22-Elec-A5 Electronics · May 2017

Question 3 of 5: Current Gain of a T-Network Feedback Stage

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2017 · 16-Elec-A5, Electronics. Closed-book, 3 hours; answer all FIVE questions (20 marks each). Op-amps ideal, supplies ±15 V (Note 7).

Reference texts (22-Elec-A5 Electronics):

A. S. Sedra & K. C. Smith, Microelectronic Circuits, 7th/8th ed. — Ch. 4 (diodes, clippers & clampers), Ch. 5–7 (MOSFET/BJT biasing & small-signal amplifiers), Ch. 2 (op-amp circuits).
R. C. Jaeger & T. N. Blalock, Microelectronic Circuit Design, 5th ed. — single-stage amplifier design and DC bias.
Boylestad & Nashelsky, Electronic Devices and Circuit Theory — diode wave-shaping (clampers/DC restorers).

Check — drawing readings. Two readings of the printed schematics drive the answers: (Q1) the MOSFET source resistor $R_S$ has no bypass capacitor — the stage is a common-source amplifier with source degeneration. (Q4) the second transistor $Q_2$ is a PNP (emitter tied through $R_{E2}$ to $V_{CC}$, collector through $R_{C2}$ to ground), not an NPN.

Question 3: Current Gain of a T-Network Feedback Stage (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 ideal op-amp with its non-inverting input grounded; a signal current source $i_I$ drives the inverting node $v_I$; the feedback is a resistive T-network — $R_2$ from $v_I$ to a central node $X$, $R_1$ from the output $v_O$ to $X$, and $R_3$ from $X$ to ground. $i_O$ is the current flowing from $v_O$ into $R_1$.

Find. the current gain $i_O/i_I$.

[Figure not reproduced: circuit. See the official exam paper or the cited reference text.]

Figure 3.1 — Current input $i_I$ into the virtual-ground node; T-network $R_1,R_2,R_3$ feedback; output current $i_O$ defined into $R_1$.

Approach. Use the ideal-op-amp virtual ground ($v_I=0$, no input current) and write KCL at the inverting node and at the central node $X$.

Virtual ground fixes the input node. $v_{+}=0\Rightarrow v_I=0$, and the op-amp draws no input current, so all of $i_I$ flows through $R_2$ into $X$.
Voltage at the central node. The current $i_I$ (from $X$ toward the virtual ground) drops across $R_2$: $\ v_X=v_I-i_I R_2=-\,i_I R_2.$
KCL at $X$. Current in from $R_2$ plus current in from $R_1$ equals current out through $R_3$: $\ i_I+\dfrac{v_O-v_X}{R_1}=\dfrac{v_X}{R_3}.$ The output current into $R_1$ is $i_O=\dfrac{v_X-v_O}{R_1}$, so rearranging, $i_O=\dfrac{v_X}{R_3}-i_I=-\dfrac{i_I R_2}{R_3}-i_I.$
Result. With $i_O$ taken in the direction of its arrow (from the T-network into the $v_O$ node), $\ i_O=i_I\!\left(1+\dfrac{R_2}{R_3}\right)$, hence $\boxed{\dfrac{i_O}{i_I}=1+\dfrac{R_2}{R_3}=\dfrac{R_2+R_3}{R_3}}.$
Note on $R_1$. The current gain is independent of $R_1$: $R_1$ only sets the output voltage $v_O=-i_I\!\left(R_1+R_2+\tfrac{R_1R_2}{R_3}\right)$ (the output-compliance / transresistance), not the current delivered.

Question 3 — result
Current gain	Independent of
$i_O/i_I = 1+R_2/R_3 = (R_2+R_3)/R_3$	$R_1$

Topic. Current amplifier built from an op-amp with a resistive T feedback network.

Key relations. Virtual ground $v_I=0$; nodal KCL; $i_O/i_I=1+R_2/R_3$.

Why this works. The input current is forced entirely through $R_2$, developing $v_X=-i_IR_2$. That node voltage pushes an extra current $v_X/R_3$ through $R_3$; the op-amp supplies the sum through $R_1$. The T-network thus multiplies the input current by $1+R_2/R_3$ while the virtual ground keeps the input a true current node (zero input voltage).

Common pitfall. Expecting $R_1$ in the answer because the question says “as a function of $R_1,R_2,R_3$” — $R_1$ cancels from the current gain (it governs only the output voltage swing).

Source. Sedra & Smith, Microelectronic Circuits, Ch. 2 (op-amp with T feedback; current amplifiers).

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