Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A5 — Semiconductor Devices & Circuits — National Exams, May 2015
3 hours duration. Closed book exam (useful constants, equations and device models are annexed to the exam paper). Any FIVE (5) of the SEVEN (7) questions constitute a complete exam paper; all seven are answered here as a complete study resource.
Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (semiconductor/diode physics Ch. 3–4, op-amp and active-filter circuits Ch. 2 & 12, MOSFET small-signal amplifiers Ch. 7, data converters Ch. 17, waveform-shaping/signal generators Ch. 13); M. M. Mano & M. D. Ciletti, Digital Design, 6th ed. (CMOS/NMOS logic-family gates Ch. 10); C. Kittel, Introduction to Solid State Physics, 8th ed. (semiconductor carrier transport and statistics Ch. 8).
Given. Figure P3: op-amp 1's non-inverting input is tied directly to its own output $V_2$ (unity-gain buffer). Its "+" node (call it A) sees a resistor $R$ to ground and a capacitor $C$ to node $V_x$; $V_x$ in turn sees the input $V_1$ through a capacitor $C$, and a feedback resistor $R$ back to $V_2$. Op-amp 2 is a plain inverting amplifier from $V_2$ to $V_3$ with $R_1=1\,\text{k}\Omega$ (input) and $R_2=10\,\text{k}\Omega$ (feedback).
Find. (a) definition + 2 advantages of active filters; (b) derive $F(s)$; (c) low/high/band-pass classification; (d) $|F|$ in dB and $\angle F$ in degrees at $\omega=0.5\,\omega_o$.
Figure — the two-stage signal path implied by Fig. P3: a unity-gain-buffered Sallen-Key high-pass section cascaded with a plain inverting amplifier.
Approach. Write KCL at node $V_x$ and at op-amp 1's "+" input node A (which is forced to equal $V_2$ by the unity-feedback buffer), eliminate $V_x$, and solve for $H_1(s)=V_2/V_1$. Cascade with op-amp 2's inverting gain $H_2=-R_2/R_1$ to get $F(s)=H_1(s)H_2$, then classify the filter from the numerator's behaviour as $s\to0,\infty$ and evaluate $F(j\omega)$ numerically at $\omega=0.5\,\omega_o$.
Part (a) — active filter, advantages. An active filter builds the desired frequency response from resistors, capacitors and an active gain element (here, op-amps) rather than from a passive R-L-C network. Two advantages: (1) it needs no inductors, which are bulky, lossy and expensive at low (audio-range) frequencies; (2) the op-amp's near-zero output impedance and near-infinite input impedance let filter stages be cascaded (as in this circuit) without one stage loading the next, and the active element can also supply real voltage gain, not just attenuation.
Part (b) — derivation of $F(s)$. At node $V_x$, KCL (currents into $V_x$ from $V_1$ through $C$, out through $R$ to $V_2$, and out through the top $C$ to node A) gives, with $V_A=V_2$ (unity buffer):
$$sC(V_1-V_x)=\frac{V_x-V_2}{R}+sC(V_x-V_2)$$
At node A (op-amp input draws no current), the current arriving through the top $C$ equals the current leaving through $R$ to ground:
$$sC(V_x-V_2)=\frac{V_2}{R}\ \ \Rightarrow\ \ V_x=V_2\left(1+\frac{1}{sCR}\right)$$
Substituting $V_x$ back into the node-$V_x$ equation and simplifying with $x=sCR$:
$$sCV_1=V_2\cdot\frac{(x+1)^2}{Rx}\ \ \Rightarrow\ \ \frac{V_2}{V_1}=\frac{x^2}{(x+1)^2}=\frac{s^2}{s^2+\dfrac{2}{RC}s+\dfrac{1}{(RC)^2}}$$
which is exactly the given denominator with $\omega_o=1/RC$. Op-amp 2 is a standard inverting amplifier, $V_3/V_2=-R_2/R_1=-10\,\text{k}\Omega/1\,\text{k}\Omega=-10$. Multiplying the two stages:
$$F(s)=\frac{V_3}{V_1}=\frac{V_3}{V_2}\cdot\frac{V_2}{V_1}=\boxed{\dfrac{-10\,s^2}{s^2+\dfrac{2}{CR}s+\dfrac{1}{(CR)^2}}}$$
— matching the given transfer function exactly.
Part (c) — filter type. As $s\to0$ (DC/low frequency), the numerator $-10s^2\to0$ while the denominator's constant term $1/(CR)^2$ stays finite, so $F\to0$: the filter blocks low frequencies. As $s\to\infty$ (high frequency), both numerator and the $s^2$ term in the denominator dominate and $F\to-10$: the filter passes high frequencies at a finite gain. This is the signature of a HIGH-PASS filter.
Part (d) — magnitude and phase at $\omega=0.5\,\omega_o$. Let $x=s/\omega_o=j0.5$. Then $x^2=-0.25$, and
$$F=\frac{-10(-0.25)}{-0.25+2(j0.5)+1}=\frac{2.5}{0.75+j1.0}$$
$$|F|=\frac{2.5}{\sqrt{0.75^2+1.0^2}}=\frac{2.5}{1.25}=\boxed{2.00}\ \ \Rightarrow\ \ 20\log_{10}(2.00)=\boxed{6.02\ \text{dB}}$$
$$\angle F=0^\circ-\tan^{-1}\!\left(\frac{1.0}{0.75}\right)=0^\circ-53.13^\circ=\boxed{-53.13^\circ}$$
