Common-Gate MOSFET Amplifier: 22-Elec-A5 Electronics, December 2018, Q3

Question 3: Common-Gate MOSFET Amplifier (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. A single NMOS in the common-gate configuration: the gate is set by the $R_1$–$R_2$ divider and held at AC ground, the source is fed by the input through $C_1$ and biased by the ideal current source $I_{bias}$, and the drain drives $R_D$ with the output taken through $C_2$.

Given data — Question 3
Quantity	Value	Quantity	Value
$V_{TH}$	1 V	$V_{DD}$	10 V
$K$	1 mA/V$^2$	$I_{bias}$	2 mA
$\lambda$	0.1 V$^{-1}$	$R_D$	2 kΩ
$R_1,R_2$ (gate bias)	10 kΩ, 5 kΩ	$C_1,C_2$	$\infty$

Find. the small-signal voltage gain $A_v=v_o/v_{in}$, the input resistance $R_{in}$ looking into the source, and the output resistance $R_o$ looking into the drain.

[Figure not reproduced: Q3: common-gate NMOS stage — gate divider $R_1$/$R_2$ (AC-grounded), source driven through $C_1$ with $I_{bias}$, drain load $R_D$, output through $C_2$. See the official exam paper or the cited reference text.]

Q3: common-gate NMOS stage — gate divider $R_1$/$R_2$ (AC-grounded), source driven through $C_1$ with $I_{bias}$, drain load $R_D$, output through $C_2$.

Approach. Fix the bias point from $I_{bias}$ to get $g_m$ and $r_o$ (neglecting $\lambda$ for the DC solution, as usual), then apply the common-gate small-signal results.

Bias point. The current source sets $I_D=I_{bias}=2\text{ mA}$. From the saturation law (dropping $\lambda$ for bias), $I_D=\tfrac12 K V_{OV}^2$, so $$V_{OV}=\sqrt{\frac{2I_D}{K}}=\sqrt{\frac{2(2)}{1}}=2\text{ V}.$$
Small-signal parameters. $$g_m=\sqrt{2KI_D}=\sqrt{2(1)(2)}=2\text{ mA/V},\qquad r_o=\frac{1}{\lambda I_D}=\frac{1}{0.1(2\times10^{-3})}=5\text{ k}\Omega.$$ Note $g_m r_o=10$.
Voltage gain (a). A common-gate stage is non-inverting; including $r_o$, $$A_v=\frac{v_o}{v_{in}}=(1+g_m r_o)\frac{R_D}{R_D+r_o}=(1+10)\frac{2}{2+5}=\frac{22}{7}=\boxed{+3.14\text{ V/V}}.$$
Input resistance (b). Looking into the source (the ideal $I_{bias}$ adds no loading), $$R_{in}=\frac{R_D+r_o}{1+g_m r_o}=\frac{2+5}{11}\text{ k}\Omega=\boxed{636\ \Omega}\ \left(\approx 1/g_m\right).$$
Output resistance (c). Setting the input to zero grounds the source through $C_1$, so looking into the drain the transistor presents $r_o$ in parallel with $R_D$: $$R_o=R_D\,\|\,r_o=\frac{2\times5}{2+5}\text{ k}\Omega=\boxed{1.43\text{ k}\Omega}.$$

Results — Question 3
Quantity	Value
$V_{OV}$, $g_m$, $r_o$	2 V, 2 mA/V, 5 kΩ
Gain $A_v=v_o/v_{in}$	$+3.14$ V/V (non-inverting)
Input resistance $R_{in}$	636 Ω
Output resistance $R_o$	1.43 kΩ

Topic: Common-gate MOSFET amplifier — a non-inverting, low-input-resistance stage.

Key relations: $g_m=\sqrt{2KI_D}$, $r_o=1/(\lambda I_D)$; $A_v=(1+g_m r_o)R_D/(R_D+r_o)$; $R_{in}=(R_D+r_o)/(1+g_m r_o)\approx 1/g_m$; $R_o=R_D\|r_o$.

Why this works: With the gate held at AC ground, the input drives the source. The source node looks back into $1/g_m$ (lowered further by $r_o$ and $R_D$), giving a small $R_{in}$ — ideal for accepting a current or matching a low-impedance line. The drain current is delivered to $R_D$ non-inverted, and because $\lambda$ is appreciable here ($r_o=5\text{ k}\Omega$ is comparable to $R_D$) it must be kept in both the gain and $R_o$.

Common pitfall: Dropping $r_o$ (which would give $A_v=g_m R_D=4$ and $R_{in}=1/g_m=500\ \Omega$) even though it is only $2.5\times R_D$; or treating the common-gate stage as inverting.

Source: Sedra & Smith, Microelectronic Circuits, 8th ed., Ch. 7 (common-gate configuration and its resistance transformations).

Question 3 of 5: Common-Gate MOSFET Amplifier

Question 3: Common-Gate MOSFET Amplifier (20 marks)