Question 13.9: Emitter-Follower Performance Compute the voltage gain, input...

First, we must find the bias point so that the value of r_π can be calculated. The dc circuit is shown in Figure 13.31(b). Because the coupling capacitors act as open circuits for dc, R_s 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 13.31(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 μA. Then, we have

I_{CQ}=\beta I_{BQ}=4.12\mathrm{~mA}

V_{CEQ}=V_{CC}-I_{EQ}R_E=11.7\mathrm{~V}

Since V_CEQ is greater than 0.2Vand I_BQ is positive, the transistor is indeed operating in the active region.
Equation 13.35 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_π ,we can proceed in finding the amplifier gains and impedances. Substituting values into Equations 13.49 and 13.50, we discover that

R_B=R_1\|R_2=\frac{1}{1/R_1+1/R_2}=50\mathrm{~k}\Omega

R^′_L=R_L\|R_E=\frac{1}{1/R_L+1/R_E}=667~\Omega

Equation 13.53 gives the voltage gain:

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

Equations 13.54 and 13.55 give the input impedance:

Z_i=\frac{1}{1/R_B+1/Z_{it}}=R_B\|Z_{it}      (13.54)

Z_{it}=\frac{v_{\mathrm{in}}}{i_b}=r_{\pi}+(1+\beta)R^′_L         (13.55)

Z_{it}=r_{\pi}+(1+\beta)R^′_L=135\mathrm{~k}\Omega

Z_i=\frac{1}{1/R_B+1/Z_{it}}=36.5\mathrm{~k}\Omega

Equations 13.58 and 13.60 yield

Z_{\mathrm{o}}=\frac{v_x}{i_x}=\frac{1}{(1+\beta)/(R^′_s+r_{\pi})+1/R_E}          (13.60)

R^′_s=\frac{1}{1/R_s+1/R_1+1/R_2}=8.33\mathrm{~k}\Omega

Z_{\mathrm{o}}=\frac{1}{(1+\beta)/(R^′_s+r_{\pi})+1/R_E}=46.6~\Omega

From Equation 11.3, we can find the current gain:

A_i=\frac{i_o}{i_i} =\frac{v_o/R_L}{v_i/R_i}=A_v\frac{R_i}{R_L}       (11.3)

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

Using Equation 11.5, we find that the power gain is

G=\frac{P_{o}}{P_i}=\frac{V_oI_o}{V_iI_i}=A_vA_i=(A_v)^2\frac{R_i}{R_L}         (11.5)

G=A_vA_i=35.8

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.

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