Emitter Follower

The input and output of an emitter follower are the base and the emitter, respectively, and the collector is at AC zero. The circuit is therefore a common-collector circuit (for AC).


The negative feedback effect due to $R_E$ can be shown qualitatively:

$\displaystyle v_{out}=v_e\uparrow \Longrightarrow v_{be} \downarrow \Longrightarrow
i_nb \downarrow \Longrightarrow i_c \downarrow \Longrightarrow
v_e\downarrow$ (107)

The DC operating point can be found as:

$\displaystyle V_{CC}=R_B I_B+V_{BE}+R_E I_E=R_B I_B+V_{BE}+R_E (\beta+1)I_B$ (108)

Solving this equation for $I_B$ we get:

$\displaystyle I_B=\frac{V_{CC}-V_{BE}}{(\beta+1)R_E+R_B}$ (109)

$\displaystyle I_E=(\beta+1) I_B=\frac{(\beta+1)(V_{CC}-V_{BE})}{(\beta+1)R_E+R_B}$ (110)

$\displaystyle V_{CE}=V_{CC}-V_E=V_{CC}-R_E I_E$ (111)

AC small-signal equivalent circuit

The AC equivalent circit (based on small signal model of the transistor) of the emitter follower can be found as:

EmitterFollower1.png TransistorModel.png emitterfollower2.gif

Based on this small signal model, the three system parameters: voltage gain, input resistance, and output resistance can be obtained as shown below.


The emitter follower is a circuit with strong negative feedback, i.e., the full output $v_{out}=v_e$ is fed back to become part of its input $v_{be}$. Due to this deep negative feedback, it has the following properties:

Not considering the effect of the source and load, i.e., when $R_S=0$ and $R_L=\infty$, we have $R_{in}=\beta R_E$ and $R_{out}=r_{be}/beta$. The emitter follower acts as an impedance transformer with a ratio of $\beta$.

Although the emitter follower circuit does not amplify the signal voltage, it drastically improves the input and output resistances, when compared with the common-emitter circuit with $r_{in}=r_{be}$ and output resistances $r_{out}=R_C$.

Due to its high input resistance $r_{in}$, it draws little current from the source, causing little internal voltage drop across the internal resistance of the source, also due to its low output resistance $r_{out}$, it can drive heavy load (low $R_L$) without lowering its output voltage. It is therefore widely used as a buffer between two stages of a multi-stage amplification circuit.


Assume $R_B=100\,k\Omega$, $V_{CC}=10\,V$, $\beta=100$. Find $R_E$ so that the DC operating point is in the middle of the load line.

For $V_{CE}$ to be in the middle of the load line, we need to have $V_{CE}=V_{CC}-V_E=10-V_E=V_{CC}/2=5\,V$, i.e., $V_E=5\,V$:

$\displaystyle V_E=I_E R_E =R_E\;\frac{(\beta+1) (V_{CC}-V_{BE})}{(\beta+1)R_E+R_B}
=\frac{101\;R_E(10-0.7)}{101\;R_E+100}=5\,V$ (120)

Solving the equation for $R_E$ we get $R_E=1.15\,k\Omega$. Now we have

$\displaystyle I_B=\frac{V_{CC}-V_{BE}}{(\beta+1)R_E+R_B}=0.043\,mA,
\;\;\;\;\;I_C=\beta\,I_B=100\,I_B=4.3\,mA$ (121)

$\displaystyle I_E=(\beta+1)I_B=4.34\,mA,\;\;\;\;V_E=I_ER_E=5\,V$ (122)

The input resistance is approximately $r_{in}=\beta R_E=100\times 1.15\,k\Omega=115\,k\Omega$, and the output resistance is approximately $r_{out}=r_{be}/\beta$.