Question 5.14: The two-stage system of Fig. 5.68 employs a transistor emitt...
The two-stage system of Fig. 5.68 employs a transistor emitter-follower
configuration prior to a common-base configuration to ensure that the maximum percentage of the applied signal appears at the input terminals of the common-base amplifier. In Fig. 5.68 , the no-load values are provided for each system, with the exception of Z_i and Z_o for the emitter-follower, which are the loaded values. For the configuration of Fig. 5.68 , determine:
a. The loaded gain for each stage.
b. The total gain for the system, A_v andA_{v_{s}}.
c. The total current gain for the system.
d. The total gain for the system if the emitter-follower configuration were removed.
a. For the emitter-follower configuration, the loaded gain is (by Eq. (5.94))
A_{v_{L}}=\frac{V_{o}}{V_{i}}=\frac{R_{L} A_{v_{ NL }}}{R_{L}+R_{o}}=\frac{R_{L}}{R_{L}+R_{o}} A_{v_{ NL }} (5.94))
V_{o_{1}}=\frac{Z_{i_{2}}}{Z_{i_{2}}+Z_{o_{1}}} A_{v_{ NL }} V_{i_{1}}=\frac{26 \Omega}{26 \Omega+12 \Omega} \text { (1) } V_{i_{1}}=0.684 V_{i_{1}}
and A_{V_{i}}=\frac{V_{o_{1}}}{V_{i_{1}}}= 0 . 6 8 4
For the common-base configuration,
V_{o_{2}}=\frac{R_{L}}{R_{L}+R_{o_{2}}} A_{v_{ NL }} V_{i_{2}}=\frac{8.2 k \Omega}{8.2 k \Omega+5.1 k \Omega}(240) V_{i_{2}}=147.97 V_{i_{2}}and A_{v_{2}}=\frac{V_{o_{2}}}{V_{i_{2}}}=1 4 7 . 9 7
b. Eq. (5.99):
A_{v_{T}}=A_{v_{1}} \cdot A_{v_{2}} \cdot A_{v_{3}} \cdots (5.99)
A_{v_{T}}=A_{v_{1}} A_{v_{2}}
= (0.684)(147.97)
= 101.20
Eq. (5.91):
V_{i}=\frac{R_{i} V_{s}}{R_{i}+R_{s}}A_{v_{s}}=\frac{Z_{i_{1}}}{Z_{i_{1}}+R_{s}} A_{v_{T}}=\frac{(10 k \Omega)(101.20)}{10 k \Omega+1 k \Omega}
= 92
c. Eq. (5.100): A_{i_{T}}=-A_{v_{T}} \frac{Z_{i_{1}}}{R_{L}}=-(101.20)\left(\frac{10 k \Omega}{8.2 k \Omega}\right)
= – 123.41
d. Eq. (5.91): V_{i}=\frac{Z_{i_{C B}}}{Z_{i_{C B}}+R_{s}} V_{s}=\frac{26 \Omega}{26 \Omega+1 k \Omega} V_{s}=0.025 V_{s}
and \frac{V_{i}}{V_{s}}=0.025 with \frac{V_{o}}{V_{i}}=147.97 from above
and A_{v_{s}}=\frac{V_{o}}{V_{s}}=\frac{V_{i}}{V_{s}} \cdot \frac{V_{o}}{V_{i}}=(0.025)(147.97)= 3 . 7
In total, therefore, the gain is about 25 times greater with the emitter-follower configuration to draw the signal to the amplifier stages. Note, however, that it is also important that the output impedance of the first stage is relatively close to the input impedance of the second stage, otherwise the signal would have been “lost” again by the voltage-divider action.