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Emitter Follower

An emitter follower circuit shown in the figure is widely used in AC amplification circuits. The input and output of the emitter follower are the base and the emitter, respectively, while the collector is at AC zero, therefore this circuit is also called common-collector circuit.

emitterfollower.gif

DC operating point


\begin{displaymath}\left\{ \begin{array}{l}
I_E=(\beta+1) I_B \\
V_{CC}=R_B ...
...)I_B \\
V_{CE}=V_{CC}-V_E=V_{CC}-R_E I_E \end{array} \right. \end{displaymath}

Solving the second equation, we get $I_B$:

\begin{displaymath}I_B=\frac{V_{CC}-V_{BE}}{R_B+(\beta+1)R_E} \end{displaymath}

and $I_E=(\beta+1)I_B$:

\begin{displaymath}I_E=(\beta+1) I_B=\frac{(\beta+1)(V_{CC}-V_{BE})}{R_B+(\beta+1)R_E} \end{displaymath}


\begin{displaymath}V_{CE}=V_{CC}-R_EI_E \end{displaymath}

Example

Assume $R_B=200 K$, $V_{CC}=10 V$, find $R_E$ so that the DC operating point is in the middle of the load line.


\begin{displaymath}V_E=I_E R_E=(\beta+1)I_B R_E=(\beta+1) R_E \frac{V_{CC}-V_{BE}}{R_B+(\beta+1)R_E} \end{displaymath}

Solving $V_E=V_{CC}/2=5$ for $R_E$, we get $R_E=2.3 K$.

AC small-signal equivalent circuit

emitterfollower2.gif

Voltage gain:

We assume $R_B \gg r_{be}+R_E$ and therefore can be ignored, and have

\begin{displaymath}\left\{ \begin{array}{l}
v_{out}=v_e=i_e (R_E\vert\vert R_L...
...+r_{be})+(\beta+1) i_b (R_E\vert\vert R_L)
\end{array} \right. \end{displaymath}

Now the voltage gain can be found to be:

\begin{displaymath}A=\frac{v_{out}}{v_{in}}
=\frac{(\beta+1) (R_E\vert\vert R_L)}{(R_S+r_{be})+(\beta+1) (R_E\vert\vert R_L)}
\approx 1 \end{displaymath}

As $R_S+r_{be} \ll (\beta+1) (R_E\vert\vert R_L)$, $A$ is smaller than but approximately equal to 1. Note that $A>0$, i.e., the output voltage is in phase with the input voltage.

Input resistance:

The input resistance is $R_B$ in parallel with the resistance $r'_{in}$ of the circuit to its right including the load $R_L$, which can be found as the ratio of the voltage $v_b$ and the current $i_b$. But as

\begin{displaymath}v_b=i_b(r_{be}+(\beta+1)(R_E\vert\vert R_L)) \end{displaymath}

we have

\begin{displaymath}r'_{in}=\frac{v_b}{i_b}=r_{be}+(\beta+1) (R_E\vert\vert R_L)\approx (\beta+1)(R_E\vert\vert R_L) \end{displaymath}

and

\begin{displaymath}r_{in}=R_B\vert\vert r'_{in} \end{displaymath}

Comparing this with the input resistance of the common-emitter circuit $r_{in}=R_1\vert\vert R_2\vert\vert r_{be} \approx r_{be}$, we see that the emitter follower has much higher input resistance.

Output resistance:

The output resistance is $R_E$ in parallel with the resistance $r'_{out}$ of the circuit to its left including the source $R_S$ but excluding $R_E$, which can be found as the ratio of the open-circuit voltage $v_{oc}$ ($R_L=\infty$) and the short-circuit current $i_{sc}$ ($R_L=0$). As the voltage gain of the emitter follower is close to unity, the open-circuit output voltage is approximately the same as the source voltage $v_{oc}\approx v_{in}$. The short-circuit current $i_{sc}$ can be found as

\begin{displaymath}i_{sc}=i_e=(\beta+1)i_b=(\beta+1)\frac{v_{in}}{r_{be}+R_S}
\approx \beta\frac{v_{in}}{r_{be}+R_S} \end{displaymath}

Therefore the output resistance is

\begin{displaymath}r'_{out}=\frac{v_{oc}}{i_{sc}}=\frac{r_{be}+R_S}{\beta+1}
\approx\frac{r_{be}+R_S}{\beta} \end{displaymath}

The overall output resistance can therefore be found to be

\begin{displaymath}r_{out} = R_E \vert\vert r'_{out}=R_E \vert\vert \frac{r_{be}+R_S}{\beta+1}
\approx \frac{r_{be}+R_S}{\beta} \end{displaymath}

Conclusion:

The emitter follower is a circuit with deep negative feedback, i.e., all of its output $v_{out}=v_e$ is fed back to become part of its input $v_{be}$. The fact that this is a negative feedback can be seen by:

\begin{displaymath}v_e \uparrow \Longrightarrow v_{be} \downarrow
<\Longrightarrow i_e \downarrow \Longrightarrow v_e \downarrow \end{displaymath}

Due to this deep negative feedback, the voltage gain of the emitter follower is smaller than unity. However, the circuit is drastically improved in terms of its input and output resistances. In fact the emitter follower acts as an impedance transformer with a ratio of $\beta+1$, i.e., the input resistance is $\beta+1$ times greater than $R_E\vert\vert R_L$ and the output resistance is $\beta+1$ times smaller than $R_S+r_{be}$.

Comparing this with the input resistance $r_{in}=r_{be}$ and output resistances $r_{out}=R_C$ of the common-emitter transistor circuit, we see that the emitter follower circuit has very favorable input/output resistances.

Although the emitter follower does not amplify voltage, due to its high input resistance drawing little current from the source, and its low output resistance capable of driving heavy load, it is widely used as both the input and output stages for a multi-stage voltage amplification circuit.


next up previous
Next: Multi-stage Amplification Up: ch4 Previous: AC Amplification
Ruye Wang 2014-04-15