23-Mechatronics-A2 Circuits and Electronics · December 2018
Question 7 of 11: Op-amp T-feedback network — output offset and small-signal gain
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams, December 2018 — 16-Mex-A2 Circuits and Electronics. Closed-book, 3-hour paper (approved Casio/Sharp calculator only). Two parts: Part A — Circuits (Q1–Q6) and Part B — Electronics (Q7–Q11); candidates normally answer 5 of the 11 questions (3+2 or 2+3 split). Full worked solutions to all eleven questions are given below so the set can be used for study regardless of which five a candidate chose.
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed.) — resistive networks, superposition, Thévenin/max-power transfer, first-order and second-order transients, AC steady-state nodal analysis; W. H. Hayt et al., Engineering Circuit Analysis (9th ed.) — Laplace-domain circuit models; A. S. Sedra & K. C. Smith, Microelectronic Circuits (8th ed.) — op-amp T-network feedback, diode rectifiers, MOSFET common-gate stages, NMOS inverters, and BJT bias-point analysis.
Reading the figures. The paper supplies the schematics but no separate figure data; every network below is redrawn element-by-element directly from the original drawing (several figures are small hand-style schematics). All node/reference choices and source polarities are stated with the solution. Q11 uses printed variable names (e.g. “VE”) exactly as labelled on the original circuit, even where the labelled terminal is physically the collector in one sub-part — this is called out explicitly where it occurs.
Part B — Electronics
Question 7: Op-amp T-feedback network — output offset and small-signal gain [8 + 12]
Find. The output dc offset voltage, and the small-signal gain $v_O/v_I$.
Figure-7: inverting op-amp with a T-feedback network. $R_2$ and $R_4$ run in series from $v^-$ to $v_O$; their midpoint feeds $R_3$ in series with a large capacitor $C_1$ to ground. Non-inverting input grounded.
Approach. This is a classic T network in the feedback path: at true dc the bypass capacitor $C_1$ blocks current through $R_3$, so the dc feedback path is simply $R_2+R_4$ in series; at signal frequencies $C_1$ acts as a short, grounding the T-junction through $R_3$ and turning the T into an equivalent large feedback resistor $R_2+R_4+R_2R_4/R_3$. This lets a modest gain-setting resistor pair deliver a very large ac gain while keeping the dc offset gain modest.
DC path (offset). With $C_1$ open at dc, no current flows into the $R_3$ branch, so $I_{R_2}=I_{R_4}$ and the T reduces to a plain series resistor $R_2+R_4=200\ \text{k}\Omega$ from $v^-$ to $v_O$. Modelling the offset as $v^-=V_{OS}$ (non-inverting input grounded, ideal op-amp forces $v^-=v^++V_{OS}$) with $v_I=0$ for this dc-only calculation: $$v_{O,dc}=V_{OS}\Big(1+\frac{R_2+R_4}{R_1}\Big)=3\,\text{mV}\times\Big(1+\frac{200\,\text{k}}{100\,\text{k}}\Big)=\boxed{9\ \text{mV}}.$$
AC path (T-network equivalent resistance). At signal frequencies $C_1$ is a short, so the T-junction is grounded through $R_3$ only. The standard T-network result (grounding a resistor at the midpoint of a feedback path) gives an equivalent single feedback resistor $$R_{F}=R_2+R_4+\frac{R_2R_4}{R_3}=100\text{k}+100\text{k}+\frac{100\text{k}\times100\text{k}}{1\text{k}}=200\,\text{k}+10{,}000\,\text{k}=\boxed{10.2\ \text{M}\Omega}.$$
Small-signal gain. The stage is an ordinary inverting amplifier with feedback resistance $R_F$: $$\frac{v_O}{v_I}=-\frac{R_F}{R_1}=-\frac{10.2\,\text{M}\Omega}{100\,\text{k}\Omega}=\boxed{-102}.$$
Interpretation. The T-network delivers a large-magnitude ac gain ($-102$) from practical 100 kΩ/1 kΩ resistors — avoiding an unwieldy 10.2 MΩ feedback resistor — while the offset is amplified only by a factor of 3, far less than the ac gain magnitude would otherwise suggest.
