EE 508 Lecture 31. Switched Current Filters

Transcription

1 EE 508 Lecture 31 Switched Current Filters

2 Review from last time Current-Mode Filters I IN I 0 I 0 s+αi0 s I OUT T s 2 IOUT I 0 I 2 2 IN s + I0s+I 0 Basic Concepts of Benefits of Current-Mode Filters: Large voltage swings difficult to maintain in integrated processes because of linearity concerns Large voltage swings slow a circuit down because of time required to charge capacitors Voltage swings can be very small when currents change Current swings are not inherently limited in integrated circuits (only voltage swings) With low voltage swings, current-mode circuits should dissipate little power

3 Review from last time Cumulative Total Current-Mode Filters Number of Papers / / / / / / / / / /09 10 Years Steady growth in research in the area since 1990 and publication rate is growing with time!!

4 Review from last time Current-Mode Filters The Conventional Wisdom: Current-Mode circuits operate at higherfrequencies than voltage-mode counterparts Current-Mode circuits operate at lower supply voltages and lower power levels than voltagemode counterparts Current-Mode circuits are simpler than voltage-mode counterparts Current-Mode circuits offer better linearity than voltage-mode counterparts This represents four really significant benefits of current-mode circuits!

5 Review from last time Some Current-Mode Integrators OTA-C I IN C I OUT I IN C I OUT Noninverting gm I OUT= IIN Cs -gm I OUT= IIN Cs Inverting Summing inputs really easy to obtain Loss is easy to add Same component count as voltage-mode integrators Many argue that since only interested in currents, can operate at lower voltages and higher frequencies

6 Review from last time Current-Mode Two Integrator Loop I IN RQ C R C R A R A R R L I OUT CM Lossy Integrator CM Integrator CM Amplifier Straightforward implementation of the two-integrator loop Simple structure

7 Review from last time Current-Mode Two Integrator Loop An Observation: I IN RQ C R C RA RA R RL I OUT VM Integrator VM Amplifier I IN RQ C R C R A R A R R L VM Integrator VM Amplifier I OUT VM Integrator This circuit is identical to another one with two voltage-mode integrators and a voltage-mode amplifier!

8 Review from last time Observation Many papers have appeared that tout the performance advantages of current-mode circuits In all of the current-mode papers that this instructor has seen, no attempt is made to provide a quantitative comparison of the key performance features of current-mode circuits with voltage-mode counterparts All justifications of the advantages of the current-mode circuits this instructor has seen are based upon qualitative

9 Review from last time Observations (cont.) It appears easy to get papers published that have the term current-mode in the title Over 900 papers have been published in IEEE forums alone! Some of the current-mode filters published perform better than other voltage-mode filters that have been published We are still waiting for even one author to quantitatively show that current-mode filters offer even one of the claimed four advantages over their voltage-mode counterparts Will return to a discussion of Current-Mode filters later

10 Review from last time Switched-Current Filters Basic idea introduced by Hughes and Bird at ISCAS 1989 Technique introduced directly in the z-domain

11 Review from last time Switched-Current Filters Basic idea introduced by Hughes and Bird at ISCAS 1989 I IN V DD I OUT If Φ 1 is a periodic signal and if I IN is also appropriately clocked, the input/output currents of this circuit can be represented with the difference equation I nt = AI nt-t OUT IN φ 1 M 1 M 2 :A AI t for φ closed IN 1 I OUT= AI IN T SW for φ 1 open This switched mirror becomes a delay element Gain A is that of a current mirror A can be accurately controlled Circuit is small and very fast Concept can be extended to implement arbitrary difference equation Difference equation characterizes filter H(z) Need only current mirrors and switches Truly a current-mode circuit

12 Review from last time Switched-Current Filters Basic idea introduced by Hughes and Bird at ISCAS 1989 I nt = AI nt-t OUT IN V DD Cp is parasitic gate capacitance on M 2 I IN Very low power dissipation φ 1 M 1 M 2 :A I OUT Potential to operate at very low voltages Potential for accuracy of a SC circuit at both low and high frequencies but without the Op Amp and large C ratios Neither capacitor or resistor values needed to do filtering! A completely new approach to designing filters that offers potential for overcoming most of the problems plaguing filter designers for decades! Before developing Switch-Current concept, need to review background information in s to z domain transformations

13 s-domain to z-domain transformations Ts? Hz For a given T(s) would like to obtain a function H(z) or for a given H(z) would like to obtain a T(s) such that preserves the magnitude and phase response Mathematically, would like to obtain the relationship: Hz jωt T s s=jω z=e

14 s-domain to z-domain transformations Ts? Hz want: Hz jωt T s s=jω z=e equivalently, want: Hz st T s z=e But if this were to happen, T(s) would not be a rational fraction in s with real coeff. Thus, it is impossible to obtain this mapping between T(s) and H(z)

15 s-domain to z-domain transformations Ts? Hz goal: consider: Case Hz st T s z e st z=e If can t achieve this goal, would like to map imaginary axis to unit circle and map stable filters to stable filters st 1 z = e st i i0 i! 1 i z = st 1 st i0 i! z s = T Termed the Forward Euler transformation

16 s-domain to z-domain transformations Ts? Hz s = z T Forward Euler transformation Doesn t map imaginary axis in s-plane to unit circle in z-plane Doesn t guarantee stable filter will map to stable filter But mapping may give stable filter with good frequency response

17 s-domain to z-domain transformations Ts? Hz consider: z e st Case 2: z = e st 1 z 1 st st e 1 1-sT -st i i! i0 s = 1 z T z Termed the Backward Euler transformation

18 s-domain to z-domain transformations Ts? Hz s = 1 z T z Backward Euler transformation Doesn t map imaginary axis in s-plane to unit circle in z-plane Does guarantee stable filter will map to stable filter

19 s-domain to z-domain transformations Ts? Hz consider: z e st Case 3: z = e st solving for s, obtain 2 z s = T z 1 i T 1 T s 2 s T 1+s e 1 i0 i! 2 2 T i -s 2 T -s T 1 e 1 -s i0 i! 2 2 Termed the Bilinear z transformation

20 s-domain to z-domain transformations Ts? Hz 2 z s = T z 1 Bilinear z transformation Maps imaginary axis in s-plane to unit circle in z-plane (preserves shape, distorts frequency axis) Does guarantee stable filter will map to stable filter Bilinear z transformation is widely used

21 s-domain to z-domain transformations Ts? consider: z e st Hz Three Popular Transformations s = z T Forward Euler s = 1 - z Tz z s= Tz Backward Euler s= 1-z T s= 2 z T z 1 Bilinear z transform 2 1-z s= T 1+z

22 These transformations preserve order s-domain to z-domain transformations Ts? s = z T Three Popular Transformations Forward Euler 1 - z s = Tz Hz z s= Tz Backward Euler s= 1-z T s= 2 z T z 1 Bilinear z transform 2 1-z s= T 1+z Transformations of standard approximations in s-domain are the corresponding transformations in the z-domain Transformations are not unique Transformations cause warping of the imaginary axis and may cause change in basic shape Transformations do not necessarily guarantee stability

23 s = z T Ts? Hz Forward Euler Three Popular Transformations z-domain integrators 1 - z s = Tz 0 Ts = I s H z = Some z-domain integrators TI0 z I0Tz z TI0 z 1 2 z Forward Euler Backward Euler Bilinear z z s= Tz Backward Euler 1-z s= T s= 2 z T z 1 Bilinear z transform 2 1-z s= T 1+z Corresponding difference equations: V nt+t TI V nt V nt Forward Euler OUT 0 IN OUT V nt+t I TV nt+t +V nt Backward Euler OUT 0 IN OUT TI0 VOUT nt+t VIN nt+t VIN nt +VOUT nt Bilinear z 2

24 z-domain lossy integrators Ts Some z-domain lossy integrators? z s = T Hz Forward Euler z s= Tz Backward Euler Three Popular Transformations 1 - z s = Tz 1-z s= T 0 Ts = I s+ H z = TI 0 z + T I0Tz z1+ T TI 0 z 1 2 T T z Forward Euler Backward Euler Bilinear z s= 2 z T z 1 Bilinear z transform 2 1-z s= T 1+z Corresponding difference equations: OUT 0 IN 1 OUT 1 V nt+t TI V nt T V nt Forward Euler T V nt+t I TV nt+t +V nt Backward Euler OUT 0 IN OUT T TI0 T 1 VOUT nt+t VIN nt+t VIN nt + 1 VOUT nt Bilinear z 2 2 2

25 z-domain lossy integrators Ts Some z-domain lossy integrators? Hz 0 Ts = I s+ Functional Form TI0 G z + T z -H Forward Euler I0Tz Gz Hz = z1+ T zh - 1 TI 0 z1 z1 G 2 T T z -H z Backward Euler Bilinear z Corresponding difference equations: V nt+t GV nt HV nt Forward Euler OUT IN OUT HV nt+t GV nt+t +V nt Backward Euler OUT IN OUT V nt+t G V nt+t V nt +HV nt Bilinear z OUT IN IN OUT

26 Consider this circuit Switched-Current Integrator V DD I IN1 I B1 :B :A M 3 φ 2 M 4 M 5 I IN2 φ 1 I OUT M 1 M 2 :1 I B2 I B3 φ 1 Clocks complimentary, nonoverlapping Phase not critical φ 2 nt T (n+1)t Assume inputs change only during phase Φ 2 (may be outputs from other like stages)

27 Switched-Current Integrator V DD i IN1 I B1 :B :A M 3 φ i 2 3 M 4 M 5 i IN2 i 1 φ 1 M 1 M 2 I OUT :1 I B2 I B3 Consider Φ 1 closed, Φ 2 open (nt-t < t < nt-t/2) i1 t = Bi3 nt-t + iin2 t Since current does not change during this interval i1 nt-t = Bi3 nt-t + iin2 nt-t

28 Switched-Current Integrator V DD i IN1 I B1 :B :A M 3 i 3 φ 2 M 4 M 5 i IN2 i 1 φ 1 M 1 M 2 i 2 i OUT :1 I B2 I B3 Consider Φ 2 closed, Φ 1 open (nt-t/2 < t < nt) i2t = i1nt-t i2 t = i3 t + iin1 t iout t = Ai3 t i1 nt-t = Bi3 nt-t + iin2 nt-t (from first phase) 1 B i t + i t = i nt-t + i nt-t A OUT IN1 OUT IN2 A

29 Switched-Current Integrator V DD i IN1 I B1 :B :A M 3 i 3 φ 2 M 4 M 5 i IN2 i 1 φ 1 M 1 M 2 i 2 i OUT :1 I B2 I B3 Consider Φ 2 closed, Φ 1 open (nt-t/2 < t < nt) 1 B iout t + iin1 t = iout nt-t + iin2 nt-t A A Evaluating at t=nt, we have 1 B iout nt + iin1 nt = iout nt-t + iin2 nt-t A A Taking z-transform, obtain Az A IOUT z = I 1 IN2 z - I 1 IN1 z 1Bz 1Bz

30 Switched-Current Integrator i IN1 i IN2 I B1 i 1 φ 1 M 1 M 2 :1 V DD M 3 i 3 φ 2 M 4 M 5 i 2 I B2 :B :A i OUT I B3 H z = Recall lossy integrators: Gz 1 -Hz G 1 - Hz 1 + z G 1 -Hz Forward Euler Backward Euler Bilinear z Az A IOUT z = I 1 IN2 z - I 1 IN1 z 1Bz 1Bz For H=1 becomes lossless If I IN1 =0, becomes Forward Euler integrator If I N2 =0, becomes Backward Euler integrator If I N1 = - I IN2, becomes Bilinear Integrator

31 Switched-Current Integrator V DD i IN1 I B1 N 2 :B :A M 3 i 3 φ 2 M 4 M 5 N 1 i IN2 i 1 φ 1 M 1 M 2 i 2 i OUT :1 I B2 I B3 Az A IOUT z = I 1 IN2 z - I 1 IN1 z 1Bz 1Bz Summing inputs can be provided by summing currents on N 1 or N 2 or both Multiple outputs can be provided by adding outputs to upper mirror Amount of loss determined by mirror gain B

32 Switched-Current Integrator Sensitivity Analysis V DD i IN1 I B1 N 2 :B :A M 3 i 3 φ 2 M 4 M 5 N 1 i IN2 i 1 φ 1 i 2 i OUT M 1 M 2 :1 I B2 I B3 Consider Forward Euler Az IOUT z = I 1 IN2 z 1 Bz A 1-B I 0= = T T S I 0 = 1 A S B -B = 1-B TI0 Hz = z + T For low loss integrator (e.g. ideal integrator), the sensitivity of α is very large!

33 Switched-Current Integrator Sensitivity Analysis V DD i IN1 I B1 N 2 :B :A M 3 i 3 φ 2 M 4 M 5 N 1 i IN2 i 1 φ 1 i 2 i OUT M 1 M 2 :1 I B2 I B3 Consider Bilinear z z 1 IOUT z = A I 1 IN z 1 Bz B I 0= A T 1+B T 1+B S -B = 1-B 1+B S I 0 = 1 B A For low loss integrator (e.g. ideal integrator), the sensitivity of α is very large! H z = TI0 2 z 1 T T z What about the sensitivity to the gain of the lower current mirror?

34 Switched-Current Integrator V DD Define A 1 to be the gain of the lower mirror Sensitivity to A 1? i IN1 I B1 :B :A M 3 i 3 φ 2 M 4 M 5 i IN2 i 1 φ 1 M 1 M 2 i 2 i OUT :1 I B2 I B3 Consider Φ 2 closed, Φ 1 open (nt-t/2 < t < nt) i2t = A1 i1nt-t i2 t = i3 t + iin1 t iout t = Ai3 t i1 nt-t = Bi3 nt-t + iin2 nt-t (from first phase) 1 A1B i t + i t = i nt-t + A i nt-t A OUT IN1 OUT A 1 IN2

35 Switched-Current Integrator V DD Define A 1 to be the gain of the lower mirror Sensitivity to A 1? i IN1 i IN2 I B1 i 1 φ 1 M 1 M 2 :1 M 3 i 3 φ 2 M 4 M 5 i 2 I B2 :B :A i OUT I B3 1 A B i nt + i nt = i nt-t + i A A 1 A nt-t OUT IN1 OUT 1 IN2 Taking z-transform, obtain A1Az A IOUT z = I 1 IN2 z - I 1 IN1 z 1BA1z 1BA1z Consider Forward Euler -BA 1-BA 1 S 1 = B = S A T 1-BA 1 1 -BA = 1-BA Sensitivity to A 1 is also large for low-loss or lossless integrator 1 1

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