# Summary and additional examples 255 example 550

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Summary and Additional Examples 255 Example 5.50 Assuming , compute the voltage gain and input impedance of the circuit shown in Fig. 5.99(a). in v Q 1 I 1 Q 2 V R CC out V CC R 1 V B R eq in v Q 1 out R eq (a) (b) v v C R C Figure 5.99 (a) Example of CE stage, (b) simplified circuit. Solution The circuit resembles a CE stage (why?) degenerated by the impedance seen at the emitter of , . Recall from Fig. 5.75 that (5.347) The simplified model in Fig. 5.99(b) thus yields (5.348) (5.349) The input impedance is also obtained from Fig. 5.75: (5.350) (5.351) Exercise Repeat the above example if is placed in series with the emitter of . Example 5.51 Calculate the voltage gain of the circuit in Fig. 5.100(a) if .

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 256 (1) 256 Chap. 5 Bipolar Amplifiers R 2 Q 1 R C V CC C B R 1 in v R S C 1 Q 1 in v R S R C R out v out v 1 (a) (b) Figure 5.100 (a) Example of CB stage, (b) simplified circuit. Solution Since the base is at ac ground, appears in parallel with and is shorted to ground on both ends [Fig. 5.100(b)]. The voltage gain is given by (5.271), but with replaced by : (5.352) Exercise What happens if is replaced by an ideal currents source? Example 5.52 Determine the input impedance of the circuit shown in Fig. 5.101(a) if . Q 1 R C V CC R out v (a) (b) Q R E R eq 2 R in Q 1 R C R R in eq B Figure 5.101 (a) Example of CB stage, (b) simplified circuit. Solution In this circuit, operates as a common-base device (why?) but with a resistance in series with its base [Fig. 5.101(b)]. To obtain , we recognize that resembles an emitter follower, e.g., the topology in Fig. 5.91(a), concluding that can be viewed as the output resistance of such a stage, as given by Eq. (5.329): (5.353)
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 257 (1) Sec. 5.4 Summary and Additional Examples 257 Now, from Fig. 5.101(b), we observe that contains two components: one equal to the resis- tance in series with the base, , divided by , and another equal to : (5.354) (5.355) The reader is encouraged to obtain through a complete small-signal analysis and compare the required “manual labor” to the above algebra. Exercise What happens if the current gain of goes to infinity? Example 5.53 Compute the voltage gain and the output impedance of the circuit depicted in Fig. 5.102(a) with . Q 1 V CC out v in v R 1 R 2 C 2 (a) X R E R S Q 1 out v in v R 1 R S R R E 2 r O (b) v R R R E 2 r O out v Thev Thev Q 1 (c) Figure 5.102 (a) Example of emitter follower, (b) circuit with shorted, (c) simplified circuit. Solution Noting that is at ac ground, we construct the simplified circuit shown in Fig. 5.102(b), where the output resistance of is explicitly drawn. Replacing , , and with their Thevenin equivalent and recognizing that , , and appear in parallel [Fig. 5.102(c)], we employ Eq. (5.330) and write (5.356)

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 258 (1) 258 Chap. 5 Bipolar Amplifiers and hence (5.357) For the output resistance, we refer to Eq. (5.332): (5.358) (5.359) Exercise What happens if ?
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