Question 5 of 7: Four-Stage RC Phase-Shift Oscillator
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-A1, Electronics — National Exams, May 2014. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource). Unless stated otherwise, diode drops $V_D=0.7\text{V}$.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters/rectifiers, MOSFET/BJT small-signal amplifiers, active loads and current mirrors, active-RC filters, BJT switches, RTL logic, CMOS inverter design, RC oscillators, ring oscillators, ADC/DAC) — the single reference text covering every question on this paper.
Given. Stage 1: inverting amp, input $R_G$, feedback $R_f=100\text{k}\Omega$. Stages 2-4: three IDENTICAL cascaded RC low-pass ($R=10\text{k}\Omega$, $C$) + unity-gain voltage-follower sections (the follower isolates each RC stage so they cascade without loading each other). Output feeds back to stage 1's input.
Given data
Quantity
Value
$R$ (each RC stage)
$10\text{k}\Omega$
$R_f$
$100\text{k}\Omega$
Target $f_0$ (part d)
$1\text{kHz}$
Find. Loop-gain expression $T(j\omega)$, Barkhausen oscillation condition and $\omega_0$, a value of $R_G$ to guarantee startup, and $C$ for $f_0=1\text{kHz}$.
Fig. Q5 — Signal-flow: an inverting stage (180° phase, gain $-R_f/R_G$) cascades with three identical unity-gain-buffered RC low-pass sections (each contributing gain $1/(1+j\omega RC)$ and up to 90° lag), fed back to the input — oscillation requires the LOOP's total phase to reach 360° and its magnitude to reach unity.
Approach. Multiply the inverting stage's gain by the three identical RC-follower stages' gain to get $T(j\omega)$; apply the Barkhausen criterion (phase $=0^\circ\!\!\mod360^\circ$, magnitude $=1$) to find $\omega_0$ and the required $R_f/R_G$ ratio.
Part (a) — loop gain. Each follower isolates its RC section, so three identical sections simply cascade in gain:
$$T(j\omega)=\left(-\frac{R_f}{R_G}\right)\left(\frac{1}{1+j\omega RC}\right)^3$$
Part (b) — oscillation condition and frequency. The inverting stage supplies a fixed $180^\circ$; each RC section supplies $-\tan^{-1}(\omega RC)$. Oscillation needs the total to reach $0^\circ$ (mod $360^\circ$):
$$180^\circ-3\tan^{-1}(\omega_0RC)=0^\circ\ \Rightarrow\ \tan^{-1}(\omega_0RC)=60^\circ\ \Rightarrow\ \omega_0RC=\tan60^\circ=\sqrt3$$
$$\omega_0=\boxed{\dfrac{\sqrt3}{RC}}$$
At this frequency each RC section's magnitude is $1/\sqrt{1+3}=1/2$, so the magnitude (Barkhausen) condition is:
$$\frac{R_f}{R_G}\left(\frac12\right)^3=1\ \Rightarrow\ \frac{R_f}{R_G}=8\ \Rightarrow\ R_G=\frac{R_f}{8}=\boxed{12.5\text{ k}\Omega\text{ (unity-loop-gain boundary)}}$$
Part (c) — choose $R_G$ to INITIATE oscillation. Startup (growth from noise) needs $|T|$ slightly >1, i.e. $R_f/R_G>8$, so $R_G$ must be chosen somewhat BELOW the 12.5k boundary:
$$R_G=\boxed{10\text{ k}\Omega\text{ (gives }R_f/R_G=10>8\text{, a practical startup margin)}}$$
Part (d) — choose $C$ for $f_0=1\text{kHz}$. From $\omega_0=\sqrt3/(RC)$:
$$C=\frac{\sqrt3}{\omega_0R}=\frac{\sqrt3}{2\pi(1000)(10\text{k}\Omega)}=\boxed{27.6\text{ nF}}$$