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Oscillators in depth: designing a Colpitts, the crystal and Leeson's phase noise

Step-by-step design of a Colpitts oscillator, the Clapp variant, the electrical model of the crystal with the series and parallel resonances proved, the three regions of phase noise in Leeson's model.

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This page goes deeper into the lesson on oscillators: it designs a Colpitts step by step, from the capacitance to the bias current, presents the Clapp variant, derives the two resonances of the crystal from its electrical model and explains where the three regions of phase noise in Leeson’s model come from.

Designing a Colpitts step by step

The formulas of the proof are enough for a complete design. Goal: a 10 MHz oscillator with a 2.2 µH inductor of quality factor Q_L = 80, driving a 2 kΩ load.

Step 1: the series capacitance. From \omega_0^2 = 1/(LC_s):

    \[ C_s = \frac{1}{\omega_0^2L} = \frac{1}{(2\pi\cdot 10^7)^2\cdot 2.2\cdot 10^{-6}} \approx 115\ \mathrm{pF} \]

With C_1 = C_2 we have C_s = C_1/2, so C_1 = C_2 = 230 pF (in practice 220 pF plus the capacitance of the transistor and of the circuit board).

Step 2: the loss resistance. An inductor with quality factor Q_L is equivalent, near resonance, to an ideal inductor with the resistance R_p = Q_L\,\omega_0L = 80\cdot 2\pi\cdot 10^7\cdot 2.2\cdot 10^{-6} \approx 11 kΩ in parallel. In parallel with the 2 kΩ load it gives R \approx 1.7 kΩ: this is the R of the formula.

Step 3: the required gain. The start-up condition requires

    \[ g_m \ge \frac{C_1}{C_2R} = \frac{1}{1700} \approx 0.59\ \mathrm{mS} \]

With a margin of 4, g_m \approx 2.4 mS. For a bipolar transistor g_m = I_C/V_T, so I_C = g_mV_T \approx 2.4\cdot 10^{-3}\cdot 0.025 = 60 µA.

Step 4: the corrections. The proof assumed that the amplifier input draws no current. A bipolar transistor instead has an input resistance r_\pi = \beta/g_m (about 40 kΩ with \beta = 100), which adds to the losses: in practice a somewhat higher current is chosen, 0.2 to 1 mA, and start-up is checked in simulation or on the bench. Too much current, however, drives the transistor into heavy compression and increases the harmonics.

Estimating the amplitude. With a large margin the transistor works almost as a switch, and the fundamental component of the collector current approaches 2I_C. The output amplitude is therefore about V_2 \approx 2I_CR: with I_C = 0.5 mA and R = 1.7 kΩ, about 1.7 V peak. This is the mechanism by which the oscillator regulates its own amplitude.

The Clapp variant

In the Colpitts the frequency depends on C_1 and C_2, to which the transistor capacitances add, and these change with temperature and bias. The Clapp adds a small capacitor C_3 in series with the inductor:

    \[ \omega_0 = \frac{1}{\sqrt{LC_{eq}}},\qquad \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

If C_3 \ll C_1, C_2, the equivalent capacitance is almost C_3 and the frequency hardly depends on the transistor any more. The price is a lower loop gain, because C_3 attenuates the signal fed back to the input.

The crystal: electrical model and two resonances

A crystal is represented by the Butterworth-Van Dyke model: a series branch L_m, C_m, R_m (the mechanical vibration seen as a circuit) in parallel with the electrode capacitance C_0.

Theorem. The crystal has a series and a parallel resonance:

    \[ f_s = \frac{1}{2\pi\sqrt{L_mC_m}},\qquad f_p = f_s\sqrt{1 + \frac{C_m}{C_0}} \approx f_s\left(1 + \frac{C_m}{2C_0}\right) \]

Proof. Neglecting R_m, the reactance of the series branch is X_s = \omega L_m - 1/(\omega C_m), zero for \omega^2L_mC_m = 1: this is the series resonance, where the crystal impedance is minimum. The series branch is in parallel with C_0, whose reactance is -1/(\omega C_0); the impedance of the parallel combination becomes infinite when the two reactances are equal and opposite, that is \omega L_m - 1/(\omega C_m) = 1/(\omega C_0). Multiplying by \omega C_m: \omega^2L_mC_m = 1 + C_m/C_0, which gives f_p. Since C_m \ll C_0, the expansion \sqrt{1 + x} \approx 1 + x/2 gives the approximate form. ∎

Reactance of a crystal versus frequency
The blue curve is capacitive, negative, below the series resonance f s, where it crosses zero; between f s and the parallel resonance f p, in the green band, it is inductive and rises steeply towards infinity; above f p it is capacitive again. Crystal oscillators work in the narrow inductive band.

The band between f_s and f_p is extremely narrow, and there the crystal behaves like an inductor that changes enormously with frequency. In a crystal Colpitts the crystal takes the place of L: the circuit finds by itself the frequency, inside that band, that satisfies the phase condition, and small phase shifts move it very little. It is the extreme version of the link between Q and stability shown in the lesson.

Why phase noise has that shape

The plot of Leeson’s model has three regions, each with a simple explanation.

  • Far from the carrier the noise is flat: it is the thermal noise of the amplifier, FkT, divided by the signal power P_s (the 2 accounts for the two sidebands).
  • Inside the resonator bandwidth, for \Delta f \mathchar"313C f_0/(2Q), the loop converts phase fluctuations into frequency fluctuations: a phase that fluctuates by \Delta\varphi shifts the frequency by \Delta\varphi f_0/(2Q), as proved in the lesson. Frequency noise, integrated, becomes phase noise with a factor 1/\Delta f in amplitude, that is 1/\Delta f^2 in power: a slope of −20 dB per decade.
  • Very close to the carrier the 1/f (flicker) noise of the transistor is added, which below the corner frequency f_c multiplies everything by f_c/\Delta f: the slope becomes −30 dB per decade.

    \[ \mathcal{L}(\Delta f) = 10\log_{10}\left[\frac{2FkT}{P_s}\left(1 + \left(\frac{f_0}{2Q\,\Delta f}\right)^2\right)\left(1 + \frac{f_c}{\Delta f}\right)\right] \]

Phase noise according to Leeson's model, on a logarithmic scale
Starting from the right, the curve is flat at the level F k T over P; below f zero over 2 Q it rises with a slope of 20 dB per decade, and below the flicker corner frequency f c the slope becomes 30 dB per decade.

Common mistakes

  • Adding C_1 and C_2: as seen from the inductor the two capacitors are in series, and the frequency is computed with C_s = C_1C_2/(C_1 + C_2).
  • Forgetting the gain margin: with g_mR exactly equal to C_1/C_2 the circuit does not start, because tolerances and temperature push it below the threshold.
  • Neglecting the loading of the transistor input: r_\pi and the base-emitter capacitance become part of the network and change both the start-up condition and the frequency.
  • Confusing f_s and f_p of the crystal: in a parallel oscillator the actual frequency depends on the load capacitance given in the datasheet, and with a different capacitance the crystal works off frequency.
  • Reading dBc/Hz as a total power: to know how much noise enters a channel one must add 10\log_{10} of its bandwidth.

Exercises on the topic, from the frequency of a Colpitts to reciprocal mixing in a receiver, are on the exercise page linked at the top.

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