diff --git a/docs/2a/ece240.md b/docs/2a/ece240.md
index 0787c76..58cb38f 100644
--- a/docs/2a/ece240.md
+++ b/docs/2a/ece240.md
@@ -117,7 +117,7 @@ $$\boxed{I_s=\frac 1 2k_nV_{ov}^2}$$
 
 <img src="https://upload.wikimedia.org/wikipedia/commons/4/4f/N-channel_JFET_common_source.svg" width=200>(Source: Wikimedia Commons)</img>
 
-Where $V_{out}=$V_{DS}$:
+Where $V_{out}=V_{DS}$:
 
 <img src="https://media.cheggcdn.com/media/b65/b65d59bd-ac35-4d28-b811-0ad1b5cf5bb6/phpCBbhn6" width=700 />
 
@@ -133,3 +133,25 @@ At a certain gate voltage:
 A_V&=\frac{\partial V_{DS}}{\partial V_{GS}} \\
 &=-g_{DS}R_D
 \end{align*}
+
+### Small signal analysis
+
+The current from the drain to the source is equal to:
+
+$$i_D=g_mV_{gs}$$
+
+For small signals, a transistor is equivalent to, where $r_0=\frac{1}{\lambda I_D}=\frac{V_A}{I_D}$:
+
+<img src="https://i.stack.imgur.com/EZK7K.png" width=600 />
+
+It can be assumed that the differential resistance is always significantly smaller than any other external resistance: $r_o << R_d$.
+
+To solve for the output resistance of the amplifier, turn off all sources and take the Thevenin resistance $R_{DS}$.
+
+### Common-drain amplifiers / source followers
+
+The input resistance of common amplifiers is infinity.
+
+<img src="https://upload.wikimedia.org/wikipedia/commons/3/30/N-channel_JFET_source_follower.svg" width=200>(Source: Wikimedia Commons)</img>
+
+As $V_{gs}$ is not necessarily zero, dependent sources must be left in when solving for output resistance, and so a small test source at the point of interest is required.