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AC equivalent circuits

As discussed before, the voltage a circuit receives from a source depends on its input impedance $R_{in}$ as well as the internal impedance $R_s$ of the source, while the voltage it delivers depends on its output impedance $r_{out}$ as well as the load impedance $R_L$. It is therefore important to consider these input and output impedances of an amplification circuit as well as its voltage gain.


In the first figure, everything inside the red box, including the amplifier as well as $V_S$ and $R_s$, is treated as the source, while everything inside the blue box, including the amplifier as well as $R_L$, is treated as the load. Given the amplifier as well as the source $V_S$ and $R_s$, and the load $R_L$, we need to find the following three parameters so that the red and blue boxes in the first figure can be modeled by the corresponding boxes in the second figure:

Consider the typical transistor AC amplification circuit below:


If the capacitances of the coupling capacitors and the emitter by-pass capacitor are large enough with respect to the frequency of the AC signal in the circuit is high enough, these capacitors can all be approximated as short circuit. Moreover, note that the AC voltage of the voltage supply $V_{CC}$ is zero, it can be treated the same as the ground. Now the AC behavior of the transistor amplification circuit can be modeled by the following small signal equivalent circuit:



As shown above, this AC small signal equivalent circuit can be modeled by as an active circuit containing three components:

Example 1:


$V_{CC}=12V$, $R_B=300 k\Omega$, $R_C=R_L=4 k\Omega$, and $\beta=40$. We further assume $R_s=0$, and the capacitances are large enough so that they can be considered as short circuit for AC signals.

DCACloadline.gif transistorBJTexample1c.gif

The circuit above can also be analyzed using the small-signal model.


Same as before, $r_{be}=1  k\Omega$, and we have the following DC variables:

$\displaystyle I_B$ $\textstyle =$ $\displaystyle (V_{CC}-V_{BE})/R_B \approx V_{CC}/R_B=40 \;\mu A$  
$\displaystyle I_C$ $\textstyle =$ $\displaystyle \beta\;I_B=40\times 40\;\mu A=1.6\;mA$  
$\displaystyle V_{CE}$ $\textstyle =$ $\displaystyle V_{CC}-R_C I_C=12-1.6\times 4=5.6\;V$  

The AC variables:

i_b=v_{in}/r_{be}, \;\;\;\;v_c=-i_c\; (R_C\vert\vert R_L)

The voltage gain is:

A_v=\frac{v_c}{v_{in}}=-\frac{\beta\; i_b\;(R_C\vert\vert R_...
... R_L)}{r_{be}}
=-\frac{(40\times 2)\;k\Omega}{1\;k\Omega}=-80

The input resistance is $r_{in}=R_B\vert\vert r_{be}\approx r_{be}=1 k\Omega$, the output resistance is $r_{out}=R_C=4\;k\Omega$.

Example 2:

Consider the circuit below with its AC small-signal model:

Example5.gif Example5a.png

We can find the voltage gain, the input and output resistances when $R_S=0$ and $R_L=\infty$ ( $v_{in}=v_b,\;\;v_{out}=v_c$).

Example 3:

Consider both the DC operating point and its AC small signal model of the circuit below:


Apply KVL to get

V_{CC}-V_{BE}=I_B R_B+(\beta+1)I_BR_E,\;\;\;\;\;


I_C=\beta I_B=\beta \frac{V_{CC}-V_{BE}}{R_B+(\beta+1)R_E},\;\;\;\;\;
V_{CE}\approx V_{CC}-I_C(R_C+R_E)

We assume $V_{CC}=12V,\;R_E=0.1 k\Omega,\;\beta=100$, and consider the following for the DC operating point to be in the middle of the linear region:

Next consider the AC equivalent circuit based on small-signal model of the transistor in the dashed line box:


As $R_B$ is significantly greater than all resistors in the circuit, it can be ignored in the AC analysis. Apply KCL to the emitter to get

i_b+\beta i_b=(\beta+1)i_b=\frac{v_e}{R_E}=\frac{v_{in}-r_{be}i_b}{R_E}

Solving for $i_b$:


In conclusion, the negative feedback introduced by $R_E$ increases the input resistance, and stabalizes the DC operating point as well as the AC voltage gain.

next up previous
Next: Emitter Follower Up: ch4 Previous: Small-Signal Model and H
Ruye Wang 2018-04-18