Two-Op-Amp Circuit — Derive v O1 and v O2: 22-Elec-A5 Electronics, December 2015, Q3

Question 3: Two-Op-Amp Circuit — Derive v_O1 and v_O2 (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. Ideal op-amps A₁ and A₂. A₁ is an inverting stage (input through R₁, non-inverting input grounded, output v_O1). A₂’s non-inverting input is tied to v_O1, its inverting input to node M (the R₂–R₃ junction), and its output drives node T = v_O2 through R₃.

Find. Closed-form v_O1 and v_O2 in terms of R₁, R₂, R₃, v_I1; then a resistor set giving |v_O2/v_I1| = 20.

Figure 3.1 — A₁ is the inverting amplifier (node N = virtual ground). A₂ senses v_O1 on its + input and forces node M to equal it; R₃ is A₂’s feedback resistor, and its output node is v_O2.

Approach. Apply the two ideal-op-amp constraints (zero input current, equal input voltages) and write KCL at the two internal nodes N (A₁’s summing node) and M (A₂’s summing node).

A₁ sets a virtual ground. Its non-inverting input is grounded, so node N (its inverting input) is held at $v_N=0$.
A₂ equates its inputs. With negative feedback through R₃, $v_M=v_+=v_{O1}$ — the node M is forced to follow the output of A₁.
KCL at N gives v_O1. The currents from v_I1 (through R₁) and from M (through R₂) sum to zero at the virtual ground: $\frac{v_{I1}}{R_1}+\frac{v_M}{R_2}=0\ \Rightarrow\ v_M=-\frac{R_2}{R_1}v_{I1}.$ Since $v_{O1}=v_M$, $\boxed{v_{O1}=-\frac{R_2}{R_1}\,v_{I1}.}$
KCL at M gives v_O2. Node M (at potential v_O1) connects to N=0 through R₂ and to T=v_O2 through R₃; A₂ draws no input current: $\frac{0-v_{O1}}{R_2}+\frac{v_{O2}-v_{O1}}{R_3}=0\ \Rightarrow\ v_{O2}=v_{O1}\!\left(1+\frac{R_3}{R_2}\right).$ Substituting v_O1: $\boxed{v_{O2}=-\frac{R_2+R_3}{R_1}\,v_{I1}.}$
Part (b) — choose the resistors. The required magnitude is $\left|\dfrac{v_{O2}}{v_{I1}}\right|=\dfrac{R_2+R_3}{R_1}=20.$ A clean choice is $\boxed{R_1=1\ \text{k}\Omega,\quad R_2=10\ \text{k}\Omega,\quad R_3=10\ \text{k}\Omega\ \Rightarrow\ \frac{R_2+R_3}{R_1}=20.}$ (This set also gives $v_{O1}=-10\,v_{I1}$.)

Question 3 — results
Quantity	Expression / value
v_O1	$-\dfrac{R_2}{R_1}\,v_{I1}$
v_O2	$-\dfrac{R_2+R_3}{R_1}\,v_{I1}$
(b) resistor set	R₁=1 kΩ, R₂=R₃=10 kΩ → gain 20

Topic: an inverting amplifier whose feedback node is buffered/level-shifted by a second op-amp.

Key relations: ideal-op-amp rules ($v_+\!=\!v_-$, zero input current); nodal KCL; inverting gain $-R_f/R_\text{in}$.

Why this works: A₁ is a textbook inverting amplifier, but its feedback resistor R₂ does not return to its own output — it returns to node M, which A₂ actively forces to equal v_O1. That makes M a low-impedance copy of v_O1, so A₁ behaves exactly as if R₂ fed back to v_O1, giving the −R₂/R₁ gain. A₂ then works as a non-inverting-style stage in which R₂ and R₃ divide, producing a larger output v_O2 that scales the same input by (R₂+R₃)/R₁.

Common pitfall: mistaking which terminal of A₂ connects to v_O1. If the derivation shorts the two op-amp outputs it is wrong; the vertical wire ties v_O1 to A₂’s + input, and A₂’s output is the separate node v_O2.

Source: Sedra & Smith, Microelectronic Circuits 8th ed., Ch. 2 (op-amp circuits, nodal analysis).

Question 3 of 5: Two-Op-Amp Circuit — Derive v O1 and v O2

Question 3: Two-Op-Amp Circuit — Derive vO1 and vO2 (20 marks)

Question 3: Two-Op-Amp Circuit — Derive v_O1 and v_O2 (20 marks)