Question 2 of 5: Instrumentation amplifier — transfer function and output waveform

Question 2: Instrumentation amplifier — transfer function and output waveform

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 classic three-op-amp instrumentation amplifier with every resistor equal to $R=10\ \text{k}\Omega$; a capacitor $C=10\ \mu\text{F}$ from the output node to ground; ideal op-amps on ±15 V rails. Inputs: v₁ a ±1 V triangle (period 2 ms) and v₂ a ±1 V square wave of much higher frequency (period ≈ 0.25 ms).

Find. (a) the closed-form $v_o(v_1,v_2)$; (b) the resulting output waveform.

Approach. Use the ideal-op-amp virtual-short rule on the A₁/A₂ input stage to find its two outputs, then apply the unity-gain difference stage A₃; finally superpose the two scaled input waveforms point-by-point.

1. Input-stage currents (part a). The virtual short forces the inverting inputs of A₁ and A₂ to v₁ and v₂, so the current in the gain resistor is $i=(v_1-v_2)/R$. That same current flows through both feedback resistors R.
2. Buffer outputs. Walking from each inverting node out through its feedback R:$$v_{o1}=v_1+R\,i=2v_1-v_2,\qquad v_{o2}=v_2-R\,i=2v_2-v_1.$$
3. Difference stage. A₃ with all four resistors equal has unity differential gain, $v_o=v_{o2}-v_{o1}$. Substituting:$$v_o=(2v_2-v_1)-(2v_1-v_2)=\boxed{3\,(v_2-v_1)}.$$ The overall gain is the standard $1+2R/R_{\text{gain}}=3$ with $R_{\text{gain}}=R$. The output capacitor C draws current but, because an ideal op-amp has zero output impedance, it does not alter $v_o$ — it is a distractor.
4. Build the output waveform (part b). Write $v_o=3v_2-3v_1$. The term $3v_2$ is a ±3 V square wave at the v₂ frequency; the term $-3v_1$ is a slow inverted triangle swinging between −3 V and +3 V. So $v_o$ is a fast ±3 V square wave whose centre-line rides the inverted triangle: its upper edge traces $-3v_1+3$ and its lower edge traces $-3v_1-3$.
5. Extremes and clipping check. The envelope reaches its widest when $|3v_1|$ is largest: at $t=0.5\ \text{ms}$ ($v_1=+1$) the square sits between 0 and −6 V, and at $t=1.5\ \text{ms}$ ($v_1=-1$) between +6 and 0 V. The peak magnitude is 6 V, well inside the ±15 V rails, so the output never saturates.