22-Elec-B5 Advanced Electronics · May 2014

Question 3 of 5: Tuned Amplifier — Centre Frequency, Mid-Band Gain and 3 dB Bandwidth

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

Notes on this paper

Paper format: National Exams, May 2014 — 07-Elec-B5 Advanced Electronics. Three hours, CLOSED BOOK, any non-communicating calculator permitted. Answer all FIVE (5) questions; all questions are worth 20 marks each. Op-amps are ideal and supply voltages are ±15 V unless stated otherwise; in schematics ground and chassis are common. Candidates are urged to state any assumption made in interpreting a question.

Reference texts (22-Elec-B5 Advanced Electronics):

A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. — Ch. 7 (Transistor Amplifiers), Ch. 8 (Differential and Multistage Amplifiers), Ch. 10 (Frequency Response & Tuned Amplifiers), Ch. 11 (Feedback), Ch. 12 (Output Stages and Power Amplifiers).
B. Razavi, Design of Analog CMOS Integrated Circuits, 2nd ed. — Ch. 3 (Single-Stage Amplifiers), Ch. 4 (Differential Amplifiers), Ch. 9 (Cascodes and Frequency Response).
P. R. Gray, P. J. Hurst, S. H. Lewis & R. G. Meyer, Analysis and Design of Analog Integrated Circuits, 5th ed. — Ch. 3–4 (single-transistor and differential stages), Ch. 8 (Feedback).

Check — two engineering assumptions used throughout this paper.

Bias currents are taken from the printed current sources, not from the drawn resistors. Questions 1, 3 and 4 all show an ideal current source setting the branch current, so each transistor's quiescent current is read directly off that source. Where a drawn resistor would imply a different current (Question 4's 1 kΩ emitter resistor), the paper's stated 1 mA governs and the resistor is treated as part of an undrawn bias network — see the Question 4 note.
Channel-length modulation is used in the small-signal model but not in the bias calculation. The tail (or bias) source fixes $I_D$ exactly, so $V_{OV}=\sqrt{2I_D/K}$ and $g_m=KV_{OV}=\sqrt{2KI_D}$ are computed with $\lambda$ set aside, while $r_o=1/(\lambda I_D)$ is carried into every gain expression. This is the standard convention and the error it introduces is well under one percent.

Question 3: Tuned Amplifier — Centre Frequency, Mid-Band Gain and 3 dB Bandwidth (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.

Quantity	Symbol	Value
Bias current through $M_1$	$I_{bias}$	2 mA
Transconductance parameter / threshold	$K$, $V_{TH}$	1 mA/V², 1 V
Channel-length modulation	$\lambda$	0 (so $r_o \to \infty$)
Device capacitances	$C_{gs}$, $C_{gd}$	10 pF, 1 pF
Tank inductance and capacitance	$L_1$, $C_1$	1 µH, 200 pF
Tank shunt resistance (drain load)	$R_L$	2 kΩ
Source resistance at the gate	$R_S$	1 kΩ
Source bypass capacitor	$C_2$	$\infty$ (source at a.c. ground)

Find. The resonant (centre) frequency of the drain tank, the complete source-to-output gain at that frequency, and the −3 dB bandwidth of the tuned response.

Check — tank topology. The May-2014 printing draws $R_L$ above the $C_1\|L_1$ pair, which read literally would place the resistor in series with a lossless parallel tank. That reading is not answerable as posed: at $\omega_o$ a lossless parallel tank is an open circuit, the drain load becomes infinite (recall $\lambda = 0$, so $r_o$ is infinite too), the gain is limited only by the $C_{gd}$ feedback path, and no −3 dB bandwidth in the sense of part (c) exists. This solution therefore takes the intended topology to be the parallel $R_L \| L_1 \| C_1$ tank, which is the standard tuned-amplifier arrangement, and notes the drafting discrepancy rather than solving an unanswerable circuit. An exam candidate should state this assumption explicitly, exactly as Note 1 on page 1 of the paper invites.

Figure 3.1 — Question 3 circuit as solved: $C_1$, $R_L$ and $L_1$ in parallel between the drain and $+V_{DD}$ (an a.c. ground), source bypassed to ground by $C_2$, gate driven through $R_S$.

Approach. Bias first to get $g_m$; the tank sets the centre frequency and, through its shunt resistance, both the mid-band gain and the bandwidth; the gate network $R_S$–$C_{in}$ then applies a separate frequency-independent-over-the-band attenuation that must be included because the question asks for $v_{OUT}/v_s$, not $v_{OUT}/v_{IN}$.

Part (a) — Bias the transistor and get $g_m$. The tail source carries the whole drain current, so $I_D = I_{bias} = 2$ mA and $$V_{OV}=\sqrt{\frac{2I_D}{K}} = \sqrt{\frac{2(2\ \text{mA})}{1\ \text{mA/V}^2}}=2.00\ \text{V}, \qquad g_m = KV_{OV} = 2.00\ \text{mA/V}.$$ With $\lambda = 0$ the device output resistance is infinite, so the tank alone loads the drain.
Locate the centre frequency. A parallel $R\|L\|C$ tank has equal and opposite inductive and capacitive susceptances at $$\omega_o = \frac{1}{\sqrt{L_1C_1}} = \frac{1}{\sqrt{(1\times10^{-6})(200\times10^{-12})}} = \frac{1}{1.414\times10^{-8}}$$ $$\boxed{\omega_o = 7.07\times10^{7}\ \text{rad/s} \quad\Longleftrightarrow\quad f_o = \frac{\omega_o}{2\pi} = 11.25\ \text{MHz}}$$ The Miller image of $C_{gd}$ at the drain, $C_{gd}(1+1/|A|)=1.25$ pF, adds to the 200 pF tank capacitor and pulls $\omega_o$ down by only 0.3 percent (to $7.05\times10^{7}$ rad/s), which is inside the tolerance of any real 1 µH inductor and is neglected in the quoted answer.
Part (b) — Evaluate the gain from the gate to the drain at resonance. At $\omega_o$ the tank's reactive branches cancel and the drain sees only the shunt resistance, so the stage is a plain common-source amplifier: $$\left.\frac{v_{OUT}}{v_{IN}}\right|_{\omega_o} = -g_mR_L = -(2\ \text{mA/V})(2\ \text{k}\Omega) = -4.00\ \text{V/V}.$$
Account for the source resistance at the gate. The question asks for $v_{OUT}/v_s$, and $R_S$ works against the transistor's input capacitance. Applying Miller's theorem to $C_{gd}$ with the gain just computed, $$C_{in} = C_{gs} + C_{gd}\left(1+g_mR_L\right) = 10 + 1(1+4) = 15\ \text{pF},$$ so the gate network is a single-pole low-pass with $\omega_p = 1/(R_SC_{in}) = 6.67\times10^{7}$ rad/s — deliberately placed by the examiner right next to $\omega_o$, so it cannot be dismissed: $$\frac{v_{IN}}{v_s}\bigg|_{\omega_o} = \frac{1}{1+j\omega_oR_SC_{in}} = \frac{1}{1+j1.061} = 0.686\,\angle\,{-46.7^\circ}.$$
Combine the two factors. Multiplying the drain gain by the input attenuation, $$\boxed{\left.\frac{v_{OUT}}{v_s}\right|_{\omega_o} = (-4.00)\left(0.686\,\angle\,{-46.7^\circ}\right) = 2.74\,\angle\,133.3^\circ \;\;\Rightarrow\; \left|\frac{v_{OUT}}{v_s}\right| = 2.74\ \text{V/V}\ (8.8\ \text{dB})}$$ A full two-node solve of the same circuit, with $C_{gd}$ kept as a true bridging element rather than replaced by its Miller image, returns $2.67\,\angle\,124.8^\circ$ — within 3 percent of the hand result, which confirms that the Miller step is a fair approximation here. The lesson is that $R_S$ costs this amplifier nearly a third of its gain; leaving it out and answering $-4$ V/V would forfeit most of the eight marks.
Part (c) — Set up the bandwidth of the tuned response. Near resonance, a parallel $R\|L\|C$ tank behaves as a second-order band-pass whose −3 dB width is set entirely by the shunt resistance and the capacitance — the inductance affects only where the band sits, not how wide it is: $$\text{BW} = \frac{\omega_o}{Q} = \frac{1}{R_LC_1}, \qquad Q = \omega_oR_LC_1 = R_L\sqrt{\frac{C_1}{L_1}}.$$
Evaluate the bandwidth and quality factor. Substituting the tank values, $$\text{BW} = \frac{1}{(2\ \text{k}\Omega)(200\ \text{pF})} = 2.50\times10^{6}\ \text{rad/s}$$ $$\boxed{\text{BW} = 2.50\times10^{6}\ \text{rad/s} = 398\ \text{kHz}, \qquad Q = \frac{7.07\times10^{7}}{2.50\times10^{6}} = 28.3}$$ so the passband runs roughly from 11.05 MHz to 11.45 MHz.
Sanity-check the neglected effects. Two of them push in opposite directions and both are small. The Miller capacitance widens $C$ by 1.25 pF, narrowing the bandwidth by 0.6 percent. The gate pole at $6.67\times10^{7}$ rad/s is only 6 percent below $\omega_o$, but across a passband that is just $\pm 1.8$ percent wide its attenuation barely changes, so it shifts the level of the response rather than its shape. Solving the complete two-node network numerically gives a peak of 2.77 at $7.03\times10^{7}$ rad/s and a true −3 dB width of $2.83\times10^{6}$ rad/s (450 kHz) — about 13 percent wider than the tank-only figure, the difference being the slight asymmetry the gate pole introduces. The tank formula is the expected exam answer and is quoted as such; the numerical width is reported here as the audit.

Part	Quantity	Result
—	Bias point: $V_{OV}$, $g_m$	2.00 V, 2.00 mA/V
(a)	Centre frequency $\omega_o$	$7.07\times10^{7}$ rad/s (11.25 MHz)
(b)	Gain at the drain, $v_{OUT}/v_{IN}$	$-g_mR_L = -4.00$ V/V
(b)	Input capacitance / gate attenuation	$C_{in} = 15$ pF; $0.686\,\angle\,{-46.7}$°
(b)	Overall gain $v_{OUT}/v_s$ at $\omega_o$	2.74 ∠133.3° (8.8 dB); exact solve 2.67
(c)	3 dB bandwidth	$2.50\times10^{6}$ rad/s = 398 kHz (exact solve 450 kHz)
(c)	Quality factor $Q$	28.3

Topic: Single-tuned common-source amplifier — resonance, mid-band gain, and the $R$–$C$ trade between gain and bandwidth.

Key relations: $\omega_o = 1/\sqrt{L_1C_1}$; at resonance the tank reduces to $R_L$ so $A_{v,\text{drain}} = -g_mR_L$; $Q = \omega_oR_LC_1 = R_L\sqrt{C_1/L_1}$; $\text{BW} = \omega_o/Q = 1/(R_LC_1)$; Miller input capacitance $C_{in} = C_{gs}+C_{gd}(1+g_mR_L)$; gate pole $\omega_p = 1/(R_SC_{in})$.

Why this works: Replacing a resistive drain load with a parallel tank does not change the amplifier — it changes only what the transistor's current sees. At d.c. the inductor shorts the load, at very high frequency the capacitor shorts it, and only in a narrow band around $\omega_o$ do the two susceptances cancel and leave the full $R_L$ in place. That is why the resonant gain is exactly the resistive-load gain and why the gain-bandwidth product is fixed: $|A|\cdot\text{BW} = g_mR_L \cdot 1/(R_LC_1) = g_m/C_1$, independent of $R_L$. Raising $R_L$ buys gain and gives it back as selectivity, one for one. The other half of the question is that the answer is asked for from $v_s$, not from the gate: at 11 MHz an 11 pF device driven from 1 kΩ is a real low-pass, and Miller multiplication of the 1 pF $C_{gd}$ by $(1+g_mR_L)$ is what makes 11 pF into 15 pF.

Common pitfall: Answering $-g_mR_L$ for part (b) and stopping — that is $v_{OUT}/v_{IN}$, not $v_{OUT}/v_s$, and it ignores $R_S$, $C_{gs}$ and $C_{gd}$, three of the given quantities. Every datum in a well-posed exam question does work somewhere, and the ones that appear unused are usually the marker's target. A second pitfall is computing $Q$ from the inductor ($Q = \omega_oL/R$, the series-circuit formula) instead of the shunt form $Q=\omega_oRC$: for a parallel tank they are reciprocal, and the wrong one here gives $Q = 0.035$ rather than 28.3.

Source: Sedra & Smith, Microelectronic Circuits, 8th ed., Ch. 10 (Frequency Response — Miller's theorem, the high-frequency MOSFET model) and the Tuned Amplifiers material (single-tuned stage, $Q$ and bandwidth); Razavi, Design of Analog CMOS Integrated Circuits, 2nd ed., Ch. 9.

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