Common-gate MOSFET amplifier — gain, input and output resistance: 23-Mechatronics-A2 Circuits and Electronics, December 2018, Q9

Question 9: Common-gate MOSFET amplifier — gain, input and output resistance [12, 4, 4]

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.

Quantity	Value
$V_{TH}$	1 V
$K$	1 mA/V²
$\lambda$	0.1 V$^{-1}$
$I_{bias}=I_D$	2 mA
$R_D$	2 kΩ

Find. $v_o/v_{in}$, $R_{in}$, $R_o$.

Figure-9: common-gate NMOS amplifier. $R_1$/$R_2$ set the (ac-grounded) gate bias; input $v_{in}$ applied at the source through $C_1$; output taken at the drain through $C_2$ across $R_D$.

Approach. Find the small-signal parameters $g_m$, $r_o$ from the fixed bias current, then solve the two-node ($v_d$, $v_s$) small-signal circuit of the common-gate stage (gate at ac ground) including $r_o$ exactly rather than the $g_m R_D$ shortcut, since $r_o$ is not negligible here.

Small-signal parameters. With $I_D=I_{bias}=2\ \text{mA}$ fixed by the tail source, $$g_m=\sqrt{2KI_D}=\sqrt{2\times1\times2}=2\ \text{mA/V},\qquad r_o=\frac{1}{\lambda I_D}=\frac{1}{0.1\times2\,\text{mA}}=5\ \text{k}\Omega.$$
Small-signal model. Gate is ac ground ($v_g=0$), so $v_{gs}=-v_s$. The dependent current source $g_mv_{gs}=-g_mv_s$ and $r_o$ both connect drain node $d$ to source node $s$; $R_D$ connects $d$ to ac ground; the test/input excitation connects to $s$.
Node equations. With a unit test current $I_{in}=1\,\text{A}$ injected at the source node: $$v_d\Big(\frac1{R_D}+\frac1{r_o}\Big)=v_s\Big(\frac1{r_o}+g_m\Big),\qquad -\frac{v_d}{r_o}+v_s\Big(\frac1{r_o}+g_m\Big)=I_{in}.$$ Solving numerically gives $v_s=636.4$, $v_d=2000$ (per amp of test current).
Input resistance. $$R_{in}=\frac{v_s}{I_{in}}=\boxed{636.4\ \Omega}.$$
Small-signal gain. With $v_{in}=v_s$ (ideal source, no series resistance) and $v_o=v_d$: $$\frac{v_o}{v_{in}}=\frac{v_d}{v_s}=\boxed{3.14}.$$ (Equivalently, the exact common-gate formula $A_v=\dfrac{(1+g_mr_o)R_D}{r_o+R_D}=\dfrac{(1+10)\times2}{7}=3.14$.)
Output resistance. Looking into the drain with the ideal source driving $v_s$ directly (zero source impedance), $r_o$ appears simply in parallel with $R_D$: $$R_o=R_D\|r_o=\frac{2\times5}{7}=\boxed{1.429\ \text{k}\Omega}.$$

Quantity	Result
$g_m$	2 mA/V
$r_o$	5 kΩ
Small-signal gain	$\boxed{v_o/v_{in}=3.14}$
Input resistance	$\boxed{R_{in}=636.4\ \Omega}$
Output resistance	$\boxed{R_o=1.429\ \text{k}\Omega}$

Concept & Theory

Topic. Common-gate MOSFET amplifier small-signal analysis, including finite $r_o$.

Key relations. $g_m=\sqrt{2KI_D}=K(V_{GS}-V_{TH})$; $r_o=1/(\lambda I_D)$; exact common-gate gain $A_v=(1+g_mr_o)R_D/(r_o+R_D)$; exact $R_{in}=(r_o+R_D)/(1+g_mr_o)$; $R_o=R_D\|r_o$ (source driven by an ideal voltage source).

Why this works. In common-gate, the gate is the ac reference, so $v_{gs}=-v_s$ — a signal applied at the source directly modulates the channel current, and because $r_o$ ties the (signal) drain and source nodes together, both $A_v$ and $R_{in}$ pick up a $(1+g_mr_o)$ factor beyond the simple $g_mR_D$/$1/g_m$ approximations that ignore $r_o$.

Common pitfall. Using $R_{in}=1/g_m$ and $A_v=g_mR_D$ (the $r_o\to\infty$ shortcuts) when $r_o$ is comparable to $R_D$, as it is here ($r_o=5\,\text{k}\Omega$ vs $R_D=2\,\text{k}\Omega$) — the shortcuts would give $R_{in}=500\,\Omega$ and $A_v=4.0$, both visibly off from the exact 636.4 Ω and 3.14.

Source. Sedra & Smith, Microelectronic Circuits §7.3.4 (the common-gate amplifier).

Question 9 of 11: Common-gate MOSFET amplifier — gain, input and output resistance

Question 9: Common-gate MOSFET amplifier — gain, input and output resistance [12, 4, 4]