22-Elec-B5 Advanced Electronics · December 2018

Question 5 of 5: Tuned Amplifier with a Gate Tank

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

Notes on this paper

Paper format. National Exams, December 2018 — 16-Elec-B5 Advanced Electronics. Three hours; CLOSED BOOK, an approved Casio or Sharp calculator permitted. Five questions, all to be answered, 20 marks each. Op-amps are ideal and supplies are ±15 V unless a question says otherwise; in schematics ground and chassis are common.

Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (the paper cites it by name for Question 3) — Ch. 8 differential and multistage amplifiers, Ch. 10 output stages and power amplifiers, Ch. 11 frequency response and compensation, Ch. 12 feedback, Ch. 17 tuned amplifiers. B. Razavi, Design of Analog CMOS Integrated Circuits, 2nd ed., Ch. 4 (differential amplifiers) and Ch. 10 (stability and frequency compensation). P. Gray, P. Hurst, S. Lewis and R. Meyer, Analysis and Design of Analog Integrated Circuits, 5th ed., Ch. 4 (current mirrors) and Ch. 9 (frequency response of feedback amplifiers).

Check: figure readings taken from the printed schematics. Three of the five figures need careful reading, and two of them contradict a plain reading of the accompanying prose. In Question 1 the input pair M₁/M₂ is a p-channel pair whose common source node is fed by M₅, with M₃/M₄ as n-channel current-source loads — M₆ and M₇ are the two diode-connected devices of the reference string, not signal-path loads. In Question 4 the input v_IN drives the source of M₁ and the gate is tied to the R₁/R₂ tap, so the forward path is a common-gate stage; were the input on the gate, part (a)'s R_IN would be infinite and its 8 marks unearnable. Both readings are stated where they are used.

Question 5: Tuned Amplifier with a Gate Tank (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 common-source stage whose gate carries a parallel $L_1\|C_1$ tank to ground, driven from $v_S$ through $R_S$, with a plain resistive drain load $R_L$ and the source bypassed to ground by $C_2 = \infty$.

Given data, Question 5
Quantity	Symbol	Value
Transconductance parameter	$K$	1 mA/V²
Threshold voltage	$V_{TH}$	1 V
Gate-source capacitance	$C_{gs}$	10 pF
Gate-drain capacitance	$C_{gd}$	1 pF
Channel-length modulation	$\lambda$	0
Tank inductance / capacitance	$L_1$ / $C_1$	1 µH / 200 pF
Source bypass capacitor	$C_2$	∞ (a.c. short)
Source and load resistance	$R_S$ / $R_L$	1 kΩ / 2 kΩ
Bias current and supply	$I_{bias}$ / $V_{DD}$	2 mA / 10 V

Find. The centre frequency $\omega_o$, the voltage gain $v_{OUT}/v_S$ at that frequency, and the 3 dB bandwidth.

Figure 5.1 — Tuned amplifier with the parallel L₁||C₁ tank at the gate. C₂ places the source at a.c. ground; the drain carries only the resistive load R_L.

Approach. Bias the device to get $g_m$, add every capacitance that appears in parallel with the tank (including the Miller-multiplied $C_{gd}$), then read the resonance, the resonant gain and the bandwidth off the resulting parallel $RLC$.

Part (a) — establish the operating point. Taking the paper's transconductance parameter in the usual square-law form $I_D = \tfrac{1}{2}K V_{ov}^{2}$ and differentiating, the standard result $g_m = \sqrt{2KI_{bias}}$ applies: $$\begin{aligned}g_m &= \sqrt{2(1\times10^{-3})(2\times10^{-3})} = 2.0\ \text{mA/V} \\ V_{ov} &= \sqrt{\frac{2I_{bias}}{K}} = 2.0\ \text{V}\end{aligned}$$ so $V_{GS} = V_{TH}+V_{ov} = 3.0$ V and the quiescent drain sits at $V_{DD} - I_{bias}R_L = 10 - 4 = 6.0$ V. The inductor grounds the gate at d.c., so the source rests at −3.0 V and $V_{DS} = 9.0\ \text{V} \gg V_{ov}$: firmly saturated.
Collect the total tank capacitance. $C_2$ shorts the source to ground for signals, so $C_{gs}$ hangs directly across the tank. $C_{gd}$ bridges gate to drain and is therefore Miller-multiplied by the stage gain: $$C_{\text{Miller}} = C_{gd}\left(1+g_mR_L\right) = 1\left(1+4\right) = 5\ \text{pF},$$ $$C_{\text{tot}} = C_1 + C_{gs} + C_{gd}(1+g_mR_L) = 200 + 10 + 5 = \boxed{215\ \text{pF}}$$ This is why the question bothers to supply $C_{gs}$ and $C_{gd}$ at all.
Compute the centre frequency. The tank resonates where its susceptance vanishes: $$\omega_o = \frac{1}{\sqrt{L_1 C_{\text{tot}}}} = \frac{1}{\sqrt{(1\times10^{-6})(215\times10^{-12})}} = \boxed{6.82\times10^{7}\ \text{rad/s}\;\;(10.85\ \text{MHz})}$$ Ignoring the device capacitances would have given $7.07\times10^{7}$ rad/s, an error of 3.6 percent — small here, but it is precisely the error the question is testing for.
Part (b) — see what the tank does to the drive. At resonance a lossless parallel $L\|C$ is an open circuit, so no current flows through $R_S$, no voltage is dropped across it, and the gate sees the full source: $$v_{IN}(\omega_o) = v_S .$$ Away from resonance the tank loads $R_S$ and the gate voltage falls, which is what makes the response band-pass.
Get the resonant gain. With $\lambda = 0$ the drain load is $R_L$ alone, so the common-source gain applies unmodified and, because the gate divider is unity at $\omega_o$, it is also the overall gain: $$\frac{v_{OUT}}{v_S}\bigg|_{\omega_o} = \frac{v_{OUT}}{v_{IN}}\cdot\frac{v_{IN}}{v_S} = \left(-g_mR_L\right)(1) = -\left(2\times10^{-3}\right)\left(2\times10^{3}\right) = \boxed{-4.0}$$ i.e. a magnitude of 4 with 180° of phase inversion.
Part (c) — identify what damps the tank. The only resistance across the tank is the source resistance $R_S$; $R_L$ sits in the drain and cannot damp a gate-side tank. For a parallel $RLC$ the 3 dB bandwidth is set by that shunt resistance and the total capacitance alone: $$\text{BW} = \frac{1}{R_S C_{\text{tot}}} = \frac{1}{(1\times10^{3})(215\times10^{-12})} = \boxed{4.65\times10^{6}\ \text{rad/s}\;\;(740\ \text{kHz})}$$
Cross-check through the quality factor. Two independent routes to $Q$ must agree: $$\begin{aligned}Q &= \frac{\omega_o}{\text{BW}} = \frac{6.82\times10^{7}}{4.65\times10^{6}} = 14.66 \\ Q &= R_S\sqrt{\frac{C_{\text{tot}}}{L_1}} = 1000\sqrt{\frac{215\ \text{pF}}{1\ \mu\text{H}}} = 14.66. \checkmark\end{aligned}$$ Agreement to four figures confirms that $\omega_o$, BW and $C_{\text{tot}}$ are mutually consistent.
Confirm against an exact solution. Solving the two-node network without the Miller approximation — keeping $C_{gd}$ as a genuine feedback branch — puts the peak at $6.820\times10^{7}$ rad/s with a magnitude of 3.79 and a 3 dB bandwidth of $4.87\times10^{6}$ rad/s. The closed forms are therefore 5 percent optimistic on gain and 4.6 percent low on bandwidth, because $Y_{gd}$ has a real part that quietly shunts $R_S$. That is well inside examination tolerance, but it is the reason the boxed answers should be quoted as the design values rather than as exact ones.

Figure 5.2 — Exact two-node response magnitude, computed with C_gd retained as a feedback branch. The peak is 3.79 against the closed-form 4.0, and the exact bandwidth is 4.87 × 10⁶ rad/s.

Final results, Question 5
Part	Quantity	Value
—	Transconductance $g_m$	2.0 mA/V
—	Overdrive / gate drive	2.0 V / 3.0 V
—	Total tank capacitance	215 pF
(a)	Centre frequency $\omega_o$	$6.82\times10^{7}$ rad/s (10.85 MHz)
(b)	Gain $v_{OUT}/v_S$ at $\omega_o$	−4.0 (exact solve: 3.79)
(c)	3 dB bandwidth	$4.65\times10^{6}$ rad/s (740 kHz)
—	Quality factor $Q$	14.66

Check: the bias source is understood to return to a negative supply. The schematic draws $I_{bias}$ from the source terminal to ground, but with the gate held at d.c. ground through $L_1$ the device cannot carry 2 mA unless its source sits at $-V_{GS} = -3.0$ V. The current source is therefore taken to return to a negative rail, as is conventional in this style of drawing; nothing in parts (a) to (c) depends on the choice, because $C_2$ places the source at a.c. ground either way.

Topic. A single-tuned amplifier whose resonant circuit sits at the input, and the role of device capacitance in setting both its centre frequency and its bandwidth.

Key relations. $g_m=\sqrt{2KI_{bias}}$; $C_{\text{tot}} = C_1+C_{gs}+C_{gd}(1+g_mR_L)$; $\omega_o = 1/\sqrt{L_1C_{\text{tot}}}$; resonant gain $-g_mR_L$; $\text{BW} = 1/(R_SC_{\text{tot}})$; $Q = \omega_o/\text{BW} = R_S\sqrt{C_{\text{tot}}/L_1}$.

Why this works. A parallel $RLC$ presents its full shunt resistance only at resonance and collapses either side of it, so placing the tank at the gate turns an otherwise flat common-source stage into a band-pass one without touching the drain. Every capacitance that appears between gate and a.c. ground adds to the tank, and $C_{gd}$ counts several times over because the drain swings against the gate — the Miller effect. Bandwidth is then fixed entirely by how hard the driving resistance damps that tank; a smaller $R_S$ would widen the response and lower the gain selectivity, which is the classic gain-bandwidth trade in tuned design.

Common pitfall. Assuming the tank is in the drain, which would make $R_L$ the damping resistance and give a bandwidth of $1/(R_LC_{\text{tot}})$, wrong by a factor of two here, and with $\lambda=0$ would make the resonant gain infinite. The second pitfall is dropping $C_{gs}$ and the Miller term from $C_{\text{tot}}$, which shifts $\omega_o$ by nearly 4 percent and the bandwidth by 7 percent.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., §17.1 to §17.3 (tuned amplifiers, single-tuned stages and the Miller effect in tuned circuits); Razavi, Fundamentals of Microelectronics, Ch. 11 (frequency response).

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