Question 2 of 6: Active Filter — Inverting Amplifier Cascaded with a Passive High-Pass Stage
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Phys-A5-B, Analog and Digital Electronic Circuits — National Exams, May 2018. 6 questions, marks as shown in the paper's marking scheme (Q1: 15; Q2: 10+10; Q3: 10+10; Q4: 15; Q5: 5+10; Q6: 10+5). Open-book exam; any non-communicating calculator permitted.
Reference texts: A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (Sedra-Smith) — used throughout for BJT/MOSFET biasing, current mirrors, op-amp and feedback-amplifier analysis. M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. (Mano-Digital) — used for the CMOS pull-up/pull-down duality construction in Question 6.
Question 2: Active Filter — Inverting Amplifier Cascaded with a Passive High-Pass Stage (10+10 marks)
Given. An inverting op-amp stage ($R_1$ in, $R_2\Vert C_1$ feedback) whose output drives a passive $C_2$–$R_4$ high-pass stage to $v_o$; $R_3$ loads the op-amp's own output node directly to ground.
Symbol
Value
$R_1$
$2\text{k}\Omega$
$R_2$
$15\text{k}\Omega$
$R_3$
$5\text{k}\Omega$
$R_4$
$1\text{k}\Omega$
$C_1$
$100\text{pF}$
$C_2$
$100\text{nF}$
Find. (a) $v_o/v_i(s)$. (b) The asymptotic Bode magnitude plot.
[Figure not reproduced: Fig. 2 as printed in the source paper. See the official exam paper or the cited reference text.]
Fig. 2 — $R_1$ into the virtual-ground summing node; $R_2\Vert C_1$ feedback sets the op-amp output $v_x$; $R_3$ merely loads $v_x$ (ideal op-amp, no effect on the transfer function); $C_2$ then $R_4$ form a passive high-pass stage to $v_o$.
Approach. Solve the inverting stage with an ideal op-amp (virtual ground, zero output impedance) for $v_x/v_i$, then cascade the passive $C_2$–$R_4$ high-pass divider for $v_o/v_x$, and multiply.
Inverting-stage gain to the op-amp's own output $v_x$. The feedback impedance is $R_2$ in parallel with $C_1$: $$Z_f(s)=\frac{R_2\cdot\frac{1}{sC_1}}{R_2+\frac{1}{sC_1}}=\frac{R_2}{1+sR_2C_1}.$$ With virtual ground at the $(-)$ input, $$\frac{v_x}{v_i}=-\frac{Z_f}{R_1}=-\frac{R_2}{R_1(1+sR_2C_1)}.$$
$R_3$ does not enter the transfer function. $R_3$ ties the op-amp's output node to ground, but an ideal op-amp output is a zero-impedance source — it supplies whatever current $R_3$ demands without changing $v_x$. $R_3$ only sets a current/power constraint, not part of the signal path.
Passive high-pass stage, $v_x\to v_o$. $C_2$ in series then $R_4$ to ground forms a simple divider: $$\frac{v_o}{v_x}=\frac{R_4}{R_4+\frac{1}{sC_2}}=\frac{sR_4C_2}{1+sR_4C_2}.$$
Combine the two stages. $$\boxed{\dfrac{v_o}{v_i}(s)=-\dfrac{R_2}{R_1}\cdot\dfrac{sR_4C_2}{(1+sR_2C_1)(1+sR_4C_2)}}$$ — a band-pass shape: one zero at the origin and two real poles.
(b) Corner frequencies and mid-band gain, numerically. $$\omega_{p2}=\frac{1}{R_4C_2}=\frac{1}{(1\text{k}\Omega)(100\text{nF})}=1.00\times10^{4}\ \text{rad/s}.$$ $$\omega_{p1}=\frac{1}{R_2C_1}=\frac{1}{(15\text{k}\Omega)(100\text{pF})}=6.67\times10^{5}\ \text{rad/s}.$$ Between the two corners the s-dependent factors both flatten toward 1, leaving the mid-band gain $R_2/R_1=7.5$, i.e. $20\log_{10}7.5=17.5\text{ dB}$.
Fig. 2b — asymptotic Bode magnitude: $+20\text{ dB/dec}$ below $\omega_{p2}=10^4\text{ rad/s}$ (zero at the origin), flat at $17.5\text{ dB}$ between $\omega_{p2}$ and $\omega_{p1}=6.67\times10^5\text{ rad/s}$, then $-20\text{ dB/dec}$ above $\omega_{p1}$.
