Tied between the input and output of an inverting

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tied between the input and output of an inverting amplifier with a gain of raises the input capacitance by an amount equal to . We say such a circuit suffers from “Miller multiplication” of the capacitor. A 0 out V in V A 0 out V in V A 0 C F ( ( 1 + A 0 C F ( ( 1 + 1 (a) (b) C F V A 0 V Figure 11.14 (a) Inverting circuit with floating capacitor, (b) equivalent circuit as obtained from Miller’s theorem. The effect of at the output can be obtained from (11.24): (11.27) (11.28) which is close to if . Figure 11.14(b) summarizes these results. The Miller multiplication of capacitors can also be explained intuitively. Suppose the input voltage in Fig. 11.14(a) goes up by a small amount . The output then goes down by . That is, the voltage across increases by , requiring that the input provide a

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 548 (1) 548 Chap. 11 Frequency Response proportional charge. By contrast, if were not a floating capacitor and its right plate voltage did not change, it would experience only a voltage change of and require less charge. The above study points to the utility of Miller’s theorem for conversion of floating capacitors to grounded capacitors. The following example demonstrates this principle. Example 11.10 Estimate the poles of the circuit shown in Fig. 11.15(a). Assume . R M 1 V DD D in V R S C out V F R M 1 V DD D in out C V V R C in S out (a) (b) Figure 11.15 Solution Noting that and constitute an inverting amplifier having a gain of , we utilize the results in Fig. 11.14(b) to write: (11.29) (11.30) and (11.31) thereby arriving at the topology depicted in Fig. 11.15(b). From our study in Example 11.8, we have: (11.32) (11.33) and (11.34) (11.35)
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 549 (1) Sec. 11.1 Fundamental Concepts 549 Exercise Calculate if k , and fF. The reader may find the above example somewhat inconsistent. Miller’s theorem requires that the floating impedance and the voltage gain be computed at the same frequency whereas Example 11.10 uses the low-frequency gain, , even for the purpose of finding high-frequency poles. After all, we know that the existence of lowers the voltage gain from the gate of to the output at high frequencies. Owing to this inconsistency, we call the procedure in Example 11.10 the “Miller approximation.” Without this approximation, i.e., if is expressed in terms of circuit parameters at the frequency of interest, application of Miller’s theorem would be no simpler than direct solution of the circuit’s equations. Another artifact of Miller’s approximation is that it may eliminate a zero of the transfer func- tion. We return to this issue in Section 11.4.3. The general expression in Eq. (11.22) can be interpreted as follows: an impedance tied be- tween the input and output of an inverting amplifier with a gain of is lowered by a factor of if seen at the input (with respect to ground). This reduction of impedance (hence in- crease in capacitance) is called “Miller effect.” For example, we say Miller effect raises the input capacitance of the circuit in Fig. 11.15(a) to .

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