Question 5 of 6: Tuned amplifier with a series-resonant trap

Question 5: Tuned amplifier with a series-resonant trap (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 source is held at signal ground by $C_{2}=\infty$ and biased by a 2 mA current source to $-V_{SS}$. Its drain load is a series combination of $L_{1}$ and $C_{1}$ placed in parallel with the drain resistor; the output is taken at the drain and the gate is driven from $v_{S}$ through the source resistance $R_{S}$.

Find. The centre frequency $\omega_{o}$ set by the reactive branch, the voltage gain $v_{OUT}/v_{S}$ at that frequency, and the asymptotic gains at very high and very low frequency.

[Figure not reproduced: Question 5 — common-source stage whose drain load is a series L 1 – C 1 branch in parallel with R L . Drawn as printed: the inductor and capacitor are in series in one branch, which makes the network a trap rather than a bandpass tank. See the official exam paper.]

Approach. Convert the bias current to $g_{m}$, write the drain load as $R_{L}$ in parallel with the series branch $j\omega L_{1}+1/(j\omega C_{1})$, and evaluate $A_{v}=-g_{m}Z$ at the three frequencies the question asks about, bringing in the device capacitances only where they matter.

1. Bias the transistor and find its transconductance. The current source sets $I_{D}=I_{bias}=2\ \text{mA}$, so from $I_{D}=\tfrac{1}{2}K\left(V_{GS}-V_{TH}\right)^{2}$, $$V_{ov}=\sqrt{\frac{2I_{D}}{K}}=\sqrt{\frac{2(2\ \text{mA})}{1\ \text{mA/V}^{2}}} =2\ \text{V},\qquad V_{GS}=3\ \text{V},$$ $$\boxed{g_{m}=KV_{ov}=(1\ \text{mA/V}^{2})(2\ \text{V})=2\ \text{mA/V}}$$ With $\lambda=0$ the output resistance is infinite, so the drain behaves as an ideal current source of strength $g_{m}v_{gs}$ and the whole frequency dependence lives in the load network. Since the MOS gate draws no current, $R_{S}$ does not attenuate the drive at low or mid frequencies; it matters only in part (c).
2. Part (a) — locate the resonance of the reactive branch. As printed, $L_{1}$ and $C_{1}$ sit in series in one branch, so their combined reactance vanishes when $\omega L_{1}=1/(\omega C_{1})$: $$\boxed{\omega_{o}=\frac{1}{\sqrt{L_{1}C_{1}}} =\frac{1}{\sqrt{(1\ \mu\text{H})(200\ \text{pF})}}=7.071\times10^{7}\ \text{rad/s} \;\Longrightarrow\; f_{o}=11.25\ \text{MHz}}$$ The characteristic impedance of the branch is $\sqrt{L_{1}/C_{1}}=70.7\ \Omega$, well below $R_{L}=2\ \text{k}\Omega$, which tells us in advance that the branch will dominate the load over a wide band around $\omega_{o}$.
3. Part (b) — evaluate the load impedance at resonance. The total drain load is $$Z\left(j\omega\right)=R_{L}\parallel\left(j\omega L_{1}+\frac{1}{j\omega C_{1}}\right) =R_{L}\parallel jX,\qquad X=\omega L_{1}-\frac{1}{\omega C_{1}} .$$ At $\omega=\omega_{o}$ the reactance $X$ is zero, so the series branch is a short circuit that ties the drain straight to $V_{DD}$ — which is signal ground. Therefore $Z(j\omega_{o})=0$ and $$\boxed{\left.\frac{v_{OUT}}{v_{S}}\right|_{\omega=\omega_{o}}=-g_{m}Z(j\omega_{o})=0}$$ This circuit is a trap: a band-reject, not a band-pass, amplifier, and $\omega_{o}$ is the centre of its notch. The rejection is complete only for an ideal inductor; any winding resistance $r$ leaves $Z=R_{L}\parallel r\approx r$ and a residual gain of $g_{m}r$, so a 2 $\Omega$ coil resistance would give a notch depth of about 52 dB rather than infinity.
4. Establish how wide the notch is. The response is within 3 dB of its plateau when $|X|=R_{L}$, i.e. when $\omega L_{1}-1/(\omega C_{1})=\pm R_{L}$. Solving the resulting quadratic gives skirts at $$f_{lo}=397\ \text{kHz}\quad\text{and}\quad f_{hi}=319\ \text{MHz},$$ whose product satisfies $\omega_{lo}\omega_{hi}=\omega_{o}^{2}$ as it must. The notch is therefore extremely broad — nearly three decades either side — because the branch impedance $70.7\ \Omega$ is so much smaller than the load it shunts.
5. Part (d) — take the low-frequency limit. As $\omega\to0$ the capacitor $C_{1}$ becomes an open circuit, the series branch disappears, and the drain sees $R_{L}$ alone. The device capacitances are open circuits as well, so $$\boxed{\left.\frac{v_{OUT}}{v_{S}}\right|_{\omega\rightarrow0}=-g_{m}R_{L} =-(2\ \text{mA/V})(2\ \text{k}\Omega)=-4\ \text{V/V}}$$ an inverting gain of 12.0 dB. There is no input coupling capacitor, so this is also the d.c. gain of the stage.
6. Part (c) — take the high-frequency limit, and be careful about which element wins. Two things happen as $\omega$ rises above the notch. The inductive reactance $\omega L_{1}$ grows without bound, so the series branch opens and the load returns towards $R_{L}$ — considered alone, the tank would restore the gain to $-g_{m}R_{L}=-4$. But $C_{gs}$ and $C_{gd}$ do not stay open. Miller-multiplying $C_{gd}$ against the recovered gain gives an input capacitance $$C_{in}=C_{gs}+C_{gd}\left(1+g_{m}R_{L}\right)=10+1(1+4)=15\ \text{pF},$$ which works against $R_{S}=1\ \text{k}\Omega$ to give a corner at $$f_{H}=\frac{1}{2\pi R_{S}C_{in}}=\frac{1}{2\pi(1\ \text{k}\Omega)(15\ \text{pF})} =10.6\ \text{MHz},$$ essentially coincident with the notch. Above it, $C_{gs}$ shorts the gate to the source — which is signal ground — and $C_{gd}$ shorts the drain to the gate, so no signal survives: $$\boxed{\left.\frac{v_{OUT}}{v_{S}}\right|_{\omega\rightarrow\infty}=0}$$ The response plot below shows both curves: the tank-only load recovers to 12 dB above the notch, while the complete model including $C_{gs}$ and $C_{gd}$ peaks near $-10\ \text{dB}$ and then falls away for good.

Read as a whole, the stage is a 12 dB amplifier with a deep notch at 11.25 MHz — the classic arrangement for suppressing a single interfering carrier or an image frequency ahead of a mixer, at the cost of the rest of the band being flat rather than selective.