Transfer Function of a Three-Op-Amp Instrumentation Amplifier: 25-Nav-B6 Ocean Engineering and Offshore Structures, Undated paper, Q5

Question 5: Transfer Function of a Three-Op-Amp Instrumentation Amplifier (equal value)

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. $E_1$ drives the non-inverting input of $U_1$ and $E_2$ the non-inverting input of $U_2$. The inverting nodes of the two amplifiers are joined by the gain-setting resistor $R_1$; feedback resistors $aR_1$ (from $x$, the output of $U_1$) and $bR_1$ (from $y$, the output of $U_2$) close the two loops. $U_3$ is a difference amplifier of gain $c$, so $E_0 = c\,(e_y - e_x)$ is given. Ideal-op-amp assumptions [i] and [ii] apply: zero differential input voltage and zero input current.

Find. $E_0$ as a function of $E_1$ and $E_2$, using superposition.

Figure 5 — two-op-amp input stage ($U_1$, $U_2$) with the gain-setting resistor $R_1$ bridging the two inverting nodes, feeding the $U_3$ difference amplifier of gain $c$.

Approach. Use assumption [i] to place the input voltages directly on the two inverting nodes, then use assumption [ii] to force a single current through the chain $aR_1 \rightarrow R_1 \rightarrow bR_1$; solve $e_x$ and $e_y$ for each input acting alone, superpose, and substitute into the given $U_3$ relation.

Locate the node potentials. By assumption [i] the inverting terminal of each amplifier sits at the same potential as its non-inverting terminal, so $$e_{-1} = E_1, \qquad e_{-2} = E_2 .$$ These are the two ends of the gain-setting resistor $R_1$, which therefore has $E_1 - E_2$ across it.
Find the single loop current. By assumption [ii] no current enters either op-amp input, so the current in $R_1$ can come only from $aR_1$ and can leave only through $bR_1$ — one current flows through all three resistors in series: $$i = \frac{E_1 - E_2}{R_1}.$$ This single-current observation is what makes the two-op-amp front end tractable.
Superposition, case $E_2 = 0$. Then $i = E_1/R_1$, and walking from the node at $E_1$ back through $aR_1$ to the output $x$, and from the node at $0$ forward through $bR_1$ to the output $y$: $$e_x\big|_{E_1} = E_1 + i\,(aR_1) = E_1(1 + a), \qquad e_y\big|_{E_1} = 0 - i\,(bR_1) = -bE_1 .$$ The contribution to the output is $$E_0\big|_{E_1} = c\left(-bE_1 - E_1(1+a)\right) = -c\,(1 + a + b)\,E_1 .$$
Superposition, case $E_1 = 0$. Now $i = -E_2/R_1$, and by the same two walks $$e_x\big|_{E_2} = 0 + i\,(aR_1) = -aE_2, \qquad e_y\big|_{E_2} = E_2 - i\,(bR_1) = E_2(1 + b),$$ so $$E_0\big|_{E_2} = c\left(E_2(1+b) + aE_2\right) = +c\,(1 + a + b)\,E_2 .$$
Add the two contributions. The circuit is linear, so the responses superpose: $$\boxed{\;E_0 = c\,(1 + a + b)\,(E_2 - E_1) = -c\,(1 + a + b)\,(E_1 - E_2)\;}$$
Cross-check with the general node potentials. Solving once with both inputs present gives $$e_x = E_1 + a\,(E_1 - E_2), \qquad e_y = E_2 - b\,(E_1 - E_2),$$ and substituting into $E_0 = c(e_y - e_x)$ reproduces the boxed result directly — confirming the superposition bookkeeping. Setting $E_1 = E_2$ gives $e_x = E_1$, $e_y = E_1$ and hence $E_0 = 0$: the circuit rejects a common-mode input exactly, which is the defining property of an instrumentation amplifier.

Quantity	Result
Loop current through $aR_1$, $R_1$, $bR_1$	$i = (E_1 - E_2)/R_1$
Potential at $x$	$e_x = E_1 + a(E_1 - E_2)$
Potential at $y$	$e_y = E_2 - b(E_1 - E_2)$
Response to $E_1$ alone	$-c(1+a+b)E_1$
Response to $E_2$ alone	$+c(1+a+b)E_2$
Overall transfer function	$\mathbf{E_0 = c(1+a+b)(E_2 - E_1)}$
Differential gain	$A_d = c(1+a+b)$
Common-mode gain	0 (ideal, matched resistors)

Topic. The three-op-amp instrumentation amplifier and superposition in ideal op-amp circuits.

Key relations. Ideal op-amp with negative feedback: $v_+ = v_-$ (virtual short) and $i_+ = i_- = 0$. Difference amplifier $E_0 = c(e_y - e_x)$. Front-end gain $e_y - e_x = -(1+a+b)(E_1 - E_2)$, so $A_d = c(1+a+b)$.

Why this works. Because no current can enter the op-amp inputs, the three resistors $aR_1$, $R_1$ and $bR_1$ form a single unbranched path, and the entire differential input voltage appears across $R_1$ alone. The two outputs therefore swing in opposite directions about their respective inputs, amplifying the difference by $(1+a+b)$ while leaving any common-mode component untouched — a common-mode shift puts zero volts across $R_1$, so no current flows and both outputs simply follow the input. The $U_3$ stage then subtracts, multiplying the difference by $c$ and removing the common-mode component that survives. In a real instrumentation amplifier the differential gain is set by the single resistor $R_1$, which is why the part is sold with one external gain resistor, and the common-mode rejection depends on matching within the $U_3$ network, not on $R_1$ at all.

Common pitfall. Treating $U_1$ and $U_2$ as independent non-inverting amplifiers with gains $(1+a)$ and $(1+b)$ to ground — that is only valid if the far end of $R_1$ is grounded, which it is not. The other frequent slip is a sign error in the walk from the inverting node through the feedback resistor: the current direction reverses between the $aR_1$ and $bR_1$ branches, which is exactly what produces the $(E_2 - E_1)$ ordering in the final answer.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., §2.4 (difference amplifiers) and §2.4.3 (the instrumentation amplifier).

Question 5 of 8: Transfer Function of a Three-Op-Amp Instrumentation Amplifier

Question 5: Transfer Function of a Three-Op-Amp Instrumentation Amplifier (equal value)