Question 5 of 7: RC-Ladder (Wien-Bridge Type) Sinusoidal Oscillator
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-A1, Electronics — National Exams, May 2016. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource, per the standing "answer all M" rule). Unless stated otherwise, diode drops $V_D=0.7\text{V}$.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters/rectifiers/zener regulators, MOSFET current-mirror biasing and CS amplifiers, active-RC filters with T-network feedback, BJT cascode differential pairs with Wilson current-mirror loads, RC-ladder (Wien-bridge type) sinusoidal oscillators, CMOS static/pass-transistor logic synthesis and sizing, switched-capacitor dual-slope integrators) — the single reference text covering every question on this paper.
Given. Passive ladder $R=10\text{k}\Omega$, $C=0.1\,\mu\text{F}$ (both values used twice); non-inverting amplifier gain $A=1+R_2/R_1$.
Find. (a) Loop gain $L(j\omega)=A\beta(j\omega)$; (b) $\omega_0$ where $\angle L=0$; (c) $R_1,R_2$ for $|L(j\omega_0)|\ge1$.
Approach. Because the op-amp draws no current into its (+) input, the leftmost node (the $(+)$ input itself, with only a shunt $C$ and a series $R$ attached) is a simple series junction — the series-$R$/shunt-$C$ pair collapses into one branch, $R+1/(j\omega C)$, from the middle node to ground. Nodal analysis on the remaining two-node ladder gives $\beta(j\omega)=V_+/V_o$; multiplying by the amplifier's real gain $A$ gives the loop gain, and the classic Barkhausen conditions (zero phase, unity magnitude) follow.
Part (a) — loop-gain expression. Nodal analysis of the $C$–(shunt $R$)–$R$–(shunt $C$) ladder (verified by symbolic nodal solution) gives the feedback factor
$$\beta(s)=\frac{V_+(s)}{V_o(s)}=\frac{RCs}{R^2C^2s^2+3RCs+1}$$
— the same closed form as the classical Wien-bridge divider, confirming the ladder is an equivalent redrawing of it. With the non-inverting amplifier gain $A=1+R_2/R_1$ (real, frequency-independent), the loop gain is
$$L(j\omega)=A\,\beta(j\omega)=\Big(1+\frac{R_2}{R_1}\Big)\cdot\frac{j\omega RC}{1-\omega^2R^2C^2+j3\omega RC}=\boxed{\Big(1+\frac{R_2}{R_1}\Big)\dfrac{j\omega RC}{(1-\omega^2R^2C^2)+j3\omega RC}}$$
Part (b) — zero loop-phase condition. Since $A$ is real and positive, $\angle L=\angle\beta$; writing $x=\omega RC$, $\beta=jx/(1-x^2+3jx)$ is purely real exactly when $x(1-x^2)=0\Rightarrow x=1$ (the only positive root), i.e.
$$\boxed{\omega_0=\frac{1}{RC}\ \Rightarrow\ f_0=\frac{1}{2\pi RC}=\frac{1}{2\pi(10\text{k}\Omega)(0.1\,\mu\text{F})}=159.2\text{ Hz}}$$
(the identical $RC$ product as Question 3, so the same corner frequency recurs). At $\omega_0$, $\beta(j\omega_0)=1/3$ exactly.
Part (c) — component choice to sustain oscillation. Barkhausen's criterion requires $|L(j\omega_0)|\ge1$ at $\angle L=0$:
$$A\cdot\frac13\ge1\ \Rightarrow\ 1+\frac{R_2}{R_1}\ge3\ \Rightarrow\ \boxed{R_2\ge2R_1}$$
Choosing $R_1=10\text{k}\Omega$ gives $R_2=20\text{k}\Omega$ at the marginal (exactly unity loop-gain) point; in practice $R_2$ is set a few percent above $2R_1$ (e.g. $R_2\approx21$–$22\text{k}\Omega$, or a trimmer) so the loop gain starts just above unity and oscillation reliably builds up from noise, after which amplifier saturation (or an AGC/limiter element) settles the amplitude and keeps distortion low.
