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22-Elec-A5 Electronics · May 2014

Question 3 of 5: Zener-Feedback Op-Amp Waveform

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.

Question 3: Zener-Feedback Op-Amp Waveform (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

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.

−+vsR₁v−D₁v₀
Q3 - Inverting op-amp whose ONLY feedback element is the Zener D1 (anode at v-, cathode at vo).
t (μs)v50100150+10−10
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.

  1. 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 \(v_o=v_- - V_F=-0.7\ \text{V}\): $$\boxed{v_o=-0.7\ \text{V}\quad(v_s\gt0).}$$
  2. 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.
  3. 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 $$t_{edge}=\frac{5.7\ \text{V}}{0.5\ \text{V/}\mu\text{s}}=11.4\ \mu\text{s}.$$
  4. 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.
t (μs)v_o75100125150+5−0.7slew
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).
Q3 output breakpoints
FeatureValue
Level for \(v_s\gt0\)−0.7 V (forward diode)
Level for \(v_s\lt0\)+5 V (Zener breakdown)
Edge amplitude5.7 V
Slew time per edge11.4 µs
Transition instantst = 0, 75, 125 µs