Shunt–Shunt Feedback Trans-Resistance Amplifier: 22-Elec-B5 Advanced Electronics, December 2016, Q3

Question 3: Shunt–Shunt Feedback Trans-Resistance Amplifier (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. A single common-emitter stage with its emitter ac-grounded, driven by a current source and enclosed by a resistive T network that samples the output voltage and returns a current to the input node — the shunt–shunt (voltage-sampling, current-mixing) topology, whose closed-loop gain is a trans-resistance.

Quantity	Symbol	Value
Feedback T network	\(R_1,\ R_2,\ R_3\)	10 kΩ each
Base bias resistor	\(R_B\)	5 kΩ
Collector resistor	\(R_C\)	5 kΩ
Emitter bias current	\(I_E\)	1 mA
Thermal voltage	\(V_T\)	25 mV
Device idealisations	\(\beta_1,\ r_o\)	both \(\infty\)

Find. The closed-loop trans-resistance \(v_{OUT}/i_S\), the resistance \(R_{if}\) presented to the source, and the resistance \(R_{of}\) at the output terminal.

Question 3. Common-emitter stage with the resistive T feedback network shaded. The emitter bypass capacitor makes the stage an undegenerated CE amplifier.

Approach. Convert the feedback T into its exact \(\Pi\) equivalent so that the topology becomes a textbook single feedback resistor, identify \(A\) and \(\beta_f\) with correct loading, and confirm the three answers against a direct nodal solution of the original network.

Part (a) — transconductance of the stage. With \(\beta_1=\infty\) the base draws no current, so the collector current equals the 1 mA supplied to the emitter: \[g_m=\frac{I_C}{V_T}=\frac{1\ \text{mA}}{25\ \text{mV}}=40\ \text{mA/V},\qquad r_\pi=\frac{\beta}{g_m}=\infty.\] The infinite \(r_\pi\) is what keeps the base node's only conductances external.
Replace the T by its exact \(\Pi\) equivalent. A three-terminal Y network transforms exactly, with \(\Sigma=R_1R_2+R_2R_3+R_3R_1=3\times 10^{8}\ \Omega^2\): \[R_f=\frac{\Sigma}{R_2}=30\ \text{k}\Omega,\qquad R_{b,g}=\frac{\Sigma}{R_3}=30\ \text{k}\Omega,\qquad R_{o,g}=\frac{\Sigma}{R_1}=30\ \text{k}\Omega.\] The network is therefore a 30 kΩ feedback resistor from output to base, plus a 30 kΩ shunt at each end. This is the step that turns an unfamiliar drawing into the standard shunt–shunt problem.
Load the basic amplifier and evaluate the open-loop trans-resistance. At the input node the shunts are \(R_B\), the 30 kΩ T shunt and (with the output shorted for loading purposes) \(R_f\); the output node sees the mirror image: \[R_i=R_B\parallel 30\parallel 30=5\parallel 15=3.75\ \text{k}\Omega,\qquad R_o=R_C\parallel 30\parallel 30=3.75\ \text{k}\Omega,\] \[A=\frac{v_o}{i_i}=-g_mR_iR_o=-40\ \text{mA/V}\times 3.75\ \text{k}\Omega\times 3.75\ \text{k}\Omega=-5.625\times 10^{5}\ \text{V/A}.\]
Feedback factor and loop gain. The feedback network returns a current \(i_f=-v_o/R_f\) to the input node, so \[\beta_f=\frac{i_f}{v_o}=-\frac{1}{R_f}=-33.33\ \mu\text{A/V},\qquad A\beta_f=\frac{5.625\times10^{5}}{30\ \text{k}\Omega}=18.75,\] \[1+A\beta_f=19.75.\] A loop gain of nearly 19 is what makes the closed-loop answers so much smaller and so much better behaved than the open-loop ones.
Closed-loop trans-resistance. \[A_f=\frac{A}{1+A\beta_f}=\frac{-5.625\times 10^{5}}{19.75}=-2.848\times 10^{4}\ \text{V/A},\] \[\boxed{\frac{v_{OUT}}{i_S}=-28.5\ \text{kV/A}\ \ (-28\,480\ \Omega)}\] An exact three-node solve of the original circuit (base, T centre, collector) returns \(-28\,480\ \text{V/A}\), so the \(\Pi\) transformation and the feedback bookkeeping agree to four figures. Note how close this is to \(-R_f=-30\ \text{k}\Omega\), the ideal-feedback limit: with \(A\beta_f\gg1\) the gain would be set by the feedback resistor alone.
Part (b) — input resistance. Shunt mixing divides the input resistance by the amount of feedback: \[R_{if}=\frac{R_i}{1+A\beta_f}=\frac{3.75\ \text{k}\Omega}{19.75}=\boxed{190\ \Omega}\] The exact nodal solve gives 190.0 Ω. A low input resistance is exactly what a current-driven amplifier wants: it makes the source behave as an ideal current source.
Part (c) — output resistance. Voltage sampling divides the output resistance by the same factor: \[R_{of}=\frac{R_o}{1+A\beta_f}=\frac{3.75\ \text{k}\Omega}{19.75}=\boxed{190\ \Omega}\] The two answers coincide here only because the loading happens to make \(R_i=R_o=3.75\ \text{k}\Omega\); it is a numerical coincidence of \(R_B=R_C\) and the symmetric T, not a general property. The exact nodal solve with \(i_S\) open again returns 190.0 Ω.

Result	Symbol	Value
Transconductance	\(g_m\)	40 mA/V
Equivalent feedback resistor (T \(\to\) \(\Pi\))	\(R_f\)	30 kΩ
Open-loop trans-resistance (loaded)	\(A\)	\(-5.625\times10^{5}\) V/A
Loop gain	\(A\beta_f\)	18.75 (\(1+A\beta_f=19.75\))
Closed-loop trans-resistance	\(v_{OUT}/i_S\)	\(-28.5\) kV/A
Input resistance	\(R_{if}\)	190 Ω
Output resistance	\(R_{of}\)	190 Ω

Topic. Shunt–shunt (voltage-sampling, current-mixing) feedback around a common-emitter stage, with a T network standing in for the single feedback resistor.

Key relations. \(A_f=A/(1+A\beta_f)\); shunt mixing gives \(R_{if}=R_i/(1+A\beta_f)\); voltage sampling gives \(R_{of}=R_o/(1+A\beta_f)\); the exact Y–\(\Delta\) transform \(R_f=\Sigma/R_2\) with \(\Sigma=R_1R_2+R_2R_3+R_3R_1\); ideal-feedback limit \(A_f\to 1/\beta_f=-R_f\).

Why this works. The T network exists because a single feedback resistor large enough to give a useful \(1/\beta_f\) would also be too large to fabricate or would load the nodes badly; three moderate resistors synthesise the same three-terminal behaviour. Because the transform is exact, the substitution costs nothing in accuracy, and the circuit reduces to the standard case in which one resistor both samples the output voltage and injects a current at the summing node. Feedback then acts in the same direction at both ports: it lowers the input resistance (good for a current input) and lowers the output resistance (good for driving a load), each by the same factor \(1+A\beta_f\).

Common pitfall. Treating the T as three separate resistors in the loop calculation, or forgetting the loading terms and using \(A=-g_mR_BR_C\), gives an open-loop gain 78 % too large and a loop gain of 33 instead of 18.75; the closed-loop answers then come out near \(-29.1\ \text{k}\Omega\) and the input resistance near 147 Ω. The correct procedure loads each port of the basic amplifier with the feedback network before opening the loop.

Source. A. S. Sedra & K. C. Smith, Microelectronic Circuits, 7th/8th ed., Ch. 11 §11.5 (the shunt–shunt topology and its loading rules); P. R. Gray, P. J. Hurst, S. H. Lewis & R. G. Meyer, Analysis and Design of Analog Integrated Circuits, 5th ed., Ch. 8.

Question 3 of 5: Shunt–Shunt Feedback Trans-Resistance Amplifier

Question 3: Shunt–Shunt Feedback Trans-Resistance Amplifier (20 marks)