Question 3 of 5: Tuned amplifier with a gate tank — centre frequency, gain and bandwidth

Question 3: Tuned amplifier with a gate tank — 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 centre (resonant) frequency ω_o of the tuned stage, the voltage gain v_OUT/v_S at that frequency, and the 3 dB bandwidth.

Approach. Bias the transistor to get g_m, add the device capacitances (C_gd after Miller multiplication) to C₁ to form the total tank capacitance, then use the standard parallel-RLC results: resonance where the susceptances cancel, gain set by the drain load once the tank is open, and bandwidth 1/(RC) with R the resistance loading the tank.

1. Part (a) — Bias the transistor and get g_m. With $I_D=\tfrac12 K V_{OV}^2$ and I_D = I_bias = 2 mA, $$V_{OV}=\sqrt{\frac{2I_D}{K}}=\sqrt{\frac{2(2\ \text{mA})}{1\ \text{mA/V}^2}}=2.0\ \text{V}, \qquad V_{GS}=V_{TH}+V_{OV}=3.0\ \text{V}$$ $$g_m=KV_{OV}=\sqrt{2KI_D}=2.0\ \text{mA/V}$$ The drain sits at $V_D=V_{DD}-I_DR_L=10-4=6\ \text{V}$, well above $V_{OV}$, so M₁ is saturated as assumed.
2. Assemble the total tank capacitance. C_gs is already from gate to (AC-grounded) source, so it adds directly. C_gd bridges the gate and a drain that swings $-g_mR_L=-4$ times harder, so by Miller's theorem it appears at the gate as $$C_M=C_{gd}\left(1+g_mR_L\right)=1\ \text{pF}\,\bigl(1+(2\ \text{mA/V})(2\ \text{k}\Omega)\bigr)=5\ \text{pF}$$ $$C_{tot}=C_1+C_{gs}+C_M=200+10+5=215\ \text{pF}$$
3. Resonate the tank. At resonance the inductive and capacitive susceptances cancel: $$\boxed{\omega_o=\frac{1}{\sqrt{L_1C_{tot}}} =\frac{1}{\sqrt{(1\ \mu\text{H})(215\ \text{pF})}}=6.82\times10^{7}\ \text{rad/s}}$$ which is $f_o=\omega_o/2\pi=10.85\ \text{MHz}$.
4. Part (b) — Take the gain at resonance. At ω_o the parallel L–C combination presents an infinite impedance, so no signal current flows in R_S and the whole source voltage appears at the gate, $v_{gs}=v_S$. The drain then drives R_L with r_o = ∞: $$\boxed{\left.\frac{v_{OUT}}{v_S}\right|_{\omega_o}=-g_mR_L =-(2\ \text{mA/V})(2\ \text{k}\Omega)=-4.0\ \text{V/V}}$$ i.e. 12.0 dB with a 180° inversion. The drain node has its own pole at $1/[2\pi R_L C_{gd}(1+1/g_mR_L)]=63.7\ \text{MHz}$, six times above ω_o, so it costs only 1.4 % of the magnitude and is neglected.
5. Part (c) — Find the bandwidth from the tank's loading. Replace the source by its Norton equivalent, $i=v_S/R_S$ in parallel with R_S: the gate node is then a parallel RLC with R = R_S. For such a circuit the 3 dB bandwidth is independent of L and is fixed by the R–C product: $$\boxed{\text{BW}=\frac{1}{R_SC_{tot}}=\frac{1}{(1\ \text{k}\Omega)(215\ \text{pF})} =4.65\times10^{6}\ \text{rad/s}=740\ \text{kHz}}$$ The corresponding quality factor is $$Q=\frac{\omega_o}{\text{BW}}=\frac{6.82\times10^{7}}{4.65\times10^{6}}=14.7 =R_S\sqrt{\frac{C_{tot}}{L_1}}$$ the two expressions agreeing exactly, which is the cheap arithmetic check on this part.