Question 2 of 5: Tuned Amplifier — Centre Frequency, Gain and Bandwidth

Question 2: Tuned Amplifier — Centre Frequency, Gain and 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.

Find. The resonant centre frequency of the drain tank, the small-signal voltage gain from the source $v_s$ to the output at that frequency, and the 3 dB bandwidth of the resulting band-pass response.

Approach. Convert the bias current into a transconductance through the saturation law, recognise the drain load as a parallel RLC whose impedance is purely $R_1$ at resonance, and read the centre frequency, mid-band gain and bandwidth off the three standard parallel-tank results.

1. Turn the bias current into a small-signal transconductance. The tail current source sets the drain current, so with $\lambda=0$ the saturation law gives the overdrive directly: $$I_D=\tfrac12 K\,(V_{GS}-V_{TH})^2=\tfrac12 K\,V_{ov}^2 \;\Rightarrow\;V_{ov}=\sqrt{\frac{2I_D}{K}}=\sqrt{\frac{2(2\ \text{mA})}{1\ \text{mA/V}^2}}=2.00\ \text{V},$$ $$g_m=K\,V_{ov}=(1\ \text{mA/V}^2)(2.00\ \text{V})=2.00\ \text{mA/V}.$$ The device therefore sits at $V_{GS}=V_{TH}+V_{ov}=3.0$ V. Two equivalent forms, $g_m=2I_D/V_{ov}$ and $g_m=\sqrt{2KI_D}$, return the same 2 mA/V, which is a free check on the algebra.
2. Part (a) — find where the tank resonates. The inductive and capacitive susceptances of the parallel tank cancel when $\omega L_1=1/(\omega C_1)$, so $$\boxed{\omega_o=\frac{1}{\sqrt{L_1C_1}} =\frac{1}{\sqrt{(1\times10^{-6})(200\times10^{-12})}}=7.07\times10^{7}\ \text{rad/s}}$$ which is $f_o=\omega_o/2\pi=11.25$ MHz. Evaluating the full tank admittance at this frequency returns exactly $1/R_1$ with zero imaginary part, confirming that the load really is resistive there.
3. Part (b) — evaluate the gain at resonance. At $\omega_o$ the drain sees only $R_1$, because $\lambda=0$ removes $r_o$ and the tank reactances cancel. The gate is driven from the ideal source $v_s$ with no series resistance and the source terminal is grounded by $C_2$, so $v_{gs}=v_s$ and $C_{gs}$ carries current from the source without attenuating it. Hence $$\boxed{\frac{v_{OUT}}{v_s}\bigg|_{\omega_o}=-g_m R_1 =-(2.00\ \text{mA/V})(2\ \text{k}\Omega)=-4.00}$$ i.e. a magnitude of 4.00 (12.04 dB) with the 180 degree inversion of a common-source stage.
4. Confirm that the feed-forward through $C_{gd}$ is negligible. $C_{gd}$ injects a little of $v_s$ straight into the drain node. Writing the drain-node equation with that path kept, $$\frac{v_{OUT}}{v_s}=\frac{j\omega_o C_{gd}-g_m}{1/R_1+j\omega_o C_{gd}},$$ and with $\omega_o C_{gd}=7.07\times10^{-5}$ S this evaluates to $3.96\,\angle\,169.9^{\circ}$ — within 1 percent of the ideal $-4.00$. The simple answer is therefore the right one to quote, and the exact form only matters if a phase specification is at stake.
5. Part (c) — get the bandwidth from the tank's damping. For a parallel RLC the 3 dB bandwidth is fixed by the resistance and capacitance alone: $$\boxed{\text{BW}=\frac{1}{R_1C_1} =\frac{1}{(2\times10^{3})(200\times10^{-12})}=2.50\times10^{6}\ \text{rad/s}}$$ which is 398 kHz. The corresponding loaded quality factor is $$Q=\frac{\omega_o}{\text{BW}}=\frac{7.07\times10^{7}}{2.50\times10^{6}}=28.3,$$ and the independent form $Q=R_1\sqrt{C_1/L_1}=28.3$ agrees exactly.
6. Quantify the correction the transistor capacitances make. Seen from the drain, $C_{gd}$ adds $C_{gd}(1+1/|A_v|)=1.25$ pF in parallel with the tank, giving $C_{tot}=201.25$ pF. That shifts the centre frequency to $7.049\times10^{7}$ rad/s (0.3 percent low) and the bandwidth to $2.484\times10^{6}$ rad/s (0.6 percent low). Both are far inside the tolerance of any real 200 pF capacitor, so the uncorrected answers stand.

Final results.