Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

Structive to consider a special case where but and we

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structive to consider a special case where but , and we wish to compute the output impedance. As illustrated in Fig. 5.69(b), is equal to in parallel with the impedance seen looking into the collector, . But is identical to the output resistance of an emitter-degenerated common emitter stage, i.e., Fig. 5.46, and hence given by Eq. (5.197): (5.276) It follows that (5.277) The reader may have recognized that the output impedance of the CB stage is equal to that of the CE stage. Is this true in general? Recall that the output impedance is determined by setting the input source to zero. In other words, when calculating , we have no knowledge of the input terminal of the circuit, as illustrated in Fig. 5.70 for CE and CB stages. It is therefore no coincidence that the output impedances are identical if the same assumptions are made for both circuits (e.g., identical values of and emitter degeneration). Example 5.38 Old wisdom says “the output impedance of the CB stage is substantially higher than that of the CE stage.” This claim is justified by the tests illustrated in Fig. 5.71. If a constant current is injected into the base while the collector voltage is varied, exhibits a slope equal to [Fig. 5.71(a)]. On the other hand, if a constant current is drawn from the emitter, displays much less dependence on the collector voltage. Explain why these tests do not represent practical situations.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 236 (1) 236 Chap. 5 Bipolar Amplifiers R out Q 1 V CC R C R E in v Q 1 V CC R C R E out v Q 1 V CC R C in v out v V b R E Q 1 V CC R C R E R out (a) (b) Figure 5.70 (a) CE stage and (b) CB stage simplified for output impedance calculation. I B V V CC 1 I C Q 1 I C V 1 V 1 I C Q 1 I C V 1 I E V B (a) (b) g m π v π v π r r O Open R out g m π v π v π r r O R out Open (c) (d) Figure 5.71 (a) Resistance seens at collector with emitter grounded, (b) resistance seen at collector with an ideal current source in emitter, (c) small-signal model of (a), (d) small-signal model of (b). Solution The principal issue in these tests relates to the use of current sources to drive each stage. From a small-signal point of view, the two circuits reduce to those depicted in Figs. 5.71(c) and (d), with current sources and replaced with open circuits because they are constant. In Fig. 5.71(c), the current through is zero, yielding and hence . On the other hand, Fig. 5.71(d) resembles an emitter-degenerated stage (Fig. 5.46) with an infinite emitter resistance, exhibiting an output resistance of (5.278) (5.279) (5.280) which is, of course, much greater than . In practice, however, each stage may be driven by a voltage source having a finite impedance, making the above comparison irrelevant. Exercise Repeat the above example if a resistor of value is inserted in series with the emitter.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 237 (1) Sec. 5.3 Bipolar Amplifier Topologies 237 Another special case of the topology shown in Fig. 5.69(a) occurs if but .
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