Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

What happens to the phase after compensation as shown

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What happens to the phase after compensation? As shown in Fig. 12.72(b), the low-frequency section of changes because is moved to , but the critical section near does not. Consequently, the phase margin increases. Example 12.42 The amplifier shown in Fig. 12.73(a) employs a cascode stage and a CS stage. Assuming that the pole at node is dominant, sketch the frequency response and explain how the circuit can be “compensated.” Solution Recall from Chapter 11 that the cascode stage exhibits one pole arising from node and an- other from node , with the latter falling much closer to the origin. We can express these poles respectively as (12.192) (12.193) where and denote the total capacitance seen at each node to ground. The third pole is associated with the output node: (12.194) where represents the total capacitance at the output node. We assume . The frequency response of the amplifier is plotted in Fig. 12.73(b). To compensate the circuit for use in a feedback system, we can add capacitance to node so as to reduce . If is sufficiently large, the gain crossover occurs well below the phase crossover, providing adequate phase margin [Fig. 12.73(c)]. An important observation here is that frequency compensation inevitably degrades the speed because the dominant pole is reduced in magnitude.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 664 (1) 664 Chap. 12 Feedback V DD M 1 M M V b1 V b2 V b3 4 3 M 2 M in V M V out V 5 6 A B X C b4 comp ω 0 ω 0 ω 90 180 270 (log scale) (log scale) ω p,B ω 20log KH KH ω 0 ω 0 ω 90 180 270 (log scale) (log scale) ω p,B ω 20log KH KH ω p,B (c) (a) (b) p,out p,A p,out p,A Figure 12.73 Exercise Repeat the above example if and are omitted, i.e., the first stage is a simple CS amplifier. We now formalize the procedure for frequency compensation. Given the frequency response of an amplifier and the desired phase margin, we begin from the phase response and determine the frequency, , at which (Fig. 12.74). We then mark on the magnitude response and draw a line having a slope of 20 dB/dec toward the vertical axis. The point at which this line intercepts the original magnitude response represents the new position of the dominant pole, . By design, the compensated amplifier now exhibits a gain crossover at . Example 12.43 A multipole amplifier exhibits the frequency response plotted in Fig. 12.75(a). Assuming the poles are far apart, compensate the amplifier for a phase margin of with .
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 665 (1) Sec. 12.8 Stability in Feedback Systems 665 ω 0 ω 0 ω ω 20 dB/dec 20log KH KH + PM 180 ω PM p1 p1 (log scale) (log scale) Figure 12.74 Systematic method for frequency compensation. 0 0 ω p1 ω p2 135 ω ω (log scale) (log scale) H H 20log ω p1 Figure 12.75 Solution Since the phase reaches at , in this example . We thus draw a line with a slope of dB/dec from toward the vertical axis [Fig. 12.75(b)]. The dominant pole must therefore be translated to . Since this phase margin is generally inadequate, in practice, .
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