Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-A1, Electronics — National Exams, May 2014. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource). Unless stated otherwise, diode drops $V_D=0.7\text{V}$.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters/rectifiers, MOSFET/BJT small-signal amplifiers, active loads and current mirrors, active-RC filters, BJT switches, RTL logic, CMOS inverter design, RC oscillators, ring oscillators, ADC/DAC) — the single reference text covering every question on this paper.
Given. Inverting op-amp, $R_1=10\text{k}\Omega$ (input), feedback $R_2=100\text{k}\Omega\|C_1=0.1\mu\text{F}$ (the classic Miller/active-RC low-pass, i.e. "lossy integrator").
Given data
Quantity
Value
$R_1$
$10\text{k}\Omega$
$R_2$
$100\text{k}\Omega$
$C_1$
$0.1\mu\text{F}$
$V_i(t)$ (part c)
$1\sin(200t)$ V
$A$ (part d)
$10^4$ V/V
Find. $H(j\omega)=V_o/V_i$, DC gain, $f_{3dB}$, unity-gain bandwidth, $V_o(t)$ for the given sinusoid, and the corrected transfer function for finite op-amp gain.
Approach. Ideal-op-amp inverting-amp analysis gives $H=-Z_f/Z_1$ directly (single pole, set by $R_2C_1$); evaluate at $\omega=200\text{rad/s}$ for part (c); re-derive with a finite-gain node equation for part (d).
Part (a) — transfer function. $Z_f=R_2\|\frac{1}{j\omega C_1}=\dfrac{R_2}{1+j\omega R_2C_1}$, $Z_1=R_1$ (ideal op-amp, virtual ground):
$$\frac{V_o}{V_i}=-\frac{Z_f}{Z_1}=\boxed{-\dfrac{R_2/R_1}{1+j\omega R_2C_1}=\dfrac{-10}{1+j\omega(0.01\text{s})}}$$
Part (b) — DC gain, $f_{3dB}$, unity-gain bandwidth.
$$A_{DC}=-\frac{R_2}{R_1}=-\frac{100\text{k}}{10\text{k}}=\boxed{-10\text{ V/V}}$$
$$\omega_{3dB}=\frac{1}{R_2C_1}=\frac{1}{(100\text{k})(0.1\mu\text{F})}=100\text{ rad/s}\ \Rightarrow\ f_{3dB}=\boxed{15.9\text{ Hz}}$$
Single-pole roll-off ($-20\text{dB/dec}$) means the gain crosses 0dB at $\omega_t=|A_{DC}|\cdot\omega_{3dB}$:
$$\omega_t=10\times100=1000\text{ rad/s}\ \Rightarrow\ f_t=\boxed{159.2\text{ Hz (unity-gain bandwidth)}}$$
Part (c) — steady-state response to $V_i=1\sin(200t)$. At $\omega=200\text{rad/s}$:
$$H(j200)=\frac{-10}{1+j(200)(0.01)}=\frac{-10}{1+j2}=-2+j4$$
$$|H|=\sqrt{2^2+4^2}=\boxed{4.47}\qquad \angle H=180^\circ-\tan^{-1}(2)=\boxed{116.6^\circ}$$
$$V_o(t)=\boxed{4.47\sin(200t+116.6^\circ)\text{ V}}$$
Part (d) — finite op-amp gain $A=10^4$. Node equation at the inverting input $v_n=-V_o/A$ (finite gain, non-inverting grounded), solved for $V_o/V_i$ in the general inverting-amp form:
$$\frac{V_o}{V_i}=\frac{-(Z_f/Z_1)}{1+\dfrac{1}{A}\left(1+\dfrac{Z_f}{Z_1}\right)}=\boxed{\dfrac{-10^4\left(\dfrac{10}{1+j\omega(0.01)}\right)}{10^4+1+\dfrac{10}{1+j\omega(0.01)}}}$$
Since $|Z_f/Z_1|_{\max}=10$ (at DC) and $A=10^4$, the correction term $(1+Z_f/Z_1)/A\le11/10^4\approx0.11\%$ — negligible, confirming the ideal result of part (a) holds to within about $0.1\%$ for this $A$.
