Low Frequency Brute Force Approach

Consider the CE amp circuit in Fig.4.2.

  

Figure 4.2: AC Coupled Common Emitter Amp.

To analyze this circuit, we can divide it into the three sections as shown in the figure. It is also useful to replace stage 2 with its amplifier equivalent circuit as shown in lab 2. The small signal equivalent circuit diagram which results is shown in Fig. 4.3.

  

Figure 4.3: AC Coupled Amp with Equivalent Voltage Amp Circuit

Since R_in, A, and R_out are already known from the Lab 2, finding the gain just becomes a simple exercise in analyzing a two voltage dividers. More specifically, we can write the voltage gain as the product of the gains of each of the three stages. So, the overall gain is $\frac{v_{out}}{v_{in}}= \frac{v_1}{v_{in}}\frac{Av_1}{v_1} \frac{v_{out}}{Av_1}$.We now take each stage individually. The voltage divider, which is the first stage, gives:

$$\frac{v_1}{v_{in}}= \frac{R_{in}}{R_{in}+\frac{1}{sC_1}}$$ (67)

Recall from Lab 2 that the gain A of CE amp is

$$A=\frac{-R_C}{R_{E}+\frac{1}{g_m}}$$ (68)

The gain of the third stage is also a voltage divider which gives

$$\frac{v_o}{Av_1}= \frac{R_L}{R_{L}+R_C+\frac{1}{sC_2}}$$ (69)

The total gain is thus the product of the individual gains of the three stages:  

$$\frac{v_{out}}{v_{in}}= \left(\frac{R_{in}}{R_{in}+\frac{1}... ...{g_m}}\right) \left(\frac{R_L}{R_{L}+R_C+\frac{1}{sC_2}}\right)$$ (70)

Where, we found from Lab 2 that $R_{in}=R_B\vert\vert(r_{\pi}+\beta R_E)$, and R_o=R_C.