# Ω a 0 db ω ω a figure 129 exercise if mhz and

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ω A 0 0 dB ω 0 ω 0 A 0 Figure 12.9 Exercise If MHz, and , calculate the unity-gain bandwidth from Eqs. (12.24)and (12.26) and compare the results. 12.2.3 Modification of I/O Impedances As mentioned above, negative feedback makes the closed-loop gain less sensitive to the load resistance. This effect fundamentally arises from the modification of the output impedance as a result of feedback. Feedback modifies the input impedance as well. We will formulate these effects carefully in the following sections, but it is instructive to study an example at this point. Example 12.7 Figure 12.10 depicts a transistor-level realization of the feedback circuit shown in Fig. 12.2.

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 611 (1) Sec. 12.2 Properties of Negative Feedback 611 Assume and for simplicity. (a) Identify the four components of the feedback system. (b) Determine the open-loop and closed-loop voltage gain. (c) Determine the open-loop and closed-loop I/O impedances. in V R R V out M 1 R D V DD 1 2 Figure 12.10 Solution (a) In analogy with Fig. 12.10, we surmise that the forward system (the main amplifier) con- sists of and , i.e., a common-gate stage. Resistors and serve as both the sense mechanism and the feedback network, returning a signal equal to to the sub- tractor. Transistor itself operates as the subtractor because the small-signal drain current is proportional to the difference between the gate and source voltages: (12.27) (b) The forward system provides a voltage gain equal to (12.28) because is large enough that its loading on can be neglected. The closed-loop voltage gain is thus given by (12.29) (12.30) We should note that the overall gain of this stage can also be obtained by simply solving the cir- cuit’s equations—as if we know nothing about feedback. However, the use of feedback concepts both provides a great deal of insight and simplifies the task as circuits become more complex. (c) The open-loop I/O impedances are those of the CG stage: (12.31) (12.32) At this point, we do not know how to obtain the closed-loop I/O impedances in terms of the open-loop parameters. We therefore simply solve the circuit. From Fig. 12.11(a), we recognize that carries a current approximately equal to because is assumed large. The drain voltage of is thus given by , leading to a gate voltage equal to . Transistor generates a drain current proportional to : (12.33)
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 612 (1) 612 Chap. 12 Feedback (12.34) Since , (12.34) yields (12.35) That is, the input resistance increases from by a factor equal to , the same factor by which the gain decreases. R R M 1 R D V DD 1 2 X v X i R R M 1 R D V DD 1 2 X X i v (a) (b) Figure 12.11 To determine the output resistance, we write from Fig. 12.11(b), (12.36) and hence (12.37) (12.38) Noting that, if , then , we obtain (12.39) It follows that (12.40) The output resistance thus decreases by the “universal” factor .

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