Question 4 of 6: Op-Amp with Zener Feedback and Slew-Rate Limit

Question 4: Op-Amp with Zener Feedback and Slew-Rate Limit (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.

Find. The output waveform $v_o(t)$ with its clamp levels and slew-limited edge timing.

Approach. With only the zener $D_1$ in the feedback, the stage is a hard clamp: identify the two clamp levels from the diode direction, then time the transitions between them with the slew-rate limit.

1. Clamp levels. The summing node is a virtual ground while $D_1$ conducts. For $v_r\gt0$ the input current $v_r/R_1$ flows out of the node through $D_1$ in the forward direction, so $v_o=0-0.7=\boxed{-0.7\ \text{V}}$. For $v_r\lt0$ the current reverses and $D_1$ conducts in zener breakdown, so $v_o=0+V_Z=\boxed{+5\ \text{V}}$.
2. Transition size. Each edge steps between the two clamp levels: $\Delta v_o=5-(-0.7)=5.7\ \text{V}$. Between clamps the diode is momentarily off (neither forward nor in breakdown) and the loop is open, so the op-amp output moves at its slew-rate limit.
3. Edge timing. $$t_{edge}=\frac{\Delta v_o}{SR}=\frac{5.7\ \text{V}}{0.5\ \text{V}/\mu\text{s}}=\boxed{11.4\ \mu\text{s}}.$$ The input triangle has slope $10/25=0.4\ \text{V}/\mu\text{s}$, so it crosses zero at $t=25,75,125,175\ \mu\text{s}$; each zero crossing triggers a slew edge.
4. Waveform. $v_o=-0.7$ V while $v_r\gt0$ and $+5$ V while $v_r\lt0$, with straight 11.4 µs ramps at every zero crossing — a trapezoidal wave between $-0.7$ V and $+5$ V.