Voltage Amplifier Equivalent Circuit

Now that we have determined the voltage gain, as well as the input and output resistances of the CE amp, it can be represented as a simple two-port circuit in Fig 2.9 The nice thing about this equivalent circuit is that once it's established, you don't have to worry about the details of the circuit operation, but can just treat it as a black box with a specific open circuit output voltage, an input resistance and an output resistance. We can therefore write the CE amp at midband frequencies as a black box with $R_{in}=(r_{\pi}+\beta R_E)\vert\vert R_B$; R_out=R_C; and the open circuit output voltage given by v_out=A_Vv_in.

  

Figure 2.9: Two-Port Equivalent Amplifier Circuit

Example:

To appreciate the usefulness of the two-port amplifier equivalent circuit, consider the following example. Suppose you have a CE amp that is driven by a voltage source with source resistance R_S, and the amp is supplying current to a load R_L. The actual circuit is drawn below.

  

Figure 2.10: CE amp driven by nonideal source of resistance R_S and driving a load R_L

If we replace the small signal part of the CE amp with its two-port amp equivalent circuit, we obtain the small signal equivalent amplifier circuit shown in Fig. 2.11.

  

Figure 2.11: Equivalent circuit of CE amp described by voltage amplifier

We can easily find the voltage gain by dividing the circuit into three stages and finding the gain of each stage. The total gain is then the product of the three individual stage gains.

The first stage is just the voltage divider consisting of v_S, R_S, and R_in.

$$A_{V1}=\frac{v_1}{v_s}=\frac{R_{in}}{R_S+R_{in}}$$ (44)

Where V₁ is the voltage seen at the input of the CE two-port amp equivalent circuit.

The second stage is the CE two-port amp equivalent circuit.

$$A_{V2}=\frac{V_2}{V_1}= \frac{-R_C}{\frac{1}{g_m}+R_E}$$ (45)

Where V₂ is the open circuit voltage at the CE amp output that would be measured if the load R_L were not there.

The final stage is the voltage divider consisting of V₂, R_out, and R_L.

$$A_{V3}=\frac{v_{out}}{V_2}=\frac{R_L}{R_L+R_{out}}$$ (46)

Finally, the total gain of the circuit A_V is the product of the three stages:  

$$A_V=\frac{v_{out}}{v_s}=A_{V1}A_{V2}A_{V3}= \left( \frac{R_{... ...rac{1}{g_m}+R_E}\right ) \left (\frac{R_L}{R_L+R_{out}}\right )$$ (47)