## Chapter 12

## Q. 12.8

**Common-Emitter Amplifier**

Find A_v, A_{voc}, Z_{in}, A_i, G, \text{ and } Z_o for the amplifier shown in Figure 12.28. If v_s(t) = 0.001 \sin(\omega t), find and sketch v_o(t) versus time. Assume that the circuit operates at a temperature for which V_T = 26 \text{ mV}.

## Step-by-Step

## Verified Solution

First, we need to know I_{CQ} to be able to find the value of r_{\pi}. Hence, we start by analyzing the dc conditions in the circuit. Only the dc supply, the transistor, and the resistors R_1 , R_2 , R_C, \text{ and } R_E need to be considered in the bias-point analysis.

The capacitors, the signal source, and the load resistance have no effect on the Q point (because the capacitors behave as open circuits for dc currents).

The dc circuit was shown earlier in Figure 12.23 and was analyzed in Example 12.7. For β = 100, the resulting Q point was found to be I_{CQ} = 4.12 \text{ mA and }V_{CE} = 6.72 \text{ V}. Substituting values into Equation 12.35, we have

r_{\pi}=\frac{\beta V_T}{I_{CQ}}=631 \ \Omega

Using Equations 12.39 and 12.40, we find that

R_B=R_1\parallel R_2=\frac{1}{1/R_1+1/R_2} =3.33 \text{ k}\Omega \\ R^\prime _L=R_L\parallel R_C=\frac{1}{R_L+1/R_C} =667 \ \Omega

Equations 12.43 through 12.48 yield

\begin{matrix} A_V &=& \frac{v_o}{v_{\text{in}}}=-\frac{R^\prime_L}{r_{\pi}}=-106 \\ A_{voc}&=&\frac{v_o}{v_{\text{in}}}=-\frac{R_C\beta}{r_{\pi}}=-158 \\ Z_{in}&=&\frac{v_{\text{in}}}{i_{\text{in}}}=\frac{1}{1/R_B+1/r_{\pi}}=531 \ \Omega \\ A_i&=&\frac{i_o}{i_{\text{in}}}=A_v\frac{Z_{\text{in}}}{R_L}=-28.1\\ G&=&A_iA_v=2980 \\ Z_o&=&R_C=1 \text{ K} \Omega \end{matrix}

The common-emitter amplifier is inverting and has large voltage gain magnitude, large current gain, and large power gain.

Notice that A_v is somewhat smaller in magnitude than A_{voc}. This is due to loading of the amplifier by R_L as discussed in Chapter 10. Power gain is quite large for the common-emitter amplifier, and primarily for this reason, it is a commonly used configuration.

The source voltage divides between the internal source resistance and the input impedance of the amplifier. Thus, we can write

V_{\text{in}}=v_s\frac{Z_{\text{in}}}{Z_{\text{in}}+R_s} =0.515v_s

Now, with the load connected, we have

v_o=A_vV_{\text{in}}=-54.6v_s

But, we are given that v_s(t) = \sin(\omega t) \text{ mV}, so we have

v_o(t)=-54.6 \sin(\omega t)\text{ mV}

The source voltage v_s(t) and the output voltage are shown in Figure 12.29. Notice the phase inversion.