Question 2 of 5: Common-gate MOSFET amplifier

Question 2: Common-gate MOSFET amplifier

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. An n-channel MOSFET whose source is the signal input (coupled through $C_1$), gate held at the fixed bias $V_G$ (a.c. ground), and output taken at the drain — a common-gate stage. Saturation model $i_{DS}=\tfrac{K}{2}(v_{GS}-V_{TH})^2(1+\lambda v_{DS})$.

Find. $A_v=v_o/v_i$, $R_{in}$, $R_o$, and the largest saturation-preserving input swing $V_{pp,\max}$.

Approach. Fix the operating point from $I_{bias}$, extract $g_m$ and $r_o$, then apply the common-gate small-signal formulas; for (c) walk the drain and source excursions to the edges of saturation.

1. Bias point. The tail source forces $I_D=I_{bias}=2\text{ mA}$. From $I_D=\tfrac{K}{2}V_{ov}^2$: $V_{ov}=\sqrt{2I_D/K}=\sqrt{2(2)/2}=1.414\text{ V}$, so $V_{GS}=2.414\text{ V}$. Then $V_D=V_{CC}-I_DR_D=10-4=6\text{ V}$, $V_S=V_G-V_{GS}=-1.414\text{ V}$, giving $V_{DS}=7.41\text{ V}\gt V_{ov}$ — saturated.
2. Small-signal parameters. $g_m=K\,V_{ov}=2\text{ mA/V}^2\times1.414\text{ V}=2.83\text{ mA/V}$; $r_o=\dfrac{1}{\lambda I_D}=\dfrac{1}{0.01\times2\text{ mA}}=50\text{ k}\Omega$.
3. (a) Voltage gain. For the common-gate stage (source driven, drain loaded by $R_D$) $$A_v=\frac{v_o}{v_i}=\frac{(1+g_mr_o)\,R_D}{R_D+r_o}=\frac{(1+141.4)(2)}{2+50}=\boxed{+5.48}.$$ It is non-inverting (source input and drain output move together).
4. (b) Input resistance (looking into the source): $R_{in}=\dfrac{R_D+r_o}{1+g_mr_o}=\dfrac{52\text{ k}\Omega}{142.4}=\boxed{365\ \Omega}\approx\dfrac{1}{g_m}$ — the hallmark low input impedance of a common-gate stage.
5. (b) Output resistance (looking into the drain, source a.c.-grounded through the ideal input source): $R_o=R_D\Vert r_o=\dfrac{2\cdot50}{52}\text{ k}\Omega=\boxed{1.92\text{ k}\Omega}$.
6. (c) Saturation limits. As the source rides $v_i$, $\Delta v_S=v_i$, $\Delta v_{GS}=-v_i$, $\Delta v_D=A_v v_i$, so $\Delta v_{DS}=(A_v-1)v_i$ and $\Delta V_{ov}=-v_i$. Triode edge ($v_{DS}=V_{ov}$): $7.41+4.48v_i=1.41-v_i\Rightarrow v_i=-1.10\text{ V}$. Cutoff edge ($I_D\to0$, drain to $V_{CC}$): $v_i=I_D/g_m=+0.71\text{ V}$.
7. (c) Result. The a.c. input (0 V d.c. through $C_1$) is symmetric about the Q-point, so it is limited by the smaller edge, $0.71\text{ V}$: $V_{pp,\max}=2(0.71)=\boxed{1.41\text{ V}_{pp}}$. (If a d.c. offset were allowed the full saturation window spans $-1.10$ to $+0.71\text{ V}$, i.e. $1.80\text{ V}$.)