Question 2 of 5: Common-source amplifier — mid-band gain with and without the source bypass, and the resulting bandwidth

Question 2: Common-source amplifier — mid-band gain with and without the source bypass, and the resulting 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 mid-band gain with the source bypassed, the mid-band gain once the bypass is deleted, and the upper 3 dB frequency of that degenerated stage.

Approach. Replace M₁ by its small-signal model, note that the ideal I_bias source is an open circuit so the only drain load is R_L in parallel with r_o, obtain the two mid-band gains from nodal analysis, then get f_H from the sum of open-circuit time constants of C_gs, C_gd and C_L.

1. Establish what the signal sees at each terminal. I_bias is an ideal current source, so it presents an open circuit to the signal and contributes nothing to the drain load. C₁ is a short at signal frequencies and the gate draws no mid-band current, so no current flows in R_i and the whole source voltage reaches the gate: $v_I = v_i$ and $v_{gs} = v_i - v_s$.
2. Part (a) — gain with C₂ in place. With C₂ = ∞ the source terminal is at signal ground, so $v_{gs}=v_i$ and the drain drives $R_L\parallel r_o$: $$A_{v,a} = -g_m\,(R_L\parallel r_o) = -(2\ \text{mA/V})\left(\frac{20\times 20}{20+20}\ \text{k}\Omega\right) = \boxed{-20.0\ \text{V/V}}$$ i.e. a gain magnitude of 20 (26.0 dB), inverting.
3. Part (b) — put the degeneration resistor back in. With C₂ gone, R_s carries the drain current and r_o is no longer in a simple parallel path, so the two node equations must be solved together. Writing $i$ for the current leaving the drain terminal, the source node gives $v_s = iR_s$ and the drain node gives $v_{OUT} = -iR_L$, while the device relation $i = g_m(v_i-v_s) + (v_{OUT}-v_s)/r_o$ closes the system.
4. Solve for the degenerated gain. Eliminating $v_s$ and $i$ from the three relations of step 3 gives the exact expression $$A_{v,b} = \frac{v_{OUT}}{v_i} = \frac{-g_m r_o R_L}{r_o + R_L + R_s + g_m r_o R_s}.$$ Substituting the data, the denominator is $D = 20 + 20 + 1.5 + (2)(20)(1.5) = 101.5\ \text{k}\Omega$ and the numerator is $-(2)(20)(20) = -800\ \text{k}\Omega$, so $$A_{v,b} = \frac{-800}{101.5} = \boxed{-7.88\ \text{V/V}}$$ which answers part (b). Note how badly the familiar textbook shortcut $-g_mR_L/(1+g_mR_s) = -40/4 = -10$ over-predicts here: it assumes $r_o \gg R_L$, whereas r_o and R_L are equal in this circuit, so the shortcut is 27 % optimistic.
5. Part (c) — set up the open-circuit time-constant method. The three device capacitors are far too small to interact with C₁ and C₂, so the upper corner is estimated from $f_H \approx 1/(2\pi\sum_k R_k C_k)$, where each $R_k$ is the resistance seen by capacitor $k$ with the other two open and the source killed. With $v_i = 0$ the gate sees R_i to ground.
6. Resistance facing C_L at the drain. Looking into the drain, the transistor with its source degeneration appears as $r_o + R_s(1+g_m r_o)$ in parallel with R_L: $$R_{C_L} = R_L \parallel \bigl[r_o + R_s(1+g_mr_o)\bigr] = 20 \parallel \bigl[20 + 1.5(41)\bigr] = 20\parallel 81.5 = 16.06\ \text{k}\Omega.$$ The degeneration has raised the intrinsic drain resistance from 20 kΩ to 81.5 kΩ, but R_L still dominates the node.
7. Resistance facing C_gd (the Miller path). Because C_gd bridges the gate and the drain, its driving-point resistance is the bridging form $$R_{C_{gd}} = R_i\bigl(1+|A_{v,b}|\bigr) + R_{C_L} = 20(1+7.882) + 16.06 = 193.7\ \text{k}\Omega,$$ an order of magnitude larger than either terminal resistance on its own. This is the Miller multiplication that makes the smallest capacitor on the page the dominant one.
8. Resistance facing C_gs. Injecting a test current into the gate and out of the source and solving the same two node equations as step 3 gives $$R_{C_{gs}} = R_i - \frac{R_s\bigl(g_m r_o R_i - R_L - r_o\bigr)}{D},$$ with the same denominator $D = 101.5\ \text{k}\Omega$ as the gain. Substituting, $R_s(g_mr_oR_i - R_L - r_o) = 1.5(800-20-20) = 1140\ \text{k}\Omega^2$, so $R_{C_{gs}} = 20 - 1140/101.5 = 20 - 11.23 = 8.77\ \text{k}\Omega$ — appreciably less than R_i, because the source terminal partly follows the gate.
9. Sum the time constants and take the corner frequency. The three contributions are $\tau_{gs} = 8.77\ \text{k}\Omega \times 20\ \text{fF} = 0.175\ \text{ns}$, $\tau_{gd} = 193.7\ \text{k}\Omega \times 5\ \text{fF} = 0.968\ \text{ns}$ and $\tau_L = 16.06\ \text{k}\Omega \times 5\ \text{fF} = 0.080\ \text{ns}$, so $\sum\tau = 1.224\ \text{ns}$ and $$f_H \approx \frac{1}{2\pi\sum\tau} = \frac{1}{2\pi(1.224\ \text{ns})} = \boxed{130\ \text{MHz}}$$ which answers part (c). C_gd alone accounts for 79 % of the total even though it is one quarter the size of C_gs.
10. Cross-check against the exact transfer function. Solving the three-node network with all three capacitors present and bisecting for the −3 dB point returns 132.4 MHz, so the open-circuit estimate is 1.8 % low — the expected direction, since the method is a conservative bound whenever one pole dominates.

Final Results