle A\beta = 0^\circ\) sets the frequency.

Part (b) — the loop gain as built

The phase condition is met (the Wien network contributes \(0^\circ\) at \(f_0\) and the non-inverting amplifier another \(0^\circ\)), so the circuit oscillates at \(f_0\). But \(|A\beta| > 1\), so the amplitude does not settle: each pass round the loop multiplies the signal by 1.067, i.e. it grows by 6.7% per cycle until the op-amp output runs into the supply rails.

\(|A\beta| = 1.067 > 1\): the circuit oscillates at \(f_0\), but the output grows until it clips into a near-square wave — heavy distortion.

Part (c) — marginal oscillation, and why not to sit there

Exactly \(R_f = 20~\text{k}\Omega\) is a knife edge. Resistor tolerance, temperature drift or the op-amp's own finite gain will push \(|A\beta|\) a fraction of a percent either way; on the low side the oscillation dies out, on the high side it clips. So a real Wien bridge is built with \(|A\beta|\) slightly greater than 1 — for example \(R_f = 21.0~\text{k}\Omega\) giving \(A = 3.1\) and \(|A\beta| = 1.0333\) — and an amplitude-limiting element (a lamp, a thermistor or back-to-back diodes across part of \(R_f\)) then pulls the gain back to exactly 3 once the amplitude has built up.

\(R_f = 20.0~\text{k}\Omega\) gives \(|A\beta| = 1\) exactly; practical designs use a slightly larger \(R_f\) plus automatic gain control.

Barkhausen is a necessary condition on a linear model, not a complete description: every real oscillator is non-linear, and it is that non-linearity which sets the final amplitude.

An RC phase-shift oscillator uses a common-emitter BJT stage whose output feeds a ladder of three identical RC sections, each \(R = 10~\text{k}\Omega\) and \(C = 0.01~\mu\text{F}\), with the last section returning to the base. The collector resistor is \(R_C = 2.2~\text{k}\Omega\).

1. Explain why three sections are used and what phase each must supply.

2. Find the frequency of oscillation, using the loaded form \(f = 1/\left(2\pi RC\sqrt{6 + 4R_C/R}\right)\), and compare it with the idealised \(f = 1/\left(2\pi RC\sqrt{6}\right)\).

3. Find the minimum \(h_{fe}\) the transistor must have, and find the \(R_C\) that minimises that requirement.

• \(R = 10~\text{k}\Omega\), \(C = 0.01~\mu\text{F}\) (three identical sections)

• \(R_C = 2.2~\text{k}\Omega\); common-emitter stage, so \(180^\circ\) from the amplifier

• The loaded formulas are used, i.e. the ladder's loading on the collector is retained but the base input resistance is assumed large.

Part (a) — why three sections

A common-emitter amplifier inverts, contributing \(180^\circ\). Barkhausen needs a total loop phase of \(360^\circ\), so the RC network must add another \(180^\circ\). A single RC section can approach \(90^\circ\) but never reach it, and at \(90^\circ\) its attenuation would be infinite. Three sections each supplying \(60^\circ\) is the usual compromise: two sections would need \(90^\circ\) each, which is unattainable.

Three sections, each contributing \(60^\circ\), to make up the \(180^\circ\) the inverting amplifier does not supply.

Part (b) — frequency of oscillation

The idealised form drops the \(4R_C/R\) loading term:

\(f = 606.8~\text{Hz}\) loaded, against 649.7 Hz idealised — the simple formula reads 7.1% high.

Part (c) — minimum \(h_{fe}\)

Requiring \(|A\beta| \ge 1\) with the loaded ladder gives the standard result

Minimise over \(R_C\): differentiating with respect to \(x = R_C/R\) gives \(-29/x^2 + 4 = 0\), so \(x = \sqrt{29/4} = 2.693\), i.e. \(R_C = 26.9~\text{k}\Omega\) and \(h_{fe(\min)} = 23 + 4\sqrt{29} = 44.5\). Even the convenient round choice \(R_C = 10~\text{k}\Omega\) (\(x = 1\)) gets most of the way there:

As drawn, \(h_{fe} \ge 156\) is needed; raising \(R_C\) to 10 k\(\Omega\) cuts this to 56, at the cost of moving \(f\) to 503.3 Hz.

The \(29/x\) term dominates: with a small collector resistor the ladder is heavily loaded and the transistor must make up the loss. If the amplifier were an op-amp instead — infinite input resistance, zero output resistance — the loading terms vanish, the network attenuation is exactly \(1/29\) and the required gain is exactly 29 at \(f = 1/(2\pi RC\sqrt{6})\).

A Wien-bridge oscillator uses a series \(RC\) arm (\(R\) in series with \(C\)) from the op-amp output to the non-inverting input, and a parallel \(RC\) arm (the same \(R\) in parallel with the same \(C\)) from that input to ground. Negative feedback is supplied by \(R_f\) from output to inverting input and \(R_1 = 10~\text{k}\Omega\) from inverting input to ground.

1. For \(R = 10~\text{k}\Omega\) and \(C = 0.01~\mu\text{F}\), find the frequency of oscillation and the required \(R_f\).

2. Show that the network ratio is \(1/3\) at \(f_0\), and evaluate it at \(f_0/2\) to demonstrate the frequency selectivity.

3. Design: retune the oscillator to 5 kHz using a standard 4.7 nF capacitor and an E24 resistor. Quote the frequency error.

• Wien network, equal components: \(R = 10~\text{k}\Omega\), \(C = 0.01~\mu\text{F}\)

• \(R_1 = 10~\text{k}\Omega\); ideal op-amp

• Amplitude stabilisation ignored; the linear gain condition is what is computed.

Part (a) — frequency and gain condition

With equal \(R\) and equal \(C\) the bridge balances where the series and parallel arms have equal and opposite reactive parts, i.e. at \(\omega_0 = 1/RC\):

\(f_0 = 1592~\text{Hz}\), \(A_v = 3\), \(R_f = 20.0~\text{k}\Omega\)

Part (b) — the network ratio and its selectivity

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

At \(f = f_0\) the bracket vanishes, leaving \(\beta = 1/3\) with zero phase — which is exactly what the amplifier's \(0^\circ\) needs. At \(f = f_0/2\), \(f/f_0 - f_0/f = 0.5 - 2 = -1.5\):

\(\beta(f_0) = 1/3 = 0.3333\) at \(0^\circ\); at \(f_0/2\), \(|\beta| = 0.2981\) at \(+26.6^\circ\) — both conditions fail, so no oscillation there.

Part (c) — design for 5 kHz

\(R = 6.8~\text{k}\Omega\), \(C = 4.7~\text{nF}\) gives \(f_0 = 4980~\text{Hz}\), 0.40% low; \(R_f\) stays at 20.0 k\(\Omega\).

The gain condition is independent of \(R\) and \(C\), which is what makes the Wien bridge the standard variable-frequency audio oscillator: gang the two resistors (or the two capacitors) and sweep the frequency without ever re-adjusting the amplifier. Both \(R\)s and both \(C\)s must track, though — unequal components move \(\beta(f_0)\) away from \(1/3\) and the gain condition is no longer 3.

A Colpitts oscillator uses a common-emitter transistor with the emitter at AC ground. The tank is an inductor \(L = 100~\mu\text{H}\) across the series combination of \(C_1 = 0.001~\mu\text{F}\), connected from the collector to ground, and \(C_2 = 0.01~\mu\text{F}\), connected from the base to ground. The output is developed across \(C_1\) and the feedback voltage is taken across \(C_2\).

1. Find the effective tank capacitance and the frequency of oscillation.

2. Find the feedback fraction and the minimum amplifier gain.

3. Retune the circuit to 1 MHz by changing \(L\) alone, and then estimate the frequency shift caused by 20 pF of stray capacitance appearing across the tank.

• \(L = 100~\mu\text{H}\), \(C_1 = 0.001~\mu\text{F} = 1000~\text{pF}\), \(C_2 = 0.01~\mu\text{F} = 10\,000~\text{pF}\)

• Common-emitter stage, \(180^\circ\); the tapped capacitive divider supplies the other \(180^\circ\).

• Coil losses neglected; tank \(Q\) assumed high enough that \(f\) is the lossless resonance.

Part (a) — tank capacitance and frequency

\(C_1\) and \(C_2\) are in series as far as the inductor is concerned (the tap is the AC ground, so the loop is \(L \to C_1 \to C_2 \to L\)):

\(C = 909.1~\text{pF}\), \(f = 527.9~\text{kHz}\)

Part (b) — feedback fraction and gain condition

The same tank current flows through both capacitors, so the voltage across each is inversely proportional to its capacitance:

\(\beta = 0.10\), so the stage needs \(|A_v| \ge 10.0\).

Part (c) — retuning, and the effect of strays

Now put the original 100 \(\mu\)H circuit back and add 20 pF of stray capacitance in parallel with the tank:

\(L = 27.9~\mu\text{H}\) for 1 MHz; 20 pF of stray shifts the original 527.9 kHz design by -1.08%, to 522.1 kHz.

Two design consequences. Making \(C_1\) small raises the required gain (\(C_2/C_1\)) but a very small \(C_1\) is also the one most polluted by device and wiring capacitance — as the 1.08% shift shows. The Colpitts is preferred over the Hartley at RF precisely because strays can be absorbed into \(C_1\) and \(C_2\), which is why it is the usual choice above a few megahertz.

A Hartley oscillator has a tapped coil across a 500 pF tuning capacitor. The section from the collector to the tap is \(L_1 = 1~\text{mH}\) and the section from the tap to the base is \(L_2 = 100~\mu\text{H}\); the tap is at AC ground. The two windings are on the same former with a mutual inductance \(M = 20~\mu\text{H}\), connected series-aiding.

1. Find the equivalent tank inductance and the frequency of oscillation.

2. Find the frequency the circuit would have if the windings were uncoupled, and the percentage error made by ignoring \(M\).

3. State the gain condition. Then choose the tuning capacitor that would place this oscillator at the 455 kHz intermediate frequency, and give the nearest E24 value and its error.

• \(L_1 = 1~\text{mH}\), \(L_2 = 100~\mu\text{H}\), \(M = 20~\mu\text{H}\) (series-aiding)

• \(C = 500~\text{pF}\); output across \(L_1\), feedback across \(L_2\)

• Lossless tank; the tap is a perfect AC ground.

Part (a) — equivalent inductance and frequency

The same current circulates through both windings, so they add in series, and series-aiding coupling adds \(2M\):

\(L_{eq} = 1140~\mu\text{H}\), \(f = 210.8~\text{kHz}\)

Part (b) — the cost of ignoring \(M\)

Ignoring \(M\) predicts 214.6 kHz against the true 210.8 kHz — an error of 1.77%.

Part (c) — gain condition and the 455 kHz design

The tank current is common to both windings, so the voltage across each is proportional to its inductance — the exact dual of the Colpitts:

For 455 kHz with the same coil:

\(|A_v| \ge 10.0\); \(C = 107~\text{pF}\) ideally, and the E24 100 pF part gives 471.4 kHz, 3.6% high.

Both parts (b) and (c) make the same point: the Hartley's frequency depends on an inductance that is not simply \(L_1 + L_2\), and a 40 \(\mu\)H coupling term — 3.51% of \(L_{eq}\) here — moves the frequency by 1.77%. A tapped coil's \(M\) is hard to control, which is why LC oscillators above a few MHz are usually Colpitts and why anything needing real accuracy uses a crystal.

A quartz crystal is modelled by a series branch of \(L = 0.52~\text{H}\), \(C_s = 0.012~\text{pF}\) and \(R = 120~\Omega\) (the motional arm, representing the mechanical resonance), all in parallel with the mounting capacitance \(C_m = 4~\text{pF}\) of the holder and electrodes.

1. Find the series resonant frequency \(f_s\).

2. Find the parallel (anti-resonant) frequency \(f_p\), and the separation between the two, in hertz and as a percentage.

3. Find the \(Q\) of the crystal and compare it with a good LC tank of \(Q = 100\). What does this mean for frequency stability?

• Motional arm: \(L = 0.52~\text{H}\), \(C_s = 0.012~\text{pF}\), \(R = 120~\Omega\)

• Mounting capacitance \(C_m = 4~\text{pF}\) in parallel with the whole arm

• \(R\) is small enough that it shifts the resonances negligibly; both are computed from \(L\) and the relevant capacitance alone.

Part (a) — series resonance

At \(f_s\) the motional \(L\) and \(C_s\) cancel and the crystal looks like the pure resistance \(R\) — its impedance minimum.

\(f_s = 2.0148~\text{MHz}\)

Part (b) — parallel resonance

Just above \(f_s\) the motional arm is net inductive, and it resonates with \(C_m\). The capacitance seen by \(L\) is now \(C_s\) in series with \(C_m\):

The same result follows more quickly from the ratio form, which shows that the spacing is governed entirely by \(C_s/C_m\):

\(f_p = 2.01780~\text{MHz}\), only 3.02 kHz (0.150%) above \(f_s\).

Part (c) — the \(Q\), and why it matters

\(Q = 5.49\times10^{4}\) — about 549 times that of a good LC tank.

Frequency stability follows directly from \(Q\). The loop oscillates where the total phase is zero, and near resonance the reactive phase of the resonator changes at a rate proportional to \(Q\). A drift in the transistor's phase shift therefore pulls the frequency by an amount inversely proportional to \(Q\): with \(Q = 5.49\times10^{4}\) the crystal holds the frequency about 549 times more tightly than the LC tank would. Note also how narrow the usable window is — the whole region between \(f_s\) and \(f_p\), where the crystal is inductive and a Pierce or Colpitts oscillator can work, is only 3.02 kHz wide, so a small load capacitance change trims the frequency but cannot move it far.