Question 3 of 7: Matched-Feedback Low-Pass Amplifier
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-A1, Electronics — National Exams, December 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, MOSFET current-mirror biasing and CS amplifiers, active-RC matched-feedback filters, BJT current-source-biased CE amplifiers, RC-ladder (Wien-bridge type) sinusoidal oscillators, CMOS static logic synthesis and sizing, flash-ADC comparator/encoder networks) — the single reference text covering every question on this paper.
Given. $R_1=R_2=10\text{k}\Omega$; $R_3=R_4=100\text{k}\Omega$; $C_1=C_2=0.1\,\mu\text{F}$ (matched feedback on both inputs); op-amp output limit $\pm15\text{V}$; $V_i(t)=A\sin(120\pi t)\text{V}$ (i.e. $f=60\text{Hz}$).
Given data
Quantity
Value
$R_1=R_2$
$10\text{ k}\Omega$
$R_3=R_4$
$100\text{ k}\Omega$
$C_1=C_2$
$0.1\,\mu\text{F}$
Supply limit
$\pm15$ V
Input frequency
$60$ Hz
Find. DC gain; AC (frequency-dependent) gain and its $3\text{dB}$ corner; maximum undistorted input amplitude $A$ at $60\text{Hz}$.
Approach. Because $R_2/R_1=R_4/R_3$ exactly ($=10$) at every frequency (both feedback impedances use identical $R\parallel C$ pairs), the circuit reduces — by the standard balanced difference-amplifier condition — to a single-pole inverting transfer function driven by $V_i$, with no separate common-mode term to track.
Part (a) — DC gain. At DC the capacitors are open, leaving the plain resistive ratio:
$$A_{DC}=-\frac{R_3}{R_1}=-\frac{100\text{k}}{10\text{k}}=\boxed{-10\text{ V/V}}$$
Part (b)/(c) — AC gain and corner frequency. The feedback impedance is $Z_3(s)=R_3\parallel\frac{1}{sC_1}=\dfrac{R_3}{1+sR_3C_1}$, giving the full transfer function
$$A(s)=-\frac{Z_3(s)}{R_1}=\frac{-10}{1+s R_3 C_1}$$
a single real pole at
$$f_{3\text{dB}}=\frac{1}{2\pi R_3 C_1}=\frac{1}{2\pi(100\text{k})(0.1\,\mu\text{F})}=\boxed{15.9\text{ Hz}}$$
Below $f_{3\text{dB}}$ the gain is flat at $|A|=10$ ($20\text{dB}$); above it, the magnitude rolls off at $-20\text{dB/decade}$, exactly the sketch in Fig. Q3(c).
Part (d) — maximum undistorted amplitude at $60\text{Hz}$. At $\omega=120\pi\text{ rad/s}$ ($f=60\text{Hz}$), well above the $15.9\text{Hz}$ corner, the gain magnitude has already rolled off:
$$|A(j\omega)|=\frac{10}{\sqrt{1+(\omega R_3C_1)^2}}=\frac{10}{\sqrt{1+3.770^2}}=\frac{10}{3.900}=\boxed{2.564}$$
The output stays within $\pm15\text{V}$ provided $|A(j\omega)|\cdot A_{\text{in}}\le15$:
$$A_{\max}=\frac{15}{2.564}=\boxed{5.85\text{ V}}$$
(Note this is much larger than the naive DC-based estimate $15/10=1.5\text{V}$ — the pole's roll-off at $60\text{Hz}$ is exactly what lets a bigger input swing through undistorted.)
Fig. Q3(c) — single-pole low-pass Bode magnitude: flat at $20\text{dB}$ ($|A|=10$) below $f_{3\text{dB}}=15.9\text{Hz}$, rolling off at $-20\text{dB/decade}$ above it.
