Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

Example 92 suppose in example 91 the early voltage of

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Example 9.2 Suppose in Example 9.1, the Early voltage of is equal to 50 V. Compare the resulting output impedance of the cascode with the upper bound given by Eq. (9.12). In integrated circuits, all bipolar transistors fabricated on the same wafer exhibit the same Early voltage. This example applies to discrete implementations.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 422 (1) 422 Chap. 9 Cascode Stages and Current Mirrors Solution Since , , , and , we have (9.13) (9.14) The upper bound is equal to 500 k , about higher. Exercise Repeat the above example if the Early voltage of is 10 V. Example 9.3 We wish to increase the output resistance of the bipolar cascode of Fig. 9.2(a) by a factor of two through the use of resistive degeneration in the emitter of . Determine the required value of the degeneration resistor if and are identical. Solution As illustrated in Fig. 9.4, we replace and with their equivalent resistance from (9.1): V b R Q 1 V R Q 1 V Q 2 b1 b2 out out R E R outA R outA Figure 9.4 . (9.15) It follows from (9.7) that (9.16) We wish this value to be twice that given by (9.7): (9.17) That is, (9.18) In practice, is typically less than , and no positive value of exists! In other words, it is impossible to double the output impedance of the cascode by emitter degeneration.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 423 (1) Sec. 9.1 Cascode Stage 423 Exercise Is there a solution if the output impedance must increase by a factor of 1.5? What does the above result mean? Comparing the output resistances obtained in Examples 9.1 and 9.2, we recognize that even identical transistors yield an ( ) that is not far from the upper bound ( ). More specifically, the ratio of (9.7) and (9.12) is equal to , a value greater than 0.5 if . For completeness, Fig. 9.5 shows a cascode, where serves as the cascode device and as the degeneration device. The output impedance is given by (9.5). V Q 1 b1 V Q b2 V CC R out 2 Figure 9.5 PNP cascode current source. While we have arrived at the cascode as an extreme case of emitter degeneration, it is also possible to view the evolution as illustrated in Fig. 9.6. That is, since provides only an output impedance of , we “stack” on top of it to raise . V R Q 1 V Q 2 b1 b2 out R V Q 2 b2 out r = O2 g r m1 O1 ( r π r ) O2 1 Figure 9.6 Evolution of cascode topology viewed as stacking atop . Example 9.4 Explain why the topologies depicted in Fig. 9.7 are not cascodes. (a) (b) R Q 1 V R Q 1 V Q 2 b1 b2 r O2 out out X g 1 m2 V R Q V Q b1 b2 out X 2 1 V CC R Q out 1 r O2 g 1 m2 V b1 V b2 V CC Figure 9.7
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 424 (1) 424 Chap. 9 Cascode Stages and Current Mirrors Solution Unlike the cascode of Fig. 9.2(a), the circuits of Fig. 9.7 connect the emitter of to the emitter of . Transistor now operates as a diode-connected device (rather than a current source), thereby presenting an impedance of (rather than ) at node . Given by (9.1), the output impedance, , is therefore considerably lower: (9.19) In fact, since , and since (why?), (9.20) (9.21) The same observations apply to the topology of Fig. 9.7(b).
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