Editing Common collector (section)

===Derivations===
[[Image:Voltage follower small-signal.svg|thumb|300px|Figure 5: Small-signal circuit corresponding to Figure 3 using the hybrid-pi model for the bipolar transistor at frequencies low enough to ignore bipolar device capacitances]]
[[Image:Voltage follower output resistance.svg|300px |thumb|Figure 6: Low-frequency small-signal circuit for bipolar voltage follower with test current at output for finding output resistance. Resistor <math>R_\text{E} = R_\text{L} \parallel r_\text{o}</math>.]]
Figure 5 shows a low-frequency hybrid-pi model for the circuit of Figure 3. Using [[Ohm's law]], various currents have been determined, and these results are shown on the diagram. Applying Kirchhoff's current law at the emitter one finds:
: <math>(\beta + 1) \frac{v_\text{in} - v_\text{out}}{R_\text{S} + r_\pi} = v_\text{out} \left(\frac{1}{R_\text{L}} + \frac{1}{r_\text{o}}\right).</math>
Define the following resistance values:
: <math>\begin{align}
  \frac{1}{R_\text{E}} &= \frac{1}{R_\text{L}} + \frac{1}{r_\text{o}}, \\[2pt]
                     R &= \frac{R_\text{S} + r_\pi}{\beta + 1}.
\end{align}</math>
Then collecting terms the voltage gain is found:
: <math>A_\text{v} = \frac{v_\text{out}}{v_\text{in}} = \frac{1}{1 + \frac{R}{R_\text{E}}}.</math>
From this result, the gain approaches unity (as expected for a [[buffer amplifier]]) if the resistance ratio in the denominator is small. This ratio decreases with larger values of current gain β and with larger values of <math>R_\text{E}</math>.
The input resistance is found as
: <math>\begin{align}
  R_\text{in} &= \frac{v_\text{in}}{i_\text{b}} = \frac{R_\text{S} + r_\pi}{1 - A_\text{v}} \\
              &= \left(R_\text{S} + r_\pi\right)\left(1 + \frac{R_\text{E}}{R}\right) \\
              &= R_\text{S} + r_\pi + (\beta + 1) R_\text{E}.
\end{align}</math>
The transistor output resistance <math>r_\text{O}</math> ordinarily is large compared to the load <math>R_\text{L}</math>, and therefore <math>R_\text{L}</math> dominates <math>R_\text{E}</math>. From this result, the input resistance of the amplifier is much larger than the output load resistance <math>R_\text{L}</math> for large current gain <math>\beta</math>. That is, placing the amplifier between the load and the source presents a larger (high-resistive) load to the source than direct coupling to <math>R_\text{L}</math>, which results in less signal attenuation in the source impedance <math>R_\text{S}</math> as a consequence of [[voltage division]].

Figure 6 shows the small-signal circuit of Figure 5 with the input short-circuited and a test current placed at its output. The output resistance is found using this circuit as
: <math>R_\text{out} = \frac{v_\text{x}}{i_\text{x}}.</math>

Using Ohm's law, various currents have been found, as indicated on the diagram. Collecting the terms for the base current, the base current is found as
: <math>(\beta + 1) i_\text{b} = i_\text{x} - \frac{v_\text{x}}{R_\text{E}},</math>
where <math>R_\text{E}</math> is defined above. Using this value for base current, Ohm's law provides
: <math>v_\text{x} = i_\text{b} \left(R_\text{S} + r_\pi\right).</math>
Substituting for the base current, and collecting terms,
: <math>R_\text{out} = \frac{v_\text{x}}{i_\text{x}} = R \parallel R_\text{E},</math>
where || denotes a parallel connection, and <math>R</math> is defined above. Because <math>R</math> generally is a small resistance when the current gain <math>\beta</math> is large, <math>R</math> dominates the output impedance, which therefore also is small. A small output impedance means that the series combination of the original voltage source and the voltage follower presents a [[Thévenin's theorem|Thévenin voltage source]] with a lower Thévenin resistance at its output node; that is, the combination of voltage source with voltage follower makes a more ideal voltage source than the original one.