Question 5 of 5: Three-Op-Amp Instrumentation Amplifier

Question 5: Three-Op-Amp Instrumentation Amplifier (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 three-op-amp instrumentation amplifier with every resistor equal to $R=10\ \text{k}\Omega$ (so the gain-setting resistor $R_g=R$), output capacitor $C=10\ \mu\text{F}$ from $v_O$ to ground, ideal op-amps on $\pm15$ V rails. Inputs (read from the exam plots): $v_1$ is a $\pm1$ V triangle of period 2 ms; $v_2$ is a $\pm1$ V square wave of period 0.25 ms.

Find. (a) $v_O$ as a function of $v_1,v_2$; (b) the resulting output waveform.

Approach. Analyse the two-op-amp input stage (which sets the differential gain) and the unity-gain difference-amplifier output stage separately, then combine; finally superpose the two scaled inputs to sketch $v_O$.

1. Input stage (A1, A2) differential gain. With ideal op-amps the two inverting inputs follow $v_1$ and $v_2$, so the current in the middle resistor is $(v_1-v_2)/R_g$ and it flows through both feedback $R$’s. Hence $$v_{o1}-v_{o2}=\left(1+\frac{2R}{R_g}\right)(v_1-v_2).$$ With $R_g=R$ the factor is $1+2=3$, so $v_{o1}-v_{o2}=3(v_1-v_2).$
2. Output stage (A3) is a unit difference amplifier. $A_3$ with four equal $R$’s forms a difference amplifier of gain one: $v_O=v_{o2}-v_{o1}$ (the $A_2$ branch drives the non-inverting side).
3. Combine. Substituting the stage-1 result: $$\boxed{\,v_O=v_{o2}-v_{o1}=-3(v_1-v_2)=3\,(v_2-v_1)\,}.$$ The overall differential gain is $1+2R/R_g=3$.
4. The output capacitor. $C$ sits directly on $A_3$’s output. Because an ideal op-amp has zero output impedance, it holds $v_O$ regardless of the capacitor, so $C$ has no effect on the ideal transfer function (it would only matter through a real op-amp’s finite output resistance / current limit).

For part (b), $v_O=3(v_2-v_1)$ superposes a $\pm3$ V, 4 kHz square wave ($3v_2$) on the inverted triangle $-3v_1$ (a $\pm3$ V ramp of period 2 ms). The instantaneous extremes are $v_O=3(\pm1-(\mp1))=\pm6$ V, well inside the $\pm15$ V rails, so there is no clipping. The two inputs and the resulting output are plotted below.

[Figure not reproduced: Q5 inputs: v_1 is a +/-1 V triangle of period 2 ms; v_2 is a +/-1 V square wave of period 0.25 ms (read from the exam plots). See the official exam paper.]