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AC Amplification

Previously we considered a simple circuit composed of a source of $V_S$ and $R_S$ and a load of $R_L$. Now we generalize this idea to a more complicated circuit containing an amplification circuit shown on top of the figure below, where everything, including the amplifier as well as $V_S$ and $R_S$, inside the red box is treated as the source, while everything, including the amplifier as well as $R_L$, inside the blue box 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 $R_{in}$, $R_{out}$ and also the voltage gain $A$.

AmplifierSourceLoad.gif

Based on the previous discussion, a single transistor AC amplification circuit is given as shown in the figure.

ACamplification1.gif

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:

ACamplification2a.gif

ACamplifierModel.gif

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

Example 1:

DCACloadlineEx.gif

Here $V_{CC}=12V$, $R_B=300 k\Omega$, $R_C=R_L=4 k\Omega$ ($R_S=0$), and the capacitors are large enough so that they can be considered as short circuit for the AC signals.

DCACloadline.gif transistorBJTexample1c.gif

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

DCACloadlineExModel.gif

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

\begin{displaymath}
I_B=(V_{CC}-V_{BE})/R_B \approx V_{CC}/R_B=40 \;\mu A
\end{displaymath}


\begin{displaymath}
I_C=\beta\;I_B=37.5\times 40\;\mu A=1.5\;mA
\end{displaymath}


\begin{displaymath}
V_{CE}=V_{CC}-R_C I_C=12-1.5\times 4=6\;V
\end{displaymath}

The AC variables:

\begin{displaymath}
i_b=v_{in}/r_{BE}, \;\;\;\;v_c=-i_c\; R_C\vert\vert R_L=-\beta\; i_b\; R'_L
\end{displaymath}

The voltage gain is:

\begin{displaymath}
A_v=\frac{v_c}{v_{in}}=-\frac{\beta\; i_b\;R'_L}{r_{BE} i_b...
...R'_L}{r_{BE}}=-\frac{(37.5\times 2)\;k\Omega}{1\;k\Omega}=-75
\end{displaymath}

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


next up previous
Next: Emitter Follower Up: ch4 Previous: Small Signal Model and
Ruye Wang 2014-11-19