MOSFET Small signal analaysis question
In DC, we can find some useful equations by using the information about the transistor.







simulate this circuit &ndash; Schematic created using CircuitLab

$$V_{GS} = 1.5V = I_{R2}\cdot R_2 \Rightarrow I_{R2} = \frac{1.5V}{R_2}$$

The current \$I_{R2}\$ flows through \$R_3\$, along with the transistor's \$I_{DS}\$, giving another linear equation:

$$V_{OUT} = R_3\cdot (I_{R2} + I_{DS}) = 2k\Omega \cdot (I_{R2} + 0.5mA)$$

Finally, we also can bring \$R_2\$ into the equations by using that the voltages across all resistors need to add up to \$V_{DD}\$. Remember that the current through \$R_1\$ is the same as \$I_{R2}\$.

$$V_{OUT} = V_{DD} - 1.5V - R_1I_{R2} = 1.5V - R_1I_{R2}$$

As we are interested in \$R_1\$ and \$R_2\$, we can eliminate \$I_{R2}\$ and \$V_{OUT}\$.

$$\begin{align}
1.5V - R_1I_{R2} &amp;= 2k\Omega\cdot (I_{R2} + 0.5mA) \\
\Rightarrow 1.5V - \frac{R_1}{R_2}1.5V &amp;= 2k\Omega\cdot \left( \frac{1}{R_2}1.5V + 0.5mA \right) \\
\Rightarrow R_2 - 3R_1 &amp;= 6k\Omega
\end{align}$$

In small signal analysis, assuming you have an input current of \$i_{in}\$, you can find the input impedance using the following KCL equations:







simulate this circuit

$$
\begin{align}
\left( \frac{1}{R_1} + \frac{1}{R_2} \right)v_{in}&amp; - \frac{1}{R_2}v_{out} &amp;=&amp; i_{in} \\
-\left( \frac{1}{R_2} + g_m \right) v_{in}&amp; + \left(\frac{1}{R_2} + \frac{1}{R_3} + g_m \right)v_{out} &amp;=&amp; 0
\end{align}
$$

(We don't need the KCL law for the output because it is "hidden" by the voltage-controlled current source, and we are only interested in \$v_{in}\$).

The input impedance can be found from these equations (while substituting \$R_3 = 2k\Omega\$ and \$g_m = 1/1k\Omega\$):

$$
\begin{align}
Z_{in} &amp;= \frac{v_{in}}{i_{in}} \\
   &amp;= \frac{R_1(R_2R_3g_m+R_3+R_2)}{R_2R_3g_m+R_3+R_2+R_1} \\
   &amp;= \frac{R_1(2k\Omega+3R_2)}{2k\Omega+3R_2+R_1}
\end{align}
$$

Leading to the final equation (\$Z_{in} = 500k\Omega\$)

$$\begin{align}
(2k\Omega+3R_2+R_1)\cdot 500k\Omega &amp;= R_1(2k\Omega+3R_2) \\
\Rightarrow 3\frac{R_1}{1k\Omega}\frac{R_2}{1k\Omega} &amp;= 1k\Omega + 1.5R_2 + 0.498R_1
\end{align}$$

We now have 2 equations for solving 2 unknowns, which leads to the following quadratic equation (eliminating \$R_1\$):

$$\frac{R_2^2}{1k\Omega} - 1672R_2 - 4k\Omega = 0$$

Taking the positive solution will finally result into

$$\begin{align}
R_2 &amp;= (10\sqrt{6989}+836)\ k\Omega \approx 1672\ k\Omega \\
R_1 &amp;= \frac{R_2 - 6k\Omega}{3} \approx 555\ k\Omega
\end{align}$$