Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

The procedure of progressively simplifying a circuit

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The procedure of progressively simplifying a circuit until it resembles a known topology proves extremely critical in our work. Called “analysis by inspection,” this method obviates the need for complex small-signal models and lengthy calculations. The reader is encouraged to attempt the above example using the small-signal model of the overall circuit to appreciate the efficiency and insight provided by our intuitive approach. Example 5.30 Determine the output resistance of the stage shown in Fig. 5.49(a). Solution Recall from Fig. 5.7 that the impedance seen at the collector is equal to if the base and emitter are (ac) grounded. Thus, can be replaced with [Fig. 5.49(b)]. From another perspective, is reduced to because its base-emitter voltage is fixed by , yielding a zero . Now, plays the role of emitter degeneration resistance for . In analogy with Fig. 5.40(a), we rewrite Eq. (5.200) as (5.210)
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 220 (1) 220 Chap. 5 Bipolar Amplifiers V b1 V b2 Q 1 Q 2 R out Q 1 R out r O2 (a) (b) Figure 5.49 (a) CE stage example, (b) simplified circuit. Called a “cascode” circuit, this topology is studied and utilized extensively in Chapter 9. Exercise Repeat the above example for a “stack” of three transistors. CE Stage with Biasing Having learned the small-signal properties of the common-emitter amplifier and its variants, we now study a more general case wherein the circuit contains a bias network as well. We begin with simple biasing schemes described in Section 5.2 and progres- sively add complexity (and more robust performance) to the circuit. Let us begin with an exam- ple. Example 5.31 A student familiar with the CE stage and basic biasing constructs the circuit shown in Fig. 5.50 to amplify the signal produced by a microphone. Unfortunately, carries no current, failing to amplify. Explain the cause of this problem. Q 1 V CC X = 2.5 V R C 1 k R B 100 k out V Figure 5.50 Microphone amplifier. Solution Many microphones exhibit a small low-frequency resistance (e.g., ). If used in this circuit, such a microphone creates a low resistance from the base of to ground, forming a voltage divider with and providing a very low base voltage. For example, a microphone resistance of 100 yields (5.211) (5.212) Thus, the microphone low-frequency resistance disrupts the bias of the amplifier.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 221 (1) Sec. 5.3 Bipolar Amplifier Topologies 221 Exercise Does the circuit operate better if is halved? How should the circuit of Fig. 5.50 be fixed? Since only the signal generated by the micro- phone is of interest, a series capacitor can be inserted as depicted in Fig. 5.51 so as to isolate the dc biasing of the amplifier from the microphone. That is, the bias point of remains inde- pendent of the resistance of the microphone because carries no bias current. The value of is chosen so that it provides a relatively low impedance (almost a short circuit) for the frequen- cies of interest. We say
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