22-Elec-A5 Electronics · May 2016

Question 5 of 5: Op-Amp / Diode Wave-Shapers — Output Sketches

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: Op-Amp / Diode Wave-Shapers — Output Sketches (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. A 10 V-amplitude sine drives five inverting-configuration op-amp stages (all with the non-inverting input grounded, so the summing node is a virtual ground) whose feedback/input networks contain ideal diodes ($V_\gamma=0.7$ V); supplies ±15 V.

Find. the output waveform of each stage over one input cycle.

Approach. Hold the inverting node at virtual ground, decide for each half-cycle which diodes conduct (a diode that must carry reverse current is off), evaluate the resulting inverting-amplifier law, and clamp any excursion at the ±15 V rails.

1. (a) T-feedback with a mid-tap diode to ground. Input 5 kΩ to the virtual ground; feedback is 5 kΩ–M–5 kΩ with a diode from mid-node M to ground (anode at M). Node balance forces $v_M=-v_{IN}$. For $v_{IN}\gt-0.7$ V the diode is off and the stage is a plain inverter $$\boxed{v_{OUT}=-2\,v_{IN}}\quad(\text{clipping at }-15\text{ V for }v_{IN}\gt7.5\text{ V}).$$ For $v_{IN}\lt-0.7$ V the diode conducts and pins M at 0.7 V, which opens the feedback loop; the amplifier saturates to $v_{OUT}=+15$ V for that whole portion. The result is an inverted, −2× scaled positive-input excursion and a +15 V rail on the negative-input excursion (Figure 5a).
2. (b) Anti-parallel diode–resistor input, 5 kΩ feedback. The input branch is a 1 kΩ path (conducts on $v_{IN}\lt0$) in parallel with a 2 kΩ path (conducts on $v_{IN}\gt0$); a 5 kΩ feedback holds the loop closed. This is a dual-slope precision rectifier with a small dead-band: $$v_{OUT}=\begin{cases}-2.5\,(v_{IN}-0.7)&v_{IN}\gt0.7\\[2pt]0&|v_{IN}|\lt0.7\\[2pt] +5\,(|v_{IN}|-0.7)&v_{IN}\lt-0.7\end{cases}$$ clipping at −15 V ($v_{IN}\gt6.7$ V) and +15 V ($v_{IN}\lt-3.7$ V) (Figure 5b).
3. (c) Diode into a biased divider node. Input 1 kΩ to the virtual ground; the feedback diode (anode at the node) reaches node Y, itself pulled to +10 V by 3 kΩ and to $v_{OUT}$ by 1 kΩ. On $v_{IN}\gt0$ the diode conducts ($v_Y=-0.7$ V) and $$\boxed{v_{OUT}=-v_{IN}-\tfrac{10.7}{3}-0.7=-(v_{IN}+4.27)}$$ (−4.27 V at $v_{IN}=0^{+}$ down to −14.27 V at $v_{IN}=10$ V, no clipping). On $v_{IN}\lt0$ the diode blocks, the loop opens and the output rails to +15 V (Figure 5c).
4. (d) Anti-parallel diode–resistor feedback (dead-zone limiter). Input 1 kΩ; feedback is a 3 kΩ-with-diode path (conducts on $v_{IN}\lt0$) in parallel with a 5 kΩ-with-diode path (conducts on $v_{IN}\gt0$). This gives two gains with a ±0.7 V cross-over offset: $$v_{OUT}=\begin{cases}-(5\,v_{IN}+0.7)&v_{IN}\gt0\\[2pt]-3\,v_{IN}+0.7&v_{IN}\lt0\end{cases}$$ clipping at −15 V ($v_{IN}\gt2.86$ V) and +15 V ($v_{IN}\lt-4.77$ V) (Figure 5d).
5. (e) Ideal battery in the feedback path. A 3 V battery (+ toward the virtual ground) sits directly across the feedback, in parallel with 3 kΩ. An ideal source has zero impedance, so it clamps the feedback voltage regardless of the resistor: $v_N-v_{OUT}=3$ V with $v_N=0$ gives $$\boxed{v_{OUT}=-3\ \text{V (constant)}}$$ a flat DC output independent of the input (Figure 5e).