Question 4 of 5: RC Oscillator — Frequency of Oscillation, Start-Up Condition and Amplitude Stabilisation

Question 4: RC Oscillator — Frequency of Oscillation, Start-Up Condition and Amplitude Stabilisation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An op-amp in the non-inverting configuration set by $R_1$ and $R_2$, with positive feedback taken from the output through a series arm $Z_s$ (a capacitor $C$ in series with a resistor $R$) to the non-inverting input, which is returned to ground by a parallel arm $Z_p$ ($2R$ in parallel with $2C$):

Find. the oscillation frequency in terms of $R$ and $C$, the condition on $R_2/R_1$ that guarantees start-up, the principal shortcoming of the circuit as drawn, and a practical fix.

Approach. Apply the Barkhausen criterion to the loop formed by the non-inverting amplifier and the $Z_s$–$Z_p$ divider: the frequency follows from the zero-phase condition on the divider, and the start-up condition from the magnitude condition at that frequency.

1. Write the transmission of the frequency-selective divider. With $s = j\omega$,$$Z_s = R + \frac{1}{sC}, \qquad Z_p = \frac{2R}{1 + s(2R)(2C)} = \frac{2R}{1+4sRC}$$and the fraction of the output that reaches the non-inverting input is $\beta(s) = Z_p/(Z_s+Z_p)$.
2. Reduce the divider to a single expression. Writing $x = sRC$ and clearing denominators,$$\beta(s) = \frac{2x}{(x+1)(1+4x) + 2x} = \frac{2x}{4x^{2}+7x+1}$$On the imaginary axis, $x = j\omega RC$, so$$\beta(j\omega) = \frac{2j\omega RC}{\left(1-4\omega^{2}R^{2}C^{2}\right) + 7j\omega RC}$$
3. Impose the phase condition. The op-amp connection is non-inverting and contributes no phase shift, so the loop can oscillate only where $\beta$ is real and positive. The numerator is purely imaginary, so the denominator must be purely imaginary too:$$1 - 4\omega_0^{2}R^{2}C^{2} = 0 \;\Longrightarrow\; \omega_0 = \frac{1}{2RC} \;\Longrightarrow\; \boxed{f_{osc} = \frac{1}{4\pi RC}}$$Note that the frequency depends only on $R$ and $C$; $R_1$ and $R_2$ do not enter it, and they never can in this topology, because they sit in the negative-feedback path that fixes gain rather than in the frequency-selective network.
4. Evaluate the attenuation at that frequency. Substituting $\omega_0 RC = 1/2$ leaves the denominator as $7j(1/2)$ against a numerator of $2j(1/2)$:$$\beta(j\omega_0) = \frac{j}{3.5\,j} = \frac{2}{7} = 0.2857$$The scaled parallel arm is what makes this differ from the textbook Wien bridge, where equal $R$ and $C$ in both arms give $\beta = 1/3$.
5. Impose the magnitude condition to obtain the start-up requirement. The loop gain is $A\beta$ with $A = 1+R_2/R_1$, and Barkhausen requires $A\beta \ge 1$ for oscillations to build up from noise:$$\left(1+\frac{R_2}{R_1}\right)\frac{2}{7} \ge 1 \;\Longrightarrow\; 1+\frac{R_2}{R_1} \ge \frac{7}{2} \;\Longrightarrow\; \boxed{\frac{R_2}{R_1} \ge 2.5}$$Strictly, sustained oscillation at constant amplitude needs $A\beta = 1$ exactly, while starting needs $A\beta$ slightly greater than unity, so a practical design uses $R_2/R_1$ a few per cent above 2.5.
6. Put numbers to it. Choosing $R = 10\ \text{k}\Omega$ and $C = 10\ \text{nF}$ gives$$f_{osc} = \frac{1}{4\pi(10^{4})(10^{-8})} = 796\ \text{Hz}$$with $R_1 = 10\ \text{k}\Omega$ and $R_2$ a little above 25 kΩ.

Main limitation. The circuit has no amplitude control. The gain that sustains the oscillation is set by a fixed resistor ratio, and the condition $A\beta = 1$ is a knife edge that no combination of real components holds. If the ratio drifts even slightly low — through resistor tolerance, temperature, or the op-amp's own finite open-loop gain at $f_{osc}$ — the poles move into the left half-plane and the oscillation dies away. If it drifts slightly high, the amplitude grows exponentially until the op-amp output runs into its supply rails, and the only thing that finally limits it is saturation. The waveform is then a clipped sinusoid with high harmonic distortion, and because the clipping level tracks the supply, the amplitude also depends on the rails rather than on anything the designer chose. In short, the circuit either stops or distorts, and the useful in-between state is not stable.

Remedy. Make the gain amplitude-dependent, so that the loop finds its own equilibrium. The simplest form is a non-linear limiter across $R_2$: two Zener diodes connected back to back (or a diode–resistor network) conduct only once the output exceeds a chosen level, shunting $R_2$ and reducing the gain at exactly the amplitude at which $A\beta$ falls back to unity. The design then sets $R_2/R_1$ a little above 2.5 so that the loop starts reliably, and lets the limiter pull it back to 2.5 once the desired output level is reached. Classical alternatives replace $R_1$ or $R_2$ with a temperature-sensitive element whose resistance rises with the current through it — a small incandescent lamp or a positive-temperature-coefficient thermistor in the $R_1$ position was the original Hewlett solution — or place a JFET operating in its triode region in the $R_1$ branch and drive its gate from a peak detector, giving a genuine automatic gain control loop. All three do the same job: they convert a marginally stable linear circuit into one with a stable limit cycle, at the price of a little distortion introduced by the non-linearity itself. The AGC approach gives the lowest distortion because the controlling element stays linear over each cycle; the Zener limiter is the cheapest and is what most exam answers are expected to describe.