(W) 2003 Analog Integrated Electronics Assignment #2

Transcription

1 (W) 2003 Analog Integrated Electronics Assignment #2 written by Leonard MacEachern, Ph.D. c 2003 by Leonard MacEachern. All Rights Reserved. 1

2 Assignment Guidelines The purpose of this assignment is to practice solving problems related to Miller s Theorem, two-stage amplifiers and the frequency response of switched-capacitor filters, and common-mode rejection ratio. This assignment is due at the start of class on March 28 th, Late assignments will NOT be accepted. Check the course web page for any further information relating to this assignment. 2

3 The mind has exactly the same power as the hands; not merely to grasp the world, but to change it. Colin Wilson Question #1 Miller s theorem 1 is useful for simplifying certain amplifier configurations. A commonly encountered situation is a small signal model with a feedback capacitance. 1.1 Miller s Theorem In 1919, J.M. Miller was trying to calculate the input impedance of vacuum tubes. He noticed that the apparent input capacitance was related to the gain of the tube under certain conditions [1]. In 1968, Cherry and Hooper successfully formalized the relation between the gain and the increased apparent capacitance [2]. The increased capacitance was called the Miller capacitance, and the effect leading to this effective capacitance was called the Miller Effect. In 1979, Millman gave the Miller Effect to the masses, in the form of the Miller Theorem, and the dual to the Miller Theorem, as described in [3]. The Miller Theorem (and its dual to a lesser extent) is a useful tool for analyzing amplifiers that have feedback elements. It is often used when explaining amplifier compensation, and the related phenomenon of polesplitting. As shown in the following sections, the application of Miller s is not always straightforward. First, the derivation of Miller s Theorem is presented. 1 Don t confuse Miller s Theorem with Millman s Theorem. Millman s Theorem concerns circuit simplification through combining of voltage sources. 3

4 Miller s Theorem Derivation Miller s Theorem is shown schematically in Figure 1.1. In Figure 1.1(a), a series impedance exists between the input port and the output port. This series impedance is replaced by two shunt impedances as shown in Figure 1.1(b). Miller s Theorem allows us to choose the correct values of the replacement impedances. i 1 (s) Z fb (s) i 2 (s) i 1 (s) i 2 (s) v 1 (s) v 2 (s) v 1 (s) v 2 (s) - Z 1 (s) Z 2 (s) - - Z 1 (s) Z 2 (s) - (a) A two port circuit representation. There is an impedance in series with the input and output nodes. (b) Using Miller s Theorem, the series impedance between the input and output ports is removed, and shunt impedances are added across the ports to compensate. Figure 1.1: The Miller Theorem transformation. In order for the two-port shown in Figure 1.1(a) to be electrically equivalent to the two-port shown in Figure 1.1(b), the current-voltage response at the input and output ports must be the same. Application of Kirchoff s current law at the positive terminal for the input port of the network shown in Figure 1.1(a) gives the expression, i 1 (s) = v 1(s) Z 1 (s) + v 1(s) v 2 (s). (1.1) Z fb (s) We expect that the same current should flow into the input port of the network shown in Figure 1.1(b) for the same v 1 (s) and v 2 (s). Writing Kirchoff s current law at the positive terminal for the input port of the network shown in Figure 1.1(b) gives the expression, i 1 (s) = v 1(s). (1.2) Z 1 (s) 4

5 Since equations (1.1) and (1.2) must give the same result for i 1 (s), we set them equal and solve for Z 1 (s), yielding, 1 Z 1(s) = 1 ( Z 1 (s) + 1 v ) 2(s) 1 v 1 (s) Z fb (s). (1.3) We can find the expression for Z 2(s) in a similar fashion. Application of Kirchoff s current law at the positive terminal for the output port of the network shown in Figure 1.1(a) gives the expression, i 2 (s) = v 2(s) Z 2 (s) + v 2(s) v 1 (s). (1.4) Z fb (s) The same current should flow into the output port of the network shown in Figure 1.1(b) for the same v 2 (s) and v 1 (s). Writing Kirchoff s current law at the positive terminal for the output port of the network shown in Figure 1.1(b) gives the expression, i 2 (s) = v 2(s). (1.5) Z 2 (s) Equations (1.4) and (1.5) must give the same result for i 2 (s); we set them equal and solve for Z 2(s), yielding, 1 Z 2(s) = 1 ( Z 2 (s) + 1 v ) 1(s) 1 v 2 (s) Z fb (s). (1.6) Taken together, equations (1.3) and (1.6) enable the replacement of the series impedance shown in Figure 1.1(a) with the two shunt impedances shown in Figure 1.1(b). The series impedance shown in Figure 1.1(a) may be entirely due to a capacitance. An alternate form of Miller s Theorem is often used when dealing with capacitances admittances are used instead of impedances. Recall that the admittance, Y (s) is given in terms of impedance as Y (s) = 1/Z(s), and equations (1.3) and (1.6) can be re-written as ( ) Y 1(s) = Y 1 (s) + Y 2 (s) = Y 2(s) + 1 v 2(s) v 1 (s) ( 1 v 1(s) v 2 (s) 5 Y fb (s) (1.7) ) Y fb (s). (1.8)

6 Alternate forms of (1.3) and (1.6)-(1.8) express the impedance and admittance transformations in terms of the two-port gain, A v (s) = v 2 (s)/v 1 (s), yielding, 1 Z 1(s) = 1 Z 1 (s) + (1 A 1 v(s)) Z fb (s) ( 1 1 ) 1 A v (s) Z fb (s) 1 Z 2 (s) = 1 Z 2 (s) + (1.9) (1.10) for the impedance relations, and Y 1 (s) = Y 1(s) + (1 A v (s))y fb (s) (1.11) ( Y 2 (s) = Y 2(s) ) Y fb (s). (1.12) A v (s) for the admittance relations. An interesting alternative derivation of the Miller Theorem equations can be found in [4]. 1.2 The Question Obtain and read a copy of [5]. You can download this paper using a computer on campus by accessing the IEEE database via ieee.org/xplore/dynwel.jsp. Once you ve read the paper, answer the following questions: 1. Figure 1(a) in the paper shows a current source labeled i i and a resistor labeled R i. What might these components represent in an actual circuit? 2. Derive equation (2) from the paper. 3. Derive equations (4a), (4b), and (4c) using Maple or Matlab, or by hand. 4. Explain pole-splitting in your own words, using the discussion in [5] as a reference. Describe the conditions under which pole-splitting occurs, and when it doesn t. 6

7 5. Explain the authors statement a careless application of the Miller effect may give inaccurate results on the bottom of page If you know the dominant pole of an amplifier is at the input node, and a Miller capacitance is added, does the dominant pole stay at the input node? Explain. 7

8 Creative minds have always been known to survive any kind of bad training. Anna Freud Question #2 Use two first order filter sections of the form shown in Figure 2.1 to realize the transfer function with Bodé plots shown in Figure 2.2. Z 2 v i Z 1 - v o + Figure 2.1: Bilinear transfer function building block. Answer the following: 1. What are the poles of the transfer function? 2. What are the zeros of the transfer function? 3. What is the DC gain of the transfer function? 4. What is the transfer function in the Laplace domain? 5. Write the total transfer function as the product of two bilinear transfer functions. 6. Choose an appropriate circuit for Z1 and Z2 for each of the first order stages. For your circuits, find the values of the R s and C s. 8

9 T(jω) db T(jω) 40dB 100 A o ω ω Figure 2.2: Bodé plots for the magnitude of the desired transfer function. 7. Draw the final two-stage filter with Z1 and Z2 for each stage replaced with the appropriate components. Write the value of each component. 8. What kind of filter is this (i.e. bandpass, lowpass, highpass, or band reject)? 9. Does it matter how the total transfer function is partitioned into two bilinear transfer functions? Explain why or why not. 10. Convert your circuit to a switched capacitor circuit. Use a switching frequency of 10MHz. All resistors must be converted to capacitors (to do so, you will need to choose the proper networks for Z1 and Z2). 9

10 When you make the finding yourself - even if you re the last person on Earth to see the light - you ll never forget it. Carl Sagan Question #3 For the cascade current mirror shown in Figure 3.1, determine the low frequency small-signal output resistance (looking into the i out node) assuming all devices operate in saturation, the MOSFET capacitances can be ignored, channel length modulation is negligible, and body effect is negligible. i in i out 1 : B M 3 M 4 1 : B M 1 M 2 Figure 3.1: A cascode current mirror. Obtain and read a copy of [6]. You can download this paper using a computer on campus by accessing the IEEE database via 10

11 ieee.org/xplore/dynwel.jsp. Once you ve read the paper, determine the improvement in CMRR attained when a cascode current mirror is used to bias a differential pair instead of a simple current mirror. The circuit configuration is shown in Figure 3.2. Express the improvement in CMRR as the ratio between the CMRR with a simple current sink and the CMRR with the cascode current sink. Hint: Use equation (8) from [6]. V DD M p1 M p2 v o v + M n1 M n2 v i in i out 1 : B M 3 M 4 1 : B M 1 M 2 Figure 3.2: A cascode current mirror used as the current sink for a differential pair. 11

12 Bibliography [1] J. Miller, Dependence of the input impedance of a three-electrode vacuum tube upon the load in the plate circuit, National Bureau of Standards Scientific Papers, vol. 15, pp , June [2] E. Cherry and D. Hooper, Amplifying Devices and Low-Pass Amplifier Design. New York: Wiley, [3] J. Millman, Microelectronics: Digital and Analog Circuits and Systems. New York: McGraw-Hill, [4] M. Davidovic, A simple proof of miller s theorem, IEEE Transactions on Education, vol. 42, no. 2, pp , May [5] W.-H. Ki, L. Der, and S. Lam, Re-examination of pole splitting of a generic single stage amplifier, IEEE Transactions on Circuits and Systems I: Fundamental Theory and Applications, vol. 44, no. 1, pp. 70 4, Jan [6] R. Jiang, H. Tang, and K. Mayaram, A simple and accurate method for calculating the low frequency common-mode gain in a mos differential amplifier with a currentmirror load, IEEE Transactions on Education, vol. 43, pp , August