Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A5 — Semiconductor Devices & Circuits — National Exams, May 2014
3 hours duration. Closed book exam (useful constants and equations annexed to the 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. (diode I–V characteristics and junction physics Ch. 3–4, op-amp circuits and active filters Ch. 2–12, BJT biasing and small-signal amplifiers Ch. 5–6, ADC architectures Ch. 17); C. Kittel, Introduction to Solid State Physics, 8th ed. (crystal structure and packing fraction Ch. 1, free-electron/semiconductor carrier statistics Ch. 8).
Stage 1: an op-amp with equal series/feedback resistors $R_0$ at the inverting input, and an $R$–$C$ low-pass ($R=1.6\,\text{k}\Omega$, $C$ unknown) feeding the non-inverting input. Stage 2: a plain inverting amplifier, gain $-R_2/R_1$.
Find. (a) $F(s)$; (b) filter order; (c) $|F(j\omega)|$ in dB; (d) $C$ for a 135° phase shift at 415 Hz.
Figure P3 — stage 1 (left op-amp) is a first-order all-pass section built from equal resistors $R_0$ and an $R$–$C$ low-pass at the non-inverting input; stage 2 (right op-amp) is a plain inverting amplifier with gain $-R_2/R_1$.
Approach. Derive stage 1's transfer function from the virtual-short condition and superposition at the non-inverting input, multiply by stage 2's fixed inverting gain, then read off order, magnitude and phase from the combined $F(s)$.
Part (a) — stage 1. The non-inverting input sees only the passive $R$–$C$ divider (no current drawn by the ideal op-amp), so $V_+=V_i\cdot\dfrac{1/sC}{R+1/sC}=\dfrac{V_i}{1+sRC}$. At the inverting input, KCL with equal resistors $R_0$ (no current into the op-amp) gives $\dfrac{V_i-V_-}{R_0}=\dfrac{V_--V_{in}}{R_0}$, i.e. $V_{in}=2V_--V_i$. The ideal op-amp forces $V_-=V_+$, so
$$V_{in}=\frac{2V_i}{1+sRC}-V_i=V_i\cdot\frac{1-sRC}{1+sRC}=-V_i\cdot\frac{s-\omega_o}{s+\omega_o},\qquad \omega_o=\frac{1}{RC}$$
(a standard first-order all-pass section). Stage 2 is a plain inverter, $V_o=-\dfrac{R_2}{R_1}V_{in}$. Multiplying:
$$\boxed{F(s)=\frac{V_o}{V_i}=\left(-\frac{R_2}{R_1}\right)\left(-\frac{s-\omega_o}{s+\omega_o}\right)=\frac{R_2}{R_1}\cdot\frac{s-\omega_o}{s+\omega_o}}$$
— exactly the target expression.
Part (b) — order. $F(s)$ has one pole ($s=-\omega_o$) and one zero ($s=+\omega_o$): a single energy-storage element ($C$) sets the denominator degree, so this is a first-order filter (specifically, a first-order all-pass network cascaded with a resistive gain stage, which adds no additional pole).
Part (c) — magnitude. On the $j\omega$ axis, $s-\omega_o=-\omega_o+j\omega$ and $s+\omega_o=\omega_o+j\omega$ are complex conjugate-symmetric points about the real axis, so $|s-\omega_o|=|s+\omega_o|=\sqrt{\omega^2+\omega_o^2}$ for every $\omega$ — the ratio is exactly 1 at all frequencies:
$$|F(j\omega)|=\frac{R_2}{R_1}\cdot\frac{|{-\omega_o+j\omega}|}{|\omega_o+j\omega|}=\frac{R_2}{R_1}=\frac{10\,\text{k}\Omega}{1\,\text{k}\Omega}=10$$
$$\boxed{|F|_{\text{dB}}=20\log_{10}(10)=20.0\ \text{dB}}\quad\text{(flat with frequency — the defining trait of an all-pass network)}$$
Part (d) — solve for C. The phase of $F(j\omega)$ is $\angle(j\omega-\omega_o)-\angle(j\omega+\omega_o)=\left[180^\circ-\tan^{-1}(\omega/\omega_o)\right]-\tan^{-1}(\omega/\omega_o)=180^\circ-2\tan^{-1}(\omega/\omega_o)$. Setting this to 135°:
$$180^\circ-2\tan^{-1}\!\left(\frac{\omega}{\omega_o}\right)=135^\circ\ \Rightarrow\ \tan^{-1}\!\left(\frac{\omega}{\omega_o}\right)=22.5^\circ\ \Rightarrow\ \frac{\omega}{\omega_o}=\tan(22.5^\circ)=0.4142$$
With $\omega=2\pi(415)=2607.5\,\text{rad/s}$: $\omega_o=\omega/0.4142=6295.1\,\text{rad/s}$, so $RC=1/\omega_o=1.589\times10^{-4}\,\text{s}$ and
$$C=\frac{RC}{R}=\frac{1.589\times10^{-4}}{1600}=\boxed{0.0993\ \mu\text{F}\ (99.3\,\text{nF})}$$
