Op-Amp with T-Network Feedback — Offset and Gain: 22-Elec-A5 Electronics, December 2018, Q1

Question 1: Op-Amp with T-Network Feedback — Offset and Gain (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. An inverting op-amp whose feedback is a T-network ($R_2$ from the summing node to a central node X, $R_4$ from X to the output, and $R_3$ in series with the large capacitor $C_1$ from X to ground). The op-amp carries an input offset voltage $V_{os}=\pm 3\text{ mV}$.

Given data — Question 1
Quantity	Value	Quantity	Value
$R_1$ (input)	100 kΩ	$R_3$	1 kΩ
$R_2$ (T left arm)	100 kΩ	$C_1$	very large
$R_4$ (T right arm)	100 kΩ	$V_{os}$	$\pm 3$ mV

Find. (a) the DC voltage that the input offset produces at the output; (b) the mid-band small-signal voltage gain $v_o/v_i$.

[Figure not reproduced: Q1: inverting op-amp with a T-network ($R_2$–$R_4$ with the $R_3$–$C_1$ leg to ground) in the feedback path. See the official exam paper or the cited reference text.]

Q1: inverting op-amp with a T-network ($R_2$–$R_4$ with the $R_3$–$C_1$ leg to ground) in the feedback path.

Approach. The capacitor $C_1$ makes the feedback frequency-dependent: at DC it is an open circuit (killing the $R_3$ leg), while for the small signal it is a short (activating the T-network). Analyse the two conditions separately.

DC condition — $C_1$ is open. With $C_1$ open, no current can flow in the $R_3$ leg, so node X simply joins $R_2$ and $R_4$ in series. The DC feedback resistance from output to the inverting node is $$R_{f,\text{dc}}=R_2+R_4=200\text{ k}\Omega.$$
Refer the offset to the output. The offset voltage appears in series with the input; the DC “noise gain” is one plus the feedback-to-input resistance ratio: $$V_{o,\text{offset}}=V_{os}\left(1+\frac{R_{f,\text{dc}}}{R_1}\right)=\pm 3\text{ mV}\left(1+\frac{200}{100}\right)=\boxed{\pm 9\text{ mV}}.$$
Small-signal condition — $C_1$ is a short. Now $R_3$ connects node X to ground and the full T-network is active. Holding the inverting input at virtual ground, the input current $i=v_i/R_1$ flows through $R_2$, so $v_X=-i\,R_2$. Applying KCL at X gives the effective feedback transresistance.
Evaluate the gain. The inverting T-network gain is $$\frac{v_o}{v_i}=-\frac{1}{R_1}\left(R_2+R_4+\frac{R_2R_4}{R_3}\right)=-\frac{1}{100\text{k}}\left(100\text{k}+100\text{k}+\frac{(100\text{k})^2}{1\text{k}}\right)=\boxed{-102\text{ V/V}}.$$ The $R_2R_4/R_3$ term is what lets a T-network reach a large gain ($\lvert A\rvert\approx 102$) with only $100\text{ k}\Omega$ resistors instead of a single $10.2\text{ M}\Omega$ feedback resistor.

Results — Question 1
Quantity	Value
Output DC offset voltage	$\pm 9$ mV
Small-signal gain $v_o/v_i$	$-102$ V/V

Topic: Op-amp input offset referred to output, and the T-network (tee) feedback trick.

Key relations: output offset $=V_{os}(1+R_f/R_1)$; T-network inverting gain $v_o/v_i=-(R_2+R_4+R_2R_4/R_3)/R_1$.

Why this works: The input offset behaves like a small DC source in series with the non-inverting input, so it is amplified by the DC noise gain $1+R_f/R_1$. Because the large capacitor blocks DC, the DC feedback is only $R_2+R_4$, giving a modest noise gain of 3. For the signal, the capacitor shorts $R_3$ to ground; the small $R_3$ divides the voltage at node X so that a large equivalent feedback resistance is synthesised, boosting the gain without any physically huge resistor (which would add noise and pick up leakage).

Common pitfall: Using the AC feedback network for the DC-offset calculation (or vice-versa). The capacitor changes the topology between the two questions, and forgetting that inflates the offset by the full signal gain.

Source: Sedra & Smith, Microelectronic Circuits, 8th ed., Ch. 2 (inverting configuration, the T-network, and DC imperfections).

Question 1 of 5: Op-Amp with T-Network Feedback — Offset and Gain

Question 1: Op-Amp with T-Network Feedback — Offset and Gain (20 marks)