Question 4 of 5: Common-Gate MOSFET Amplifier

Question 4: 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 common-gate stage: R₁/R₂ bias the gate (AC-grounded), the input drives the source through C₁, the drain load is R_D, and the drain bias current is set by I_bias so I_D = 1 mA.

Find. (a) A_v = v_o/v_in; (b) R_in looking into the source; (c) R_o looking into the drain.

Approach. Get the operating point from I_D = 1 mA to find g_m and r_o, then apply the standard common-gate small-signal formulas (gate at AC ground, input at source, output at drain).

1. Overdrive and transconductance. From $I_D=\tfrac12 K V_{ov}^2$, the overdrive is $V_{ov}=\sqrt{2I_D/K}=\sqrt{2(1)/1}=1.414$ V, and
   $$g_m=\sqrt{2K I_D}=\sqrt{2(1\,\text{mA/V}^2)(1\,\text{mA})}=1.414\ \text{mA/V}.$$
2. Output resistance of the device.
   $$r_o=\frac{1}{\lambda I_D}=\frac{1}{(0.1)(1\,\text{mA})}=10\ \text{k}\Omega.$$
   Because $r_o$ is comparable to R_D, it must be kept in all three results.
3. (a) Small-signal gain. For the common-gate stage (non-inverting),
   $$A_v=\frac{v_o}{v_{in}}=(g_m r_o+1)\,\frac{R_D}{R_D+r_o}=(1.414\times10+1)\,\frac{5}{5+10}.$$
   With $g_m r_o=14.14$,
   $$\boxed{A_v=(15.14)(0.3333)=+5.05\ \text{V/V}.}$$
4. (b) Input resistance (into the source).
   $$R_{in}=\frac{r_o+R_D}{1+g_m r_o}=\frac{10+5}{1+14.14}\ \text{k}\Omega\ \Rightarrow\ \boxed{R_{in}=0.99\ \text{k}\Omega\ (991\ \Omega).}$$
   This is close to $1/g_m=707\ \Omega$, raised by the $R_D/(g_m r_o)$ term because $r_o$ is finite.
5. (c) Output resistance (into the drain). With the ideal input source setting the source terminal at AC ground, the resistance looking into the drain is just r_o, in parallel with the drain load:
   $$R_o=R_D\,\|\,r_o=5\,\|\,10=\boxed{3.33\ \text{k}\Omega.}$$