Question 11 of 11: Part B, Question 5: Op-amp/diode circuits — output waveform sketches

Part B, Question 5: Op-amp/diode circuits — output waveform sketches [4,4,4,4,4]

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. Ideal diode drop 0.7 V; ideal op-amps, supplies $\pm15$ V; $v_{IN}=10\sin\theta$ V in every sub-circuit.

Find. $v_{OUT}(\theta)$ for one input cycle, for each of (a)–(e), with every clamp/saturation level identified.

Approach. In every sub-circuit the inverting input is a virtual ground (0 V) as long as the negative-feedback resistor is present; find the diode’s conduction condition at that virtual ground, evaluate the linear (diode-off) transfer function, and check the result against the $\pm15$ V supply rails.

1. (a) Diode clamps the negative output. Diode anode is at the virtual ground (0 V), cathode at $v_{OUT}$; it conducts once $v_{OUT}<-0.7$ V. Diode off: $v_{OUT}=-\dfrac{5\text{k}}{5\text{k}}v_{IN}=-v_{IN}$, valid while $v_{IN}\le0.7$ V. $$\boxed{v_{OUT}=-v_{IN}\ (v_{IN}\le0.7\text{V}),\qquad v_{OUT}=-0.7\text{V (clamped)}\ (v_{IN}>0.7\text{V}).}$$ Peaks reach $+10$ V (well inside the rails), no saturation.
2. (b) Diode clamps the positive output; op-amp also saturates. The 2 kΩ resistor sits between the virtual ground and true ground — both its ends are at 0 V, so it carries no current and has no effect. Diode (cathode at virtual ground, anode at $v_{OUT}$) conducts once $v_{OUT}>0.7$ V, i.e. once $v_{IN}<-0.14$ V (from the diode-off gain $-5$). For $v_{IN}\ge-0.14$ V the unclamped $-5v_{IN}$ would run past $-15$ V once $v_{IN}>3$ V, so the op-amp itself saturates there: $$\boxed{v_{OUT}=+0.7\text{V}\ (v_{IN}<-0.14\text{V}),\quad -5v_{IN}\ (-0.14\le v_{IN}\le3\text{V}),\quad -15\text{V (saturated)}\ (v_{IN}>3\text{V}).}$$
3. (c) Summing amp, diode clamps almost the whole cycle. Diode off gives $v_{OUT}=-3v_{IN}-30$; setting this to the $-0.7$ V clamp threshold gives $v_{IN}=-9.767$ V — only in the narrow window $-10\le v_{IN}\le-9.767$ V (near the input’s own trough) is the diode off. $$\boxed{v_{OUT}=-0.7\text{V for essentially the whole cycle}\ (v_{IN}>-9.767\text{V}),\ \text{rising briefly to }0\text{V at the trough}\ (v_{IN}=-10\text{V}).}$$
4. (d) Summing amp, diode clamps the negative excursion, op-amp saturates the positive. Diode off: $v_{OUT}=-5v_{IN}-5$; conducts (clamps at $+0.7$ V) once $v_{IN}<-1.14$ V. The linear branch would exceed $-15$ V once $v_{IN}>2$ V, saturating the op-amp there: $$\boxed{v_{OUT}=+0.7\text{V}\ (v_{IN}<-1.14\text{V}),\quad -5v_{IN}-5\ (-1.14\le v_{IN}\le2\text{V}),\quad -15\text{V}\ (v_{IN}>2\text{V}).}$$
5. (e) Plain inverting amp, gain limited only by the rails. $v_{OUT}=-3v_{IN}$ ideally, but this reaches $\pm30$ V for a $\pm10$ V input, saturating at $\pm15$ V once $|v_{IN}|>5$ V: $$\boxed{v_{OUT}=-3v_{IN}\ (|v_{IN}|\le5\text{V}),\qquad v_{OUT}=\mp15\text{V (saturated)}\ (v_{IN}\gtrless\pm5\text{V}).}$$ The output is a $\times3$-amplified sine flat-topped at both rails.