Question 3 of 7: Inverting Op-Amp with Zener-Diode Output Limiter

Question 3: Inverting Op-Amp with Zener-Diode Output Limiter (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 amplifier: $R_1=1\text{k}\Omega$ input resistor, $R_F=10\text{k}\Omega$ feedback resistor, paralleled by two back-to-back Zener diodes ($V_Z=5\text{V}$ each, common-anode) from $V_{out}$ to the virtual-ground summing node. $R_L=1\text{k}\Omega$ load, $\pm15\text{V}$ supplies drawing $I_L=100\text{mA}$ each, $V_{in}=1\text{V}$ peak at 1kHz.

Find. $A_v$, $A_p$ (dB); efficiency; the equivalent circuit; the max $\pm V_{in}$ before clipping.

Approach. The linear (small-signal) gain is set by $-R_F/R_1$ as usual; the back-to-back Zener pair sits directly across $V_{out}$ (since the summing node is virtual ground), so it clamps $|V_{out}|$ to $V_Z+V_F$. Compare the linear prediction against that clamp to see whether the stated $1\text{V}$ input actually keeps the stage linear (it does not), then compute power/efficiency from the real, clipped output.

1. Part (a) — gain. $A_v=-R_F/R_1=-10\text{k}/1\text{k}=\boxed{-10\text{ V/V}=20\text{ dB}}$. Because $R_1=R_L=1\text{k}\Omega$ (equal source/load resistances), the $10\log(R_1/R_L)$ term in the power-gain formula vanishes, so $A_p(\text{dB})=A_v(\text{dB})=\boxed{20\text{ dB}}$.
2. Clamp voltage. The clamp sits from $V_{out}$ to the virtual-ground node ($\approx0\text{V}$), so it limits $V_{out}$ directly: $V_{clamp}=V_Z+V_F=5+0.7=\boxed{5.7\text{ V}}$ (symmetric, both polarities, since the two Zeners are identical).
3. Is the stage actually linear here? With $V_{in}=1\text{V}$ peak, the linear prediction is $|V_{out}|=|A_v|V_{in}=10\text{V}$, which exceeds the $5.7\text{V}$ clamp — the amplifier IS driven into its limiter by this input. $V_{out}(t)=\text{clip}(-10\,v_{in}(t),\,\pm5.7\text{V})$, a flat-topped sine (Fig. Q3-b).
4. Part (b) — efficiency. Numerically integrating the clipped waveform's power over one cycle gives $P_{out,avg}=\overline{V_{out}^2}/R_L\approx\boxed{24.3\text{ mW}}$. Supply power: $P_{supply}=2\,V_S\,I_L=2(15)(0.1)=\boxed{3.0\text{ W}}$. Efficiency: $$\eta=\frac{P_{out,avg}}{P_{supply}}=\frac{24.3\text{mW}}{3.0\text{W}}=\boxed{0.81\%}$$ — characteristically poor, since the $100\text{mA}$-per-supply spec is far larger than what this low-level signal actually needs; the supplies mostly power headroom/quiescent current, not the small output signal.
5. Parts (d), (e) — clipping thresholds. The stage stays linear only while $|V_{out}|=|A_v|V_{in}<5.7\text{V}$, i.e. $$V_{in,max}=\frac{V_{clamp}}{|A_v|}=\frac{5.7}{10}=\boxed{0.57\text{ V}}$$ Because the clamp is symmetric, the maximum POSITIVE input (which drives $V_{out}$ toward $-5.7\text{V}$) and the maximum NEGATIVE input (which drives $V_{out}$ toward $+5.7\text{V}$) have the same magnitude: $V_{in,max}=+0.57\text{V}$ and $V_{in,max}=-0.57\text{V}$ respectively.