Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

Solution since and we have 5266 exercise compare this

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Solution Since and , we have (5.266) Exercise Compare this value with that obtained for the CE stage. From Eqs. (5.256) and (5.266), we conclude that the CB stage exhibits a set of trade-offs similar to those depicted in Fig. 5.33 for the CE amplifier. It is instructive to study the behavior of the CB topology in the presence of a finite source resistance. Shown in Fig. 5.67, such a circuit suffers from signal attenuation from the input to node , thereby providing a smaller voltage gain. More specifically, since the impedance seen looking into the emitter of (with the base grounded) is equal to (for ), we have Q 1 R C V CC out in R S g m 1 v v X in R S v g m 1 X Figure 5.67 CB stage with source resistance. (5.267) (5.268) We also recall from Eq. (5.254) that the gain from the emitter to the output is given by (5.269)
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 234 (1) 234 Chap. 5 Bipolar Amplifiers It follows that (5.270) (5.271) a result identical to that of the CE stage (except for a negative sign) if is viewed as an emitter degeneration resistor. Example 5.37 A common-base stage is designed to amplify an RF signal received by a 50- antenna. De- termine the required bias current if the input impedance of the amplifier must “match” the impedance of the antenna. What is the voltage gain if the CB stage also drives a 50- load? Assume . Solution Figure 5.68 depicts the amplifier and the equivalent circuit with the antenna modeled by a Q 1 R C V CC out v Antenna in R S v V B Q 1 R C V CC out v V B Antenna Figure 5.68 (a) CB stage sensing a signal received by an antenna, (b) equivalent circuit. voltage source, , and a resistance, . For impedance matching, it is necessary that the input impedance of the CB core, , be equal to , and hence (5.272) (5.273) If itself is replaced by a 50- load, then Eq. (5.271) reveals that (5.274) (5.275) The circuit is therefore not suited to driving a 50- load directly. The dots denote the need for biasing circuitry, as described later in this section.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 235 (1) Sec. 5.3 Bipolar Amplifier Topologies 235 Exercise What is the voltage gain if a 50- resistor is also tied from the emitter of to ground? Another interesting point of contrast between the CE and CB stages relates to their cur- rent gains. The CB stage displays a current gain of unity because the current flowing into the emitter simply emerges from the collector (if the base current is neglected). On the other hand, as mentioned in Section 5.3.1, for the CE stage. In fact, in the above example, , which upon flowing through , yields . It is thus not surprising that the voltage gain does not exceed 0.5 if . As with the CE stage, we may desire to analyze the CB topology in the general case: with emitter degeneration, , and a resistance in series with the base [Fig. 5.69(a)]. Outlined in Problem 64, this analysis is somewhat beyond the scope of this book. Nevertheless, it is in- Q 1 R C V CC out in R v v V B R B E r O Q 1 R E r O R out1 R C R out2 (a) (b) Figure 5.69 (a) General CB stage, (b) output impedance seen at different nodes.
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