Question 3 of 7: Differential Active High-Pass Amplifier
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-A1, Electronics — National Exams, December 2015. 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/clampers, MOSFET/BJT small-signal amplifiers, active loads and current mirrors, active-RC filters, differential pairs with current-mirror loads, op-amp astable multivibrators, CMOS inverter design and delay, clocked cross-coupled latches, current-steering DACs) — the single reference text covering every question on this paper.
Question 3: Differential Active High-Pass Amplifier (20 marks)
Given. Ideal op-amp; matched impedances $Z_1=R_1+1/(j\omega C_1)$ ($R_1=10\text{k}\Omega$, $C_1=0.1\,\mu\text{F}$) on both differential inputs; matched $Z_2=R_2=40\text{k}\Omega$ (feedback on the $-$ side, reference-to-ground on the $+$ side). $V_i(t)=10\sin(120\pi t)$ V for parts (c),(d).
Given data
Quantity
Value
$R_1$
$10\text{ k}\Omega$
$C_1$
$0.1\,\mu\text{F}$
$R_2$
$40\text{ k}\Omega$
$V_i(t)$
$10\sin(120\pi t)$ V
Find. $H(j\omega)=V_o/V_i$; $f_{3dB}$ and Bode sketch; $V_o(j\omega)$ and $V_o(t)$ at $f=60\text{Hz}$.
Approach. With matched $Z_1,Z_2$ on both inputs this is a standard difference amplifier: $V_o=(Z_2/Z_1)(V_i^+-V_i^-)=(Z_2/Z_1)V_i$. Substituting the series $R_1$-$C_1$ impedance gives a single-pole high-pass shape whose corner comes directly from $R_1C_1$ and whose passband gain is the plain resistor ratio $R_2/R_1$.
Part (a) — transfer function. $Z_1=R_1+\dfrac{1}{j\omega C_1}=\dfrac{1+j\omega R_1C_1}{j\omega C_1}$, $Z_2=R_2$:
$$H(j\omega)=\frac{V_o(j\omega)}{V_i(j\omega)}=\frac{Z_2}{Z_1}=\frac{R_2}{R_1}\cdot\frac{j\omega R_1C_1}{1+j\omega R_1C_1}=\boxed{\dfrac{j\omega R_2C_1}{1+j\omega R_1C_1}}$$
a single-pole high-pass with DC gain $0$ and high-frequency gain $R_2/R_1$.
Part (b) — frequency response. Corner frequency (where the pole sits, $\omega R_1C_1=1$):
$$f_{3dB}=\frac{1}{2\pi R_1C_1}=\frac{1}{2\pi(10^4)(10^{-7})}=\boxed{159.2\text{ Hz}}$$
Passband (high-frequency) gain $R_2/R_1=40\text{k}/10\text{k}=\boxed{4\text{ V/V}}=12.0\text{ dB}$. Below $f_{3dB}$ the response rises at $+20\text{dB/decade}$; above it, it flattens at $12.0\text{dB}$.
Fig. Q3(b) — Bode magnitude sketch. Rises at $+20\text{dB/decade}$ below $f_{3dB}=159\text{Hz}$, flattens to the passband gain $20\log_{10}(4)=12.0\text{dB}$ well above it.
Part (c) — $V_o(j\omega)$ at 60Hz. $\omega=120\pi=377.0\text{ rad/s}$, so $\omega R_1C_1=377.0\times10^{-3}=0.377$:
$$H(j\omega)=4\cdot\frac{j0.377}{1+j0.377}$$
$$|H|=4\cdot\frac{0.377}{\sqrt{1+0.377^2}}=1.411\qquad\angle H=90^\circ-\tan^{-1}(0.377)=90^\circ-20.6^\circ=69.4^\circ$$
$$V_o(j\omega)=10\times1.411\angle69.4^\circ=\boxed{14.11\angle69.4^\circ\text{ V}}$$
Part (d) — $V_o(t)$. A sinusoidal steady-state phasor result converts back directly:
$$V_o(t)=\boxed{14.11\sin(120\pi t+69.4^\circ)\text{ V}}$$
