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:

$$v_{out}=v_e\uparrow \Longrightarrow v_{be} \downarrow \Longrightarrow i_b \downarrow \Longrightarrow i_c \downarrow \Longrightarrow v_e\downarrow$$ (110)

The DC operating point can be found by KVL:

$$V_{CC}=R_B I_B+V_{BE}+R_E I_E=R_B I_B+V_{BE}+R_E (\beta+1)I_B$$ (111)

Solving this equation for $I_B$ we get:

$$I_B=\frac{V_{CC}-V_{BE}}{(\beta+1)R_E+R_B}$$ (112)

$$I_E=(\beta+1) I_B=\frac{(\beta+1)(V_{CC}-V_{BE})}{(\beta+1)R_E+R_B}$$ (113)

$$V_{CE}=V_{CC}-V_E=V_{CC}-R_E I_E$$ (114)

Example 1

$V_{CC}=10\;V$, $R_B=100\;K$, $R_E=1.5\;k\Omega$, $\beta=100$, then

$$I_B = \frac{V_{CC}-V_{BE}}{(\beta+1)R_E+R_B} \approx \frac{10}{150+100}=0.04\;mA$$

$$I_E = (\beta+1)I_B\approx 4\;mA$$

$$V_E = I_E R_E=4\times 1.5=6\;V$$

$$V_{CE}=V_{CC}-V_E=10-6=4\;V$$ (115)

Example 2

In the same circuit, find $R_E$ so that the DC operating point is in the middle of the linear region, i.e., $V_{CE}=V_{CC}-V_E=10-V_E=V_{CC}/2=5\,V$. For this to be the case, we need $V_E=5\,V$:

$$V_E=I_E R_E =R_E\;\frac{(\beta+1) (V_{CC}-V_{BE})}{(\beta+1)R_E+R_B} =\frac{101\;R_E(10-0.7)}{101\;R_E+100}=5\,V$$ (116)

Solving the equation for $R_E$ we get $R_E=1.15\,k\Omega$. We can further verify this result:

$$I_B=\frac{V_{CC}-V_{BE}}{(\beta+1)R_E+R_B}=0.043\,mA, \;\;\;\;\;I_C=\beta\,I_B=100\,I_B=4.3\,mA$$ (117)

$$I_E=(\beta+1)I_B=4.34\,mA,\;\;\;\;V_E=I_ER_E=5\,V$$ (118)

AC small-signal equivalent circuit

Based on small signal model of the transistor:

we find the AC equivalent circit of the emitter follower:

and we can further find the three system parameters: voltage gain, input resistance, and output resistance, as shown below.

• AC voltage gain:

  As $R_B$ is significantly greater than $R_S$ and $R_{be}$, it is neglected in the analysis below.

  $$\left\{ \begin{array}{l} v_{out}=v_e=(R_E\vert\vert R_L)i_e =(R_E... ...b +v_{out}=(R_s+r_{be})i_b+(R_E\vert\vert R_L)(\beta+1) i_b \end{array} \right.$$ (119)

  The voltage gain can be found to be:

  $$A=\frac{v_{out}}{v_{in}} =\frac{(\beta+1) (R_E\vert\vert R_L)}{(R_s+r_{be})+(\beta+1) (R_E\vert\vert R_L)} \approx 1$$ (120)

  where the approximation is due to $R_s+r_{be} \ll (\beta+1) (R_E\vert\vert R_L)$.

  We note that the voltage gain $A$ is smaller than but approximately equal to 1. Note that $A$ is positive, i.e., the output voltage is in phase with the input voltage.

• Input resistance:

  The input resistance is the parallel combination of $R_B$ and $r'_{in}$, the resistance of the circuit to right of the base of the transistor, including the load $R_L$, which can be found as the ratio of $v_b$ and $i_b$. We first get:

  $$v_b=i_b r_{be}+v_{out}=i_b\, [ r_{be}+(\beta+1)(R_E\vert\vert R_L) ]$$ (121)

  and then find

  $$r'_{in}=\frac{v_b}{i_b}=r_{be}+(\beta+1) (R_E\vert\vert R_L) \approx \beta (R_E\vert\vert R_L)$$ (122)

  The overall input resistance is

  $$r_{in}=R_B\vert\vert r'_{in} \approx r'_{in}\approx \beta (R_E\vert\vert R_L)$$ (123)

  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 the parallel combination of $R_E$ and $r'_{out}$, the resistance of the circuit to the left of the emitter of the transistor (including $R_s$), which can be found as the ratio of the open-circuit voltage $v_{oc}$ and the short-circuit current $i_{sc}$.

  • Find open-circuit voltage $v_{oc}$ ( $R_L=\infty$):

    Due to the unity gain of the voltage gain follower, $v_{oc}$ is approximately the same as the source voltage $v_{oc}\approx v_{in}$.

  • Find short-circuit current $i_{sc}$ ($R_L=0$):

    $$i_{sc}=i_e=(\beta+1)i_b=(\beta+1)\frac{v_{in}}{r_{be}+R_s} \approx \beta \frac{v_{in}}{r_{be}+R_s}$$ (124)

  We therefore get

  $$r'_{out}=\frac{v_{oc}}{i_{sc}} \approx\frac{r_{be}+R_s}{\beta}$$ (125)

  and the overall output resistance can therefore be found to be

  $$r_{out}= R_E \vert\vert r'_{out}=R_E \vert\vert \left(\frac{r_{be}+R_s}{\beta}\right) \approx \frac{r_{be}+R_s}{\beta}$$ (126)

  Comparing this with the output resistance of the common-emitter circuit $r_{out}=R_C$, we see that the emitter follower has much lower output resistance.

Conclusion:

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:

• The voltage gain is smaller than but close to unity with $A\approx 1$.
• The input resistance is large $r_{in}=\beta\,R_E$ (assuming open-circuit with $R_L=\infty$).
• The output resistance is small $r_{out}=r_{be}/\beta$ (assuming ideal source with $R_s=0$).

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.

Example

For the emiter follower in the example above, 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 $r_{out}=r_{be}/\beta$.