AC Amplification

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

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:

AC Input Impedance: For AC signals, the input of the amplification circuit is shown below, where $R_s$ is the internal resistance of the signal source, and the input impedance of the circuit is the three resistances $R_1$, $R_2$ and $r_{be}$ in parallel:

$$R_{in}=R_1\vert\vert R_2\vert\vert r_{be} =(\frac{1}{R_1}+\f... ...frac{R_1 R_2 r_{be}}{R_1R_2+R_1r_{be}+R_2r_{be}}\approx r_{be}$$

AC Output Impedance: This is simply the resistance of the resistor

$$R_{out}=R_C$$

AC Amplification Gain: Given the AC input voltage $v_{in}$, the base voltage can be found (voltage divider) to be

$$v_B=v_{in}\; \frac{(R_1\vert\vert R_2\vert\vert r_{be})}{(R_... ...rt\vert r_{be})+R_s} \approx v_{in}\;\frac{r_{be}}{r_{be}+R_s}$$

and the base current is

$$i_B=\frac{v_B}{r_{be}}=v_{in}\;\frac{(R_1\vert\vert R_2\vert... ...r_{be})+R_s}\frac{1}{r_{be}} \approx v_{in}\frac{1}{r_{be}+R_s}$$

The collector current is $i_C=\beta \;i_B$ and the output voltage is

$$v_C=-i_C\;(R_C\vert\vert R_L)=-\beta\;i_B\;(R_C\vert\vert R_L)$$

Here the negative sign indicates the fact that $v_C$ is $180^\circ$ out of phase with $v_B$, as

$$v_B \uparrow \Longrightarrow i_B \uparrow \Longrightarrow i_C \uparrow \Longrightarrow v_C \downarrow$$

The voltage gain is therefore

$$G=\frac{v_{out}}{v_{in}}=\frac{v_C}{v_{in}}= -\beta \frac{(R... ...t\vert R_L)\approx -\beta\;\frac{R_C\vert\vert R_L}{r_{be}+R_S}$$

In particular, if the input resistance is much larger than the internal resistance of the voltage source, i.e.,

$$R_{in}=R_1\vert\vert R_2\vert\vert r_{be} \approx r_{be} \gg R_s$$

and the output resistance is much smaller than the load resistance, i.e.,

$$R_{out}=R_C \ll R_L$$

then the gain can be approximated as

$$G=\frac{v_{out}}{v_{in}}=-\beta \frac{R_C}{r_{be}}$$

Example 1:

This figure shows a common-emitter amplification circuit of npn transistor. Assume $V_{CC}=12V$, $R_B=300K\Omega$, $R_C=4K\Omega$, $R_L=4K\Omega$ (assuming $R_S=0$). Also we assume the capacitors are large enough so that they can be considered as short circuit for the AC signals.

• Find base current:
  
  $$I_B=(V_{BB}-V_{BE})/R_B \approx V_{BB}/R_B=12/(300\times 10^3) =40\times 10^{-6}\; A=40\;\mu A$$
  
  

• Find DC load line: when $I_C=0$, $V_{CE}=12V$, when $V_{CE}=0$,
  
  $$I_C=V_{CC}/R_C=12V/(4\times 10^3)=3\times 10^{-3}A=3 \;mA$$
  
  
  The DC load line is determined by these two points
• Find DC operating point $Q$: The intersection of the DC load line and the $I_C$ curve corresponding to $I_B=40 \;\mu A$ is the DC operating point with $V_{CE}=6V$ and $I_C=1.5 mA$ (i.e., $\beta=1.5ma/40 \mu A=37.5$).
• Find AC load line: The AC load is $R'_L=R_C//R_L=2\;K\Omega$. The AC load line is a straight line passing the DC operating point $Q$ with its slope equal to $-1/R'_L=-1/2$. The intersections of AC load line with $V_{CE}$ and $I_C$ axes can be found by
  
  $$\frac{1.5}{6-v_0}=-\frac{1}{2}\;\;\;\;\Longrightarrow v_0=9V$$
  
  
  
  
  $$\frac{i_0-1.5}{-6}=-\frac{1}{2}\;\;\;\;\Longrightarrow i_0=4.5\;mA$$
  
  

• Find input voltage and current: Assume input voltage is $v_i(t)=20\; sin\;\omega\;t\;mV$ and $V_{BE}=0.6V$, the overall base voltage is $v_{be}(t)=V_{BE}+v_i=0.6+0.02\;\sin\;\omega\;t\;V$, and the corresponding base current can be found from the input characteristics to be $i_B=I_B+i_i=40+20\;\sin\;\omega\;t\;\mu A$ between 20 and 60 $\mu A$. Also note that $r_{be}=\Delta v_{be}/\Delta i_b =20\mu V/20\mu A=1K\Omega$.
• Find AC output voltage: This can be found graphically from the output i-v characteristics, based on $I_B$, to be $V_o=v_{CE}+v_o=6-1.5 \;sin\;\omega\; t\; V$, and the current is $I_o=i_C+i_o=1.5+ 0.75 \;sin\omega \;t\; mA$. Note that the output is in opposite phase (180 phase shift) with the input.
• Find the voltage gain:
  
  $$A_v=\frac{\vert v_o\vert}{\vert v_i\vert}=\frac{1.5}{0.02}=75$$
  
  

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:

$$I_B=(V_{BB}-V_{BE})/R_B \approx V_{BB}/R_B=40 \;\mu A$$

$$I_C=\beta\;I_B=37.5\times 40\;\mu A=1.5\;mA$$

$$V_{CE}=V_{CC}-R_C I_C=12-1.5\times 4000=6\;V$$

The AC variables:

$$v_i=i_b r_{be}, \;\;\;\;v_o=-i_c R_C//R_L=-\beta i_b R_C//R_L$$

The voltage gain is:

$$A_v=\frac{v_o}{v_i}=-\frac{\beta R_C//R_L}{r_{be}} =-\frac{37.5\times 2K\Omega}{1K\Omega}=-75$$

The input resistance is $R_i=R_B//r_{be}\approx 1K\Omega$, the output resistance is $R_o=R_C//r_{ce}\approx R_C=4\;K\Omega$.