Question 3 of 7: Matched Difference Amplifier — Gain and Clipping Limit

Question 3: Matched Difference Amplifier — Gain and Clipping 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. Standard 4-resistor difference amplifier: $R_1=10\text{k}\Omega$ ($V_i^+\to$ inverting input), $R_2=10\text{k}\Omega$ ($V_i^-\to$ non-inverting input), $R_3=100\text{k}\Omega$ (feedback, output to inverting input), $R_4=100\text{k}\Omega$ (non-inverting input to ground). Matched ratio $R_3/R_1=R_4/R_2$. Supplies $\pm15\text{V}$; $V_{in}=1\text{V}$ (peak) at 1kHz.

Find. $|A_v|$ in dB; the amplifier's equivalent (gain-block) circuit; the largest differential input that avoids output clipping.

Approach. With $R_2/R_1=R_4/R_3$ (here $R_2=R_1$ and $R_4=R_3$, ratios both 1, still matched at the ratio $R_3/R_1=10$), superposition on the ideal op-amp gives a pure differential amplifier — common-mode is fully rejected and the gain is set entirely by $R_3/R_1$. Clipping is then a simple headroom calculation against the supply rails.

1. Part (a) — gain. For the matched difference amplifier, $V_o=(R_3/R_1)(V_i^+-V_i^-)$: $$A_v=\frac{R_3}{R_1}=\frac{100\text{k}\Omega}{10\text{k}\Omega}=\boxed{10.0\text{ V/V}}$$ $$|A_v|_{\text{dB}}=20\log_{10}(10.0)=\boxed{20.0\text{ dB}}$$
2. Part (b) — equivalent circuit. Because the network is perfectly matched, the whole 4-resistor amplifier collapses (from the source's point of view) to a single ideal difference block: a dependent source $V_o=10(V_i^+-V_i^-)$ with input resistance $R_{id}\approx 2R_1=20\text{k}\Omega$ (differential) and output resistance $\approx0$ (ideal op-amp output). See Fig. Q3.
3. Part (c) — clipping limit. The output can swing to at most the supply rails, $\pm15\text{V}$ (ideal op-amp assumption — real devices saturate a volt or two inside the rails, which only makes the true limit slightly more conservative). Setting $|V_o|_{\max}=15\text{V}=A_v\,V_{in,\max}$: $$V_{in,\max}=\frac{15\text{V}}{10.0}=\boxed{1.5\text{ V (peak, differential)}}$$ At the stated operating point $V_{in}=1\text{V}$, $V_o=10\text{V}$ peak — comfortably inside $\pm15\text{V}$, so no clipping occurs at that input level; clipping would begin only if the differential input exceeded 1.5V peak.