Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 · 07-Elec-A5, Electronics. Closed-book; non-communicating calculator permitted; 3 hours. Five questions, 20 marks each; op-amps ideal with ±15 V supplies unless stated. All five are solved below.
Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 7th/8th ed. (diodes & wave-shaping Ch. 4; MOS amplifiers Ch. 7; BJT amplifiers Ch. 6–7; op-amp imperfections Ch. 2); R. C. Jaeger & T. N. Blalock, Microelectronic Circuit Design.
Given. Inverting configuration in which the Zener \(D_1\) is the only feedback element (anode at the \(v_-\) node, cathode at \(v_o\)); \(R_1=10\ \text{k}\Omega\), \(V_Z=5\ \text{V}\), \(V_F=0.7\ \text{V}\), slew rate \(SR=0.5\ \text{V/}\mu\text{s}\); the input triangle reaches \(+10\ \text{V}\) at 50 µs and \(-10\ \text{V}\) at 100 µs.
Find. The output waveform \(v_o(t)\) with its clamp levels and slew-limited edges.
Q3 - Inverting op-amp whose ONLY feedback element is the Zener D1 (anode at v-, cathode at vo).
Q3 - Input v_s: triangular wave, +-10 V, period 100 μs (zero crossings at t = 0, 75, 125 μs).
Approach. Because a diode always conducts in one direction, negative feedback is always present, so \(v_-\) stays a virtual ground; determine which way \(D_1\) conducts for each input polarity, read off the clamp level, then apply the slew-rate limit to the transitions.
Positive input half. For \(v_s\gt0\) the current \(i=v_s/R_1\) flows from the virtual ground toward \(v_o\), i.e. anode→cathode — \(D_1\) is forward. Then \(v_o=v_- - V_F=-0.7\ \text{V}\): $$\boxed{v_o=-0.7\ \text{V}\quad(v_s\gt0).}$$
Negative input half. For \(v_s\lt0\) the current reverses and \(D_1\) conducts in Zener breakdown, holding \(v_o-v_-=V_Z\): $$\boxed{v_o=+5\ \text{V}\quad(v_s\lt0).}$$ The output is therefore a hard clamp toggling between \(-0.7\ \text{V}\) and \(+5\ \text{V}\); \(R_1\) only sets the (irrelevant) diode current, not the levels.
Slew-limited transitions. Each edge spans \(\Delta v_o=5-(-0.7)=5.7\ \text{V}\); at \(0.5\ \text{V/}\mu\text{s}\) the op amp needs $$t_{edge}=\frac{5.7\ \text{V}}{0.5\ \text{V/}\mu\text{s}}=11.4\ \mu\text{s}.$$
Locate the edges. \(v_s\) crosses zero at \(t=0\) (going +), \(t=75\ \mu\text{s}\) (going −) and \(t=125\ \mu\text{s}\) (going +). So \(v_o\) ramps from \(+5\) down to \(-0.7\) over \(0\!-\!11.4\ \mu\text{s}\), holds \(-0.7\) to \(75\ \mu\text{s}\), ramps up to \(+5\) over \(75\!-\!86.4\ \mu\text{s}\), holds to \(125\ \mu\text{s}\), then ramps down again — a slew-rounded rectangular wave.
Q3 - Output v_o: hard Zener clamp. v_o = −0.7 V while v_s>0, +5 V while v_s<0; each edge slew-rate limited (5.7 V / 0.5 V·μs⁻¹ = 11.4 μs).