MOS Amplifiers

We want to take some voltage input signal and get an amplified voltage output signal. However, we know that MOSFETs turn a voltage to current instead. In order to get a voltage output, we can pass that current through some resistor:

This topology is called a common source amplifier. Taking the drain current at saturation, equating that with the current through the resistor, and solving for \(V_{\text{out}}\), we get the following large signal relationship:

\begin{align} V_{\text{out}} = -R_D\left[\frac{1}{2}\mu_nC_{\text{ox}}\frac{W}{L}(V_{\text{in}}-V_T)^2\right]+V_{DD} \notag \end{align}

However, we want the small signal gain \(A_v\), so we have

\begin{align} A_v = \frac{\partial V_{\text{out}}}{\partial V_{\text{in}}} = -R_D\left[\mu_nC_{\text{ox}}\frac{W}{L}(V_{\text{in}}-V_T)\right] \notag \end{align}

Notice the part inside the brackets is just \(g_m\). Therefore, our small signal gain (ignoring channel length modulation) is:

\begin{align} \boxed{A_v = -g_mR_D} \end{align}

This result is also immediately apparent if we draw the small signal model, and realizing that \(R_D\) connects across drain and ground. We can use this technique to derive the small signal gain when we also consider channel length modulation:

Then, the small signal gain is:

\begin{align} \boxed{A_v = -g_m(R_D\parallel r_o) } \end{align}

Moving the resistor and output to the source side gives us a common drain topology:

We can use the T small signal model to analyze this amplifier. It can be seen that the small signal model can be simplified to become:

This is just a voltage divider! The small signal gain becomes

\begin{align} A_v = \frac{R_s \parallel r_{mb} \parallel r_{o}}{r_m + R_s \parallel r_{mb} \parallel r_{o}} \end{align}

Notice that the gain is always less than one. However, let’s take a look at its impedances. Looking into the input (gate), we see infinite input impedance; looking from the output, we see four resistances in parallel: \(r_o \parallel r_m \parallel r_{mb} \parallel R_s\). Usually, \(r_o\) and \(r_{mb}\) are orders of magnitude larger than the other resistances, so \(r_m \parallel R_s\) dominates. This means that we can actually make the output impedance small.

The reason why we care is because the gain changes due to the impedance of your load. If the load impedance is very small, than the gain of the amplifier will drop significantly if the output impedance is high. This means that the common drain amplifier is good for designs that drive a low impedance load.

The following is a common gate amplifier configuration:

Again, we can use a small signal model to analyze this:

The gain can be derived from the small signal model:

\begin{align} A_v = \frac{R_D \parallel r_o}{r_o \parallel r_m \parallel r_{mb}} \end{align}