Question 4 of 5: Tuned amplifier with a gate tank — centre frequency, resonant gain and bandwidth

Question 4: Tuned amplifier with a gate tank — centre frequency, resonant 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 input tank, the overall source-to-output gain at that frequency, and the −3 dB bandwidth of the resulting band-pass response.

Approach. Recover g_m from the bias, collect every capacitance that appears in parallel with C₁ at the gate node (including the Miller image of C_gd), then treat the gate as a parallel RLC tank damped only by R_S and read off ω_o, the resonant gain and the bandwidth.

1. Recover the operating point. M₁ carries I_bias in saturation, so from the square law $I_D=\tfrac12 K(V_{GS}-V_{TH})^2$ the overdrive is $$V_{OV}=\sqrt{\frac{2I_{bias}}{K}}=\sqrt{\frac{2(2\ \text{mA})}{1\ \text{mA/V}^2}}=2.00\ \text{V}, \qquad g_m = K V_{OV} = 2.00\ \text{mA/V},$$ with $V_{GS}=V_{TH}+V_{OV}=3.00$ V. Since $\lambda = 0$, $r_o$ is infinite and the drain load is R_L alone.
2. Read the topology off the drawing. C₁ and L₁ each run from the gate node to their own ground symbol, i.e. they form a parallel tank at the input, and the drain carries only R_L up to V_DD. The damping resistance of the tank is therefore R_S, not R_L, and the drain node is a plain resistive load.
3. Part (a) — collect the total tank capacitance. Both device capacitances sit at the gate node in parallel with C₁: C_gs directly, and C_gd through Miller's theorem multiplied by the stage gain, $$C_M = C_{gd}\bigl(1+g_mR_L\bigr) = 1\ \text{pF}\bigl(1+(2)(2)\bigr)=5\ \text{pF},$$ so that $$C_{tot}=C_1+C_{gs}+C_M = 200+10+5 = 215\ \text{pF}.$$ That the paper supplies C_gs and C_gd at all is the signal that they are meant to be included.
4. Evaluate the centre frequency. The tank resonates when the inductive and capacitive susceptances cancel: $$\omega_o=\frac{1}{\sqrt{L_1C_{tot}}} =\frac{1}{\sqrt{(1\ \mu\text{H})(215\ \text{pF})}} =\boxed{6.82\times 10^{7}\ \text{rad/s}}$$ which is 10.85 MHz, answering part (a). Ignoring the device capacitances would give $1/\sqrt{L_1C_1} = 7.07\times 10^{7}$ rad/s (11.25 MHz), 3.7 % high.
5. Part (b) — the gate voltage at resonance. At ω_o the parallel combination of L₁ and C_tot is an open circuit (both are lossless and λ = 0), so no current flows in R_S and the whole source voltage appears at the gate: $v_{gs}=v_s$ at $\omega=\omega_o$.
6. Convert to the output. With the source terminal bypassed by C₂ the drain simply loads g_mv_gs with R_L, so $$\left.\frac{v_{OUT}}{v_s}\right|_{\omega_o}=-g_mR_L =-(2\ \text{mA/V})(2\ \text{k}\Omega)=\boxed{-4.00\ \text{V/V}}$$ — a magnitude of 4.00 (12.0 dB) at 180°, answering part (b). Note the question asks for v_OUT/v_S, referred to the source behind R_S; at resonance that happens to coincide with v_OUT/v_IN, but away from resonance the divider formed by R_S and the tank makes the two quite different.
7. Part (c) — damping and bandwidth. The only loss in the tank is R_S (the gate draws no real current, the inductor and capacitor are ideal, and λ = 0 leaves nothing at the drain to reflect back). For a parallel RLC the −3 dB bandwidth in rad/s equals the conductance divided by the capacitance: $$\text{BW}=\frac{1}{R_SC_{tot}} =\frac{1}{(1\ \text{k}\Omega)(215\ \text{pF})} =\boxed{4.65\times 10^{6}\ \text{rad/s}}$$ that is 740 kHz, answering part (c).
8. Confirm through the quality factor. Two independent routes to Q must agree: $Q=\omega_o/\text{BW}=6.82\times10^{7}/4.65\times10^{6}=14.66$, and $Q=R_S\sqrt{C_{tot}/L_1}=1000\sqrt{215\ \text{pF}/1\ \mu\text{H}}=14.66$. They do, so the centre frequency and the bandwidth are mutually consistent.
9. Check that the drain node is not a second bandwidth limit. The output-side Miller image is $C_{gd}(1+1/g_mR_L)=1.25$ pF against R_L, placing a pole at $1/(2\pi R_LC) = 63.7$ MHz — almost six times ω_o/2π, so the drain contributes nothing to the shape of the passband and the single-tank analysis stands.
10. Cross-check against an exact two-node solution. Solving the network exactly, with C_gd kept as a true bridging element rather than a Miller image, puts the peak at 6.82×10⁷ rad/s (identical) with a magnitude of 3.79 instead of 4.00, i.e. the Miller estimate is 5.5 % optimistic because the feed-forward through C_gd partly cancels the drain current.

Final Results