22-Mec-A5 Electrical and Electronics Engineering · May 2018
Question 5 of 8: Instrumentation Amplifier — Transfer Function by Superposition
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. PEO/Engineers Canada National Examination
16-Mec-A5 Electrical & Electronics Engineering, May 2018 — 3 hours,
closed book, Casio or Sharp approved calculator only. Eight questions
of equal value; any five constitute a complete paper. All eight are solved
here, since the set is intended as a study resource.
Constants printed on the front page.
$\pi = 3.14159$, $1\ \text{hp} = 746\ \text{W}$, and the permeability of free space
$\mu_0 = 4\pi \times 10^{-7}\ \text{H}\,\text{m}^{-1}$.
Reference texts for this subject.
Sedra & Smith, Microelectronic Circuits, 8th ed. — transistor
current mirrors, biasing and operational-amplifier circuits (Ch. 7, 8, 2).
Mano & Ciletti, Digital Design, 6th ed. — Boolean algebra, DeMorgan's
theorems, universal-gate synthesis (Ch. 2–3).
Chapman, Electric Machinery Fundamentals, 5th ed. — the linear dc machine,
magnetic circuits, transformers and induction machines (Ch. 1–2, 6–8).
Sadiku & Alexander, Fundamentals of Electric Circuits, 7th ed. —
first-order transients, phasors, ac power, frequency response (Ch. 7, 9–11, 14).
Glover, Sarma & Overbye, Power System Analysis and Design, 6th ed. —
power-factor correction and transmission loss.
Question 5: Instrumentation Amplifier — Transfer Function by Superposition (20 marks)
Given. Two ideal buffers $U_1$ and $U_2$ receive $E_1$ and $E_2$ at their non-inverting inputs; their inverting inputs are joined through the single gain-setting resistor $R_1$, and each is fed back from its own output through $aR_1$ and $bR_1$ respectively. The output stage $U_3$ is a difference amplifier of gain $c$, so that $E_0 = c(e_y - e_x)$.
Find. The overall transfer function $E_0$ as a function of the two inputs $E_1$ and $E_2$ and the constants $a$, $b$ and $c$.
Figure 5 — circuit schematic. This is the classical three-operational-amplifier instrumentation amplifier: $U_1$ and $U_2$ form the differential input stage, $U_3$ the difference stage.
Approach. The two ideal assumptions do all the work. A zero differential input voltage forces each buffer's inverting terminal to sit at its own source potential, which puts a known voltage across $R_1$; zero input current then forces that entire current to continue through $aR_1$ and $bR_1$, which fixes both output potentials. The results are combined by the given difference-stage relation, and the superposition route suggested by the hint is carried out afterwards as an independent check.
Apply the virtual-short assumption. Since the differential input voltage of each ideal amplifier is zero and the non-inverting terminals are driven directly by the sources, the inverting terminals follow: $$v^{-}_{U_1} = E_1, \qquad v^{-}_{U_2} = E_2.$$ These two nodes are the ends of $R_1$, so the full input difference appears across it.
Find the current in the gain-setting resistor. The voltage across $R_1$ is $E_1 - E_2$, so $$i = \frac{E_1-E_2}{R_1}.$$ Because no current enters either inverting terminal, this same $i$ must have arrived through $aR_1$ from the output of $U_1$ and must continue through $bR_1$ into the output of $U_2$. One current flows through all three resistors in series — this is the central insight of the topology.
Work out the potential at x. Travelling from the inverting node of $U_1$ back through $aR_1$ to its output, $$e_x = E_1 + i\,(aR_1) = E_1 + a(E_1-E_2).$$
Work out the potential at y. Travelling from the inverting node of $U_2$ through $bR_1$ to its output, in the direction the current leaves, $$e_y = E_2 - i\,(bR_1) = E_2 - b(E_1-E_2).$$
Combine through the difference stage. Substituting into the given relation $E_0 = c(e_y-e_x)$, $$E_0 = c\Big[\,E_2 - b(E_1-E_2) - E_1 - a(E_1-E_2)\,\Big] = c\Big[-(E_1-E_2) - (a+b)(E_1-E_2)\Big],$$ so the transfer function of the complete circuit is $$E_0 = \boxed{c\,(1+a+b)\,(E_2-E_1)}$$
Confirm by superposition, as the hint directs. Setting $E_2 = 0$ gives $e_x = E_1(1+a)$ and $e_y = -bE_1$, hence a partial output $-c(1+a+b)E_1$. Setting $E_1 = 0$ gives $e_x = -aE_2$ and $e_y = E_2(1+b)$, hence a partial output $+c(1+a+b)E_2$. Adding the two partial responses reproduces $E_0 = c(1+a+b)(E_2-E_1)$, confirming the direct derivation.
Note the common-mode behaviour. The result depends only on the difference $E_2-E_1$. If both inputs are raised together by any amount, no current flows in $R_1$, both buffer outputs rise by that same amount, and the difference stage rejects them entirely — the ideal common-mode gain is exactly zero.
