Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2019, 16-Elec-A5 Electronics. Closed book; a Casio or Sharp approved calculator permitted. Three hours. Answer all five (5) questions; each worth 20 marks. Unless stated, op-amps are ideal with ±15 V supplies, and ground and chassis are common.
Reference texts. A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (Oxford, 2020) — diode piecewise-linear models & limiters (Ch. 4), MOSFET biasing & the common-source amplifier (Ch. 7), op-amp dynamics/slew rate (Ch. 2), CMOS inverter VTC and noise margins (Ch. 13), and precision/limiting op-amp–diode circuits (Ch. 4 & 18). B. Razavi, Fundamentals of Microelectronics, 2nd ed., is an equivalent secondary reference.
Given. Slew rate SR = 1 V/µs, unity-gain bandwidth ft = 1 MHz, so the small-signal closed-loop pole is at $\omega_t=2\pi f_t$; the follower’s step response is $v_{OUT}(t)=V\!\left(1-e^{-\omega_t t}\right)$.
Find. (a) the largest step V that keeps the response a pure exponential (not slew-limited), plus the rise/fall sketches; (b) the 10%–90% rise time for that step; (c) the rise time when the step is 10× larger.
(a) Slew boundary. The exponential’s maximum slope is its initial slope, $V\omega_t$. It stays a pure exponential only while this does not exceed the slew rate:
$$V_{\max}=\frac{\text{SR}}{\omega_t}=\frac{\text{SR}}{2\pi f_t}=\frac{1\ \text{V/}\mu\text{s}}{2\pi(1\ \text{MHz})}=\frac{1}{2\pi}=\boxed{0.159\ \text{V}}.$$
This tiny step is far inside the ±10 V rails, so the supply does not limit it.
Sketch. Rising edge: $v_{OUT}=V(1-e^{-\omega_t t})$; falling edge: $v_{OUT}=V e^{-\omega_t t}$; both with time constant $\tau=1/\omega_t=0.159\ \mu\text{s}$.
Q3(a): for V = 0.159 V the follower responds as a clean single-pole exponential — a symmetric rise and fall, each with τ = 0.159 µs.
(b) 10%–90% rise time of the exponential. For a single pole,
$$t_r = \frac{\ln 9}{\omega_t}=\frac{2.2}{\omega_t}=\frac{2.2}{2\pi(1\ \text{MHz})}=\boxed{0.35\ \mu\text{s}}.$$
This is independent of amplitude (it is set by the pole alone).
(c) Ten-times-larger step — check for slewing. $V=10\times0.159=1.59\ \text{V}$. The demanded initial slope $V\omega_t=1.59\times6.28\ \text{V/}\mu\text{s}\approx10\ \text{V/}\mu\text{s}$ is ten times the slew rate, so the output slew-limits: it ramps linearly at SR for almost the whole swing.
New rise time (slew-limited). Between 10% and 90% the output travels $0.8V$ at the constant slope SR:
$$t_r=\frac{0.8\,V}{\text{SR}}=\frac{0.8(1.59\ \text{V})}{1\ \text{V/}\mu\text{s}}=\boxed{1.27\ \mu\text{s}}.$$
The rise time is now set by SR, not by $\omega_t$, and it scales with amplitude.
Q3(c): the 10× step forces a straight-line slew at 1 V/µs (solid); the small-signal exponential it would otherwise follow is shown dashed. The 10%–90% rise time stretches to 1.27 µs.
