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Chapter 12

Q. 12.9

Emitter-Follower Performance
Compute the voltage gain, input impedance, current gain, power gain, and output impedance for the emitter-follower amplifier shown in Figure 12.31. Assume that the circuit operates at a temperature for which V_T = 26 \text{ mV}.

12.31

Step-by-Step

Verified Solution

First, we must determine the bias point so the value of r_{\pi} can be calculated.

First, we must find the bias point so that the value of r_{\pi} can be calculated.
The dc circuit is shown in Figure 12.31(b). Because the coupling capacitors act as open circuits for dc, R_s \text{ and } R_L do not appear in the dc bias circuit.
Replacing the base bias circuit by its Thévenin equivalent, we obtain the equivalent circuit shown in Figure 12.31(c). Now, if we assume operation in the active region, we can write the following voltage equation around the base loop:

V_B=R_BI_{BQ}+V_{BEQ}+R_E(1+ \beta)I_{BQ}

Substituting values, we find I_{BQ} = 20.6 \text{ mA}. Then, we have

\begin{matrix} I_{CQ}&=&\beta I_{BQ}=4.12 \text{ mA} \\ V_{CEQ}&=&V_{CC}-I_{EQ}R_E=11.7 \text{ V} \end{matrix}

Since V_{CEQ} is greater than 0.2 V and I_{BQ} is positive, the transistor is indeed operating in the active region.
Equation 12.35 yields

r_{\pi}=\frac{\beta V_T}{I_{CQ}}=1260 \ \Omega

Now that we have established that the transistor operates in the active region and found the value of r_\pi, we can proceed in finding the amplifier gains and impedances. Substituting values into Equations 12.49 and 12.50, we discover that

R_B=R_1\parallel R_2=\frac{1}{1/R_1+1/R_2} = 50 \text{ k}\Omega \\ R^\prime _L=R_L\parallel R_E=\frac{1}{1/R_L+1/R_E}=667 \ \Omega

Equation 12.53 gives the voltage gain:

A_v=\frac{(1+ \beta)R^\prime _L}{r_{\pi}+(1+ \beta)R^\prime_L}=0.991

Equations 12.54 and 12.55 give the input impedance:

Z_{it}=r_{\pi}+(1+\beta)R^\prime_L=135 \text{ k}\Omega \\ Z_i=\frac{1}{1/R_B+1/Z_{it}}=36.5 \text{ k}\Omega

Equations 12.58 and 12.60 yield

R^\prime_s=\frac{1}{1/R_s+1/R_1+1/R_2}=8.33 \text{ k}\Omega \\ Z_o=\frac{1}{(1+ \beta)/(R^\prime_s+r_{\pi})+1/R_E}=46.6 \ \Omega

From Equation 10.3, we can find the current gain:

A_i=A_v\frac{Z_i}{R_L}=36.2

Using Equation 10.5, we find that the power gain is

G=A_vA_i=35.8

Even though the voltage gain of the emitter follower is less than unity, the current gain and power gain can be large.

Notice that even though the voltage gain is less than unity, the current gain is large (compared with unity). Thus, the output power is larger than the input power, and the circuit is effective as an amplifier.