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Frequency Response of Frequency Response of Amplifiers

Frequency Response of Frequency Response of Amplifiers

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1. 1. Chapter 6 Frequency Response of Frequency Response of Amplifiers Amplifiers
2. 2. MOSFET Capacitances-L22 MOSFET Capacitances L22
3. 3. Device Capacitances Device Capacitances There is some capacitance between every There is some capacitance between every pair of MOSFET terminals. Only exception: We neglect the capacitance between Source and Drain.
4. 4. Gate to Substrate capacitance Gate to Substrate capacitance C1 is the oxide capacitance (between Gate and channel) C1 = WLCOX C2 is the depletion capacitance between channel and Substrate C2 = WL(qεsiNsub/(4ΦF))1/2 = Cd
5. 5. Gate-Drain and Gate-Source Overlap Capacitance C and C are due to overlap between the gate C3 and C4 are due to overlap between the gate poly-silicon and the Source and Drain regions. No simple formulas for C3 and C4: It is incorrect to write C3=C4=WLDCOX because of fringing electric field lines. We denote the overlap capacitance per unit width We denote the overlap capacitance per unit width as Cov
6. 6. Source-Substrate and Drain-Substrate Junction Capacitances A i i l f l f C d C Again, no simple formulas for C5 and C6: See right figure: Each capacitance should be decomposed into two components: Bottom-plate capacitance, denoted as Cj, which j depends on the junction area. Side-wall capacitance Cjsw which depends on the p jsw p perimeter of the junction.
7. 7. Source-Substrate and Drain-Substrate Junction Capacitances I MOSFET d l C d C ll i In MOSFET models Cj and Cjsw are usually given as capacitance-per-unit-area, and capacitance-per-unit- length respectively length, respectively. Cj = Cj0 / [1+VR/ΦB]m where VR is the junction’s reverse voltage Φ is the junction built in potential and m voltage, ΦB is the junction built-in potential, and m typically ranges from 0.3 to 0.4
8. 8. Examples (both transistors have the same Cj and Cjsw parameters) Calculate CDB and CSB for each structure Calculate CDB and CSB for each structure
9. 9. Left structure Left structure C =C = WEC + 2(W+E)C CDB=CSB = WECj + 2(W+E)Cjsw
10. 10. Layout for Low Capacitance (see folded structure on the right) Two parallel transistors with one Two parallel transistors with one common Drain
11. 11. Layout for Low Capacitance (see folded structure on the right) CDB=(W/2)ECj+2((W/2)+E)Cjsw C =2((W/2)EC+2((W/2)+E)C = CSB=2((W/2)ECj+2((W/2)+E)Cjsw= = WECj +2(W+2E)Cjsw j jsw Compare to the left structure: CDB=CSB = WECj + 2(W+E)Cjsw CSB slightly larger, CDB a lot smaller. Both transistors have same W/L
12. 12. C C at Cutoff CGS,CGD at Cutoff CGD=CGS=CovW
13. 13. C at Cutoff CGB at Cutoff C1 = WLCOX,C2= WL(qεsiNsub/(4ΦF))1/2 =Cd CGB=C1C2/(C1+C2)= WLCOXCd/(WLCOX+Cd ) L is the effective length of the channel.
14. 14. C and C at all modes CDB and CSB at all modes Both depend on the reverse voltages Drain to Substrate and Source to Substrate Recall Cj = Cj0 / [1+VR/ΦB]m
15. 15. C C at Triode Mode CGS,CGD at Triode Mode If VDS is small enough, we can assume VGS≈VGD Æ Channel uniform Æ Gate-Channel VGS VGD Æ Channel uniform Æ Gate Channel capacitance WLCOX is distributed uniformly CGD=CGS≈ (WLCOX)/2+CovW
16. 16. C C at Saturation Mode CGS,CGD at Saturation Mode CGD≈CovW because there isn’t much of a channel near Drain channel near Drain. (It can be proved): (It can be proved): CGS=2WLeffCOX/3+WCov
17. 17. CGB at Triode and Saturation G Modes CGB is negligible, because channel acts as a “shield” between Gate and Substrate. That is, if VG varies, change of charge comes from Source and Drain, and not from the Substrate!
18. 18. MOS Small Signal Model with Capacitance
19. 19. C V of NMOS C-V of NMOS I “W k I i ” C i i In “Weak Inversion” CGS is a series combination of COX-capacitance and Cdep- it Æ L it k capacitance Æ Low capacitance peak near VGS≈VTH
20. 20. High-Frequency Response of High-Frequency Response of Amplifiers-L23 p Basic Concepts Basic Concepts
21. 21. Miller’s Theorem Miller s Theorem Replacement of a bridging component (typically a capacitor) with equivalent input and output capac to ) t equ a e t put a d output impedances, for the sake of facilitating high- frequency response analysis.
22. 22. Miller’s Theorem Statement Miller s Theorem Statement • Let AV = VY/VX be known Let AV VY/VX be known. • Then (a) and (b) above are equivalent if: Z Z/(1 A ) d Z Z/(1 A 1) • Z1=Z/(1-AV) and Z2=Z/(1-AV -1) • Equivalence in VX,VY and currents to Z’s.
23. 23. Proof of Miller’s Theorem Proof of Miller s Theorem • Current equivalence: (VX-VY)/Z=VX/Z1 Current equivalence: (VX VY)/Z VX/Z1 • That is: Z1=Z/(1-VY/VX) etc. Si il l (V V )/Z V /Z Æ Z Z/ (1 V /V ) • Similarly: (VY –VX)/Z=VY/Z2 Æ Z2=Z/ (1-VX/VY) • A simple-looking result, but what does it mean?
24. 24. The most common application The most common application • Mid-band gain of amplifier is known to be –A Mid band gain of amplifier is known to be A. • We want to know (at least approximately) how does the bridging capacitor C influence the does the bridging capacitor CF influence the amplifier’s bandwidth.
25. 25. Exact vs Approximate Analysis Exact vs. Approximate Analysis • For exact analysis, replace CF with its impedance y , p F p ZC=1/sCF and find transfer function VY(s)/VX(s) of circuit. • Approximate analysis: Use Miller’s theorem to break capacitor into t o capacitances and then st d the inp t capacitor into two capacitances, and then study the input circuit’s time constant and the output circuit’s time constant.
26. 26. Miller Effect input capacitance Miller Effect – input capacitance • Input circuit’s added capacitance due to CF: Input circuit s added capacitance due to CF: Since Z1 is smaller than Z by (1-AV), and since CF appears in denominator of Z, it means that F input capacitance C1=CF(1-AV). If |AV| is large, C1 may become pretty large.
27. 27. Miller Effect output capacitance Miller Effect – output capacitance • Output circuit’s added capacitance due to CF: Output circuit s added capacitance due to CF: Since Z2 is smaller than Z by (1-AV -1), and since CF appears in denominator of Z, it means that F output capacitance C2=CF(1-AV -1). If |AV| is large, C2 will be almost equal to CF .
28. 28. Bridging component restriction Bridging component restriction • Both sides of the bridging component must be fully floating, for Miller’s theorem to be valid. • If voltage, on either side, if fixed then there is no Miller effect. • Later we’ll use Miller’s theorem to deal with CGD. Therefore in CG and Source Follower amplifiers, Therefore in CG and Source Follower amplifiers, in which one side of CGD is ground, there will be no Miller effect. no Miller effect. • Main impact of Miller is on CS amplifiers.
29. 29. Meaning of “approximated analysis” Meaning of approximated analysis Th i littl f th l t hi h • There is little use for the complete high- frequency transfer function (all its poles and zeros). • We are interested in the amplifier only at p y mid-band frequencies (at which all MOSFET capacitances are considered OS capac ta ces a e co s de ed open circuit), and may be a little bit beyond, near high-frequency cutoff (-3db) beyond, near high frequency cutoff ( 3db) frequency.
30. 30. Approximate Analysis Meaning Approximate Analysis Meaning • We calculate input circuit’s and output circuit’s We calculate input circuit s and output circuit s time constant for the sake of determining the amplifier’s bandwidth. • We use the known mid-band voltage gain in Miller’s theorem.
31. 31. Validity of Miller’s analysis at HF Validity of Miller s analysis at HF • We use the known mid-band voltage gain in We use the known mid band voltage gain in Miller’s theorem. • At high frequencies, much higher than the -3db t g eque c es, uc g e t a t e 3db frequency, gain is much lower Æ Miller’s theorem, using mid-band gain, is not valid!
32. 32. Bandwidth Calculation Bandwidth Calculation • Typically, the two time constants are rarely of the same order of magnitude. g • The larger time constant determines the amplifier’s -3db frequency amplifier s 3db frequency. • The smaller time constant (if much smaller than other) should be discarded as invalid other) should be discarded as invalid.
33. 33. Poles and Zeros Poles and Zeros • A time constant τ = 1/RC means that the transfer A time constant τ 1/RC means that the transfer function has a pole at s=-1/τ. • Typically every capacitor contributes one pole to • Typically, every capacitor contributes one pole to the transfer function.
34. 34. Example below: All poles are real; No zeros ) ( A A s V ) 1 )( 1 )( 1 ( ) ( ) ( 2 1 2 1 P N in S in out C sR C sR C sR A A s V s V + + + =
35. 35. Complicated example Complicated example • There are three poles due to the three There are three poles due to the three capacitors. It’s hard to determine poles location by inspection (some poles may be complex) • There may be some zeros. • Difficult due to Vin,Vout interaction (via R3,C3) in, out ( 3, 3)
36. 36. Example: Feedback capacitor over finite gain • V t(s)/VX(s)= -A Vout(s)/VX(s) A • (Vin(s)-VX(s))/RS=sCF(VX(s)-Vout(s)) E t R lt V ( )/V ( ) A/[1 R C (1 A)] • Exact Result: Vout(s)/Vin(s)=-A/[1+sRSCF(1+A)]
37. 37. Example: CG Amplifier Example: CG Amplifier • CS=CGS1+CSB1 Capacitance “seen” from Source CS CGS1+CSB1 Capacitance seen from Source to ground. • C =C +C Capacitance “seen” from Drain to • CD=CDG+CDB Capacitance seen from Drain to ground.
38. 38. Example: CG Amplifier Example: CG Amplifier • Need to determine (by inspection) the two poles Need to determine (by inspection) the two poles contributed, one by CS and another by CD. • This is easy to do only if we neglect r01. s s easy to do o y e eg ect 01 • Strategy: Find equivalent resistance “seen” by each capacitor, when sources are nulled. p ,
39. 39. Example: CG Amplifier Example: CG Amplifier • τS=CSRS =CS{RS||[1/(g 1+g b1)] τS CSRS,eq CS{RS||[1/(gm1+gmb1)] • τD=CDRD,eq=CDRD A ( + )R /(1+( + )R ) • A=(gm1+gmb1)RD/(1+(gm1+gmb1)RS) • Vout(s)/Vin(s)=A/[(1+sτS)(1+sτD)]
40. 40. High-Frequency Response- High-Frequency Response- L24 Common-Source Amplifier Common Source Amplifier
41. 41. High Frequency Model of CS Amplifier • CGD creates a Miller Effect. • It is essential to assume a nonzero signal g source resistance RS, otherwise signal voltage changes appear directly across CGS g pp y GS
42. 42. C Miller Effect on input CGD Miller Effect on input L A b h l f i l i f • Let AV be the low frequencies voltage gain of the CS amplifier, such as AV≈-gmRD. C i dd d C i C (1 A ) • Capacitance added to CGS is CGD(1-AV). • τin=RS[CGS+CGD(1-AV)]
43. 43. C Miller Effect on output CGD Miller Effect on output C i dd d C i C (1 A 1) C • Capacitance added to CDB is CGD(1-AV -1)≈CGD. • τout = RD(CGD+CDB) • If τin and τout are much different, then the largest one is valid and the smaller one is invalid.
44. 44. Common Source (must be in Saturation Mode) Neglecting input/output interaction, fp,in = 1 2πRS CGS + (1+ gmRD )CGD [ ] S GS ( gm D ) GD [ ] 1 fp,out = 1 2π CGD + CDB ( )RD [ ]
45. 45. CS Amplifier’s Bandwidth if RS S large ( ) ( ) • τout = RD(CGD+CDB) ; τin=RS[CGS+CGD(1-AV)] ; AV=-gmRD • If all transistor capacitances are of the same order of magnitude and if R is of the same order of magnitude of R magnitude, and if RS is of the same order of magnitude of RD (or larger) then AV(s)≈ -gmRD/(1+sτin) : Input capacitance dominates. BW=fh=1/2πτin
46. 46. CS Amplifier’s Bandwidth if Vin is almost a perfect voltage source • τout = RD(CGD+CDB) ; τin=RS[CGS+CGD(1-AV)] ; AV=-gmRD • If all transistor capacitances are of the same order of i d d if R i ll A ( ) R /(1 ) magnitude, and if RS is very small : AV(s)≈-gmRD/(1+sτout) : Output capacitance dominates. BW=fh=1/2πτout
47. 47. High Frequency Response of CS Amplifier – “Exact” Analysis • Use small-signal diagram, and replace every capacitor by its impedance 1/sC ; Neglect ro • Result: Only two poles, and one zero in Vout(s)/Vin(s)! out( ) in( )
48. 48. CS HF Response: Exact Result CS HF Response: Exact Result ) ( ) ( − D m GD out R g sC s V s ⎛ ⎜ ⎞ ⎟ s ⎛ ⎜ ⎞ ⎟ s2 s [ ] 1 ) ( ) 1 ( ) ( ) ( ) ( ) ( 2 + + + + + + + + = DB GD D GS S GD D m S DB GD SB GS GD GS D S D m GD in out C C R C R C R g R s C C C C C C R R s g s V Assume D = s ωp1 +1 ⎝ ⎜ ⎜ ⎠ ⎟ ⎟ s ωp2 +1 ⎝ ⎜ ⎜ ⎠ ⎟ ⎟ = s ωp1 ωp 2 + s ωp1 +1 , ωp2 >> ωp1 fp,in = 1 2π RS CGS + (1+ gmRD)CGD [ ]+ RD(CGD + CDB ) ( ) [ ] ( )
49. 49. CS Exact Analysis (cont ) CS Exact Analysis (cont.) RS (1+ g RD )CGD + RSCGS + RD (CGD + CDB ) fp,out = RS (1+ gmRD )CGD + RSCGS + RD (CGD + CDB ) 2πRS RD(CGSCGD + CGSCDB + CGDCDB ) fp,out ≈ 1 2πR (C + C ) , for large CGS 2πRD (CGD + CDB ) C R R g DB GD DB GS GD GS D S GD D S m out p g C C C C C C R R C R R g f ) ( 2 , + + ≈ π GD DB GS m C C C g large for , ) ( 2 + ≈ π
50. 50. CS Amplifier RHP Zero CS Amplifier RHP Zero Right half plane zero, from the numerator of Vout(s)/Vin(s) [ ] 1 ( ) 1 ( ) ( ) ( ) ( ) ( 2 + + + + + + + + − = D m GD out C C R C R C R g R s C C C C C C R R s R g sC s V s V C [ ] 1 ( ) 1 ( ) ( ) ( ) + + + + + + + + DB GD D GS S GD D m S DB GD SB GS GD GS D S in C C R C R C R g R s C C C C C C R R s s V sCGD − gm . . . → fz = +gm 2πCGD
51. 51. Zero Creation Zero Creation • A Zero indicates a frequency at which output A Zero indicates a frequency at which output becomes zero. • The two paths from input to output may create e t o pat s o put to output ay c eate signals that perfectly cancel one another at one specific frequency.
52. 52. Zero Calculation Zero Calculation • At frequency at which output is zero (s=sz) , the At frequency at which output is zero (s sz) , the currents through CGD and M1 are equal and opposite. Can assume that at this frequency (only) output node is shorted to ground. • V1/(1/szCGD)=gmV1 Æ Zero expression results.
53. 53. RHP Zero Effect on the Frequency Response • RHP Zero (like LHP zero) contributes positively RHP Zero (like LHP zero) contributes positively to the slope of magnitude frequency response curve Æ reduces the roll-off slope. • RHP Zero (contrary to LHP zero) adds NEGATIVE phase shift.
54. 54. RHP Zero Impact RHP Zero Impact • It is often negligible, as it happens in very high g g , pp y g frequencies. • It becomes important if CS amplifier is part of multi-stage p p p g amplifiers creating a very large open-loop gain. It may affect stabilization compensation.
55. 55. Input Impedance of CS stage Input Impedance of CS stage
56. 56. High-Frequency Response- High-Frequency Response- L25 Source Follower Source Follower
57. 57. High-Frequency Response of Source Follower – No Miller • No Miller Effect – CGD is not a bridge capacitor No Miller Effect CGD is not a bridge capacitor (as Drain side is grounded). • C is a combination of several capacitances: • CL is a combination of several capacitances: CSB1, CDB,SS, CGD,SS and Cin of next stage.
58. 58. High-Frequency Response of Source Follower – No solution “by inspection” • Because of CGS that ties input to output it is Because of CGS that ties input to output, it is impossible to find various time constants, one capacitor at a time – combined effects of CGS capacitor at a time combined effects of CGS and CL.
59. 59. Source Follower HF Response Derivation Assumptions • Use small-signal diagram and 1/sC terms Use small signal diagram and 1/sC terms. • Simplifying assumptions: We neglect effects of ro and γ (body effect) and γ (body effect).
60. 60. Source Follower HF Response Derivation • Sum currents at output node (all functions of s): Sum currents at output node (all functions of s): • V1CGSs+gmV1=VoutCLs , yielding: V [C /( C )]V • V1=[CLs/(gm+CGSs)]Vout • KVL: Vin=RS[V1CGSs+(V1+Vout)CGDs]+V1+Vout
61. 61. Source Follower Transfer Function ( ) GS GD GD S L GD GD GS L GS S GS m i out g C C C R g s C C C C C C R s sC g s V s V + + + + + + + = ) ( ) ( ) ( 2 ( ) m GS GD GD S m L GD GD GS L GS S in g C C C R g s C C C C C C R s s V + + + + + + ) ( ) (
62. 62. Idea for detecting “dominant pole”: Idea for detecting dominant pole : 1 1 1 1 1 ) ( ) 1 )( 1 ( 1 2 2 1 2 1 2 2 1 2 1 + + ≈ + + + = + + s s s s s s τ τ τ τ τ τ τ τ τ If Æ C fi d b i i f If τ1 >> τ2 Æ Can find τ1 by inspection of the s coefficient
63. 63. Source Follower Dominant Pole (if it exists…) ( ) GS m out sC g s V + = ) ( 2 ( ) m GS GD GD S m L GD GD GS L GS S in g C C C R g s C C C C C C R s s V + + + + + + ) ( ) ( 2 fp1 ≈ gm 2π g R C + C + C ( ) , assuming fp2 >> fp1 p 2π gmRSCGD + CL + CGS ( ) p p = 1 C + C ⎛ ⎜ ⎞ ⎟ 2π RSCGD + CL + CGS gm ⎝ ⎜ ⎜ ⎠ ⎟ ⎟
64. 64. Source Follower Dominant Pole (if RS is very small) ( ) GS m out sC g s V + = ) ( 2 ( ) m GS GD GD S m L GD GD GS L GS S in g C C C R g s C C C C C C R s s V + + + + + + ) ( ) ( 2 1 ( ) 2 1 2 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = + ≈ GS L GS L m p g C C C C g f π π ⎠ ⎝ m g
65. 65. Source Follower Input Impedance Source Follower Input Impedance • At low frequencies we take Rin of CS, Source Follower, Cascode and Differential , amplifiers to be practically infinite. • At higher frequencies we need to study • At higher frequencies, we need to study the Input Impedance of the various f amplifiers. • Specifically for a Source Follower: Is Zin Specifically for a Source Follower: Is Zin purely capacitive?
66. 66. Source Follower Input Impedance Derivation C Laying CGD aside, as it appears in parallel to the i i f Z remaining part of Zin: ) 1 || 1 ( s C g sC I g I sC I V X m X X X ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + ≈ s C g sC sC L mb GS GS ⎠ ⎝
67. 67. Source Follower Input Impedance Neglecting C Neglecting CGD , Zin ≈ 1 + 1+ gm ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 1 in sCGS sCGS ⎝ ⎜ ⎠ ⎟ gmb + sCL | sC >>| g s, frequencie low At L mb ( ) / 1 ) / ( 1 1 g g g sC Z mb mb m GS in + + ≈ ) ( ) /( Miller as same C g g g C C GD mb m mb GS in + + = ∴
68. 68. “Miller Effect” of C Miller Effect of CGS Low frequency gain is gm/(gm+gmb) m m mb CG,Miller=CGS(1-AV) | sC >>| g s, frequencie low At L mb ( ) / 1 ) / ( 1 1 g g g sC Z mb mb m GS in + + ≈ ) ( ) /( Miller as same C g g g C C GD mb m mb GS in + + = ∴ Effect not important: Only a fraction of C Effect not important: Only a fraction of CGS is added to Cin
69. 69. Source Follower HF Z Source Follower HF Zin At high frequencies g << |sC | At high frequencies, gmb << |sCL | Zin ≈ 1 C + 1 C + gm 2 C C sCGS sCL s CGSCL At a particular frequency, input impedance includes CGD in parallel with series combination of CGS and CL and a negative GS L resistance equal to -gm/(CGSCLω2).
70. 70. Source Follower Output Impedance Derivation Assumptions Body effect and CSB contribute an contribute an impedance which is a parallel portion of a parallel portion of Zout – we’ll keep it in mind We also mind. We also neglect CGD. ? / = = X X OUT I V Z
71. 71. Source Follower Output Impedance Derivation I V g s C V − = + X S GS X m GS V V sR C V I V g s C V − = + − = + 1 1 1 1 X S GS 1 1 X X OUT I V Z = / GS S sC g C sR + + = 1 GS m sC g +
72. 72. Source Follower Output Impedance - Discussion I V Z / = C C sR I V Z GS S X X OUT 1 / + = = s frequencie low at g sC g m GS m , / 1 ≈ + s frequencie high at RS , ≈ We already know that at low frequencies We already know that at low frequencies Zout≈ 1/gm At high frequencies, CGS short-circuits the Gate-Source, and that’s why Zout depends on the Source Follower’s driving signal source.
73. 73. Source Follower Output Impedance - Discussion I V Z / = C C sR I V Z GS S X X OUT 1 / + = = s frequencie low at g sC g m GS m , / 1 ≈ + s frequencie high at RS , ≈
74. 74. Source Follower Output Impedance - Discussion Typically RS > 1/gm Therefore |Zout(ω)| is increasing with ω increasing with ω.
75. 75. Source Follower Output Impedance - Discussion If |Zout(ω)| is increasing with ω, then it must have an INDUCTIVE component!
76. 76. Source Follower Output Impedance – Passive Network Modeling? Can we find values for R1, R2 and L such that Z1=Zout of the Source Follower? that Z1 Zout of the Source Follower?
77. 77. Source Follower Output Impedance – Passive Network Modeling Construction Clues: At ω=0 Zout=1/gm ; At ω=∞ Zout=RS At ω=0 Z1=R2 ; At ω=∞ Z1=R1+R2 Let R2=1/g and R1=RS-1/g (here is where Let R2 1/gm and R1 RS 1/gm (here is where we assume RS>1/gm !)
78. 78. Source Follower Output Impedance – Passive Network Modeling Final Result R 1/g R2 =1/gm R1 = RS −1/gm L = CGS gm RS −1/gm ( ) Output impedance inductance dependent on source impedance dependent on source impedance, RS (if RS large) !
79. 79. Source Follower Ringing Source Follower Ringing O tp t ringing d e to t ned circ it formed Output ringing due to tuned circuit formed with CL and inductive component of output impedance It happens especially if output impedance. It happens especially if load capacitance is large.
80. 80. High Frequency Response- High Frequency Response- L26 Common Gate Amplifier Common Gate Amplifier
81. 81. CG Amplifier neglecting r CG Amplifier neglecting ro • CS=CGS1+CSB1 Capacitance “seen” from Source CS CGS1+CSB1 Capacitance seen from Source to ground. • C =C +C Capacitance “seen” from Drain to • CD=CDG+CDB Capacitance seen from Drain to ground.
82. 82. CG Amplifier if r negligible CG Amplifier if ro negligible • Can determine (by inspection) the two poles Can determine (by inspection) the two poles contributed, one by CS and another by CD. • This is easy to do only if we neglect r01. s s easy to do o y e eg ect 01 • Strategy: Find equivalent resistance “seen” by each capacitor, when sources are nulled. p ,
83. 83. CG Amplifier Transfer Function if ro o is negligible • τS=CSRS =CS{RS||[1/(g 1+g b1)] τS CSRS,eq CS{RS||[1/(gm1+gmb1)] • τD=CDRD,eq=CDRD A ( + )R /(1+( + )R ) • A=(gm1+gmb1)RD/(1+(gm1+gmb1)RS) • Vout(s)/Vin(s)=A/[(1+sτS)(1+sτD)]
84. 84. CG Bandwidth taking ro into o account Can we use Miller’s theorem for the bridging ro resistor?
85. 85. CG Bandwidth taking ro into o account Can we use Miller’s theorem for the bridging ro resistor? Well, no. Effect of ro at input: Parallel resistance ro/(1-AV). Recall: AV is large and positive Æ Negative V g p g resistance! Æ How to compute time constants?
86. 86. CG Bandwidth taking ro into o account Need to solve exactly, using impedances 1/sC, and Kirchhoff’s laws.
87. 87. Example: ro effect if RD is replaced o with a current source load • Current from Vi through RS is: V tCLs+(-V1)Ci s Current from Vin through RS is: VoutCLs+( V1)Cins • Adding voltages: (-VoutCLs+V1Cins)RS+Vin=-V1 V [ V C R V ]/(1 C R ) • V1= -[-VoutCLsRS+Vin]/(1+CinRSs) • Also: ro(-VoutCLs-gmV1)-V1=Vout
88. 88. Example: ro effect if RD is replaced with o a current source load – final result V t(s)/Vi (s)=(1+g r )/D(s) where: Vout(s)/Vin(s) (1+gmro)/D(s), where: D(s)=roCLCinRSs2+[roCL+CinRS+(1+gmro)CLRS]s+1 Th ff t f b dd d i l b The effect of gmb can now be added simply by replacing gm by gm+gmb.
89. 89. CG with R load Input Impedance CG with RD load Input Impedance Recall the exact Rin formula at low frequencies: Rin=[RD/((gm+gmb)ro)]+[1/(gm+gmb)]. Rin [RD/((gm gmb)ro)] [1/(gm gmb)]. Simply replace Rin by Zin and RD with Z =R ||(1/sC ) ZL=RD||(1/sCD)
90. 90. CG with current source load Input Impedance Recall the exact Rin formula at low frequencies: Rin=[RD/((gm+gmb)ro)]+[1/(gm+gmb)]. Rin [RD/((gm gmb)ro)] [1/(gm gmb)]. Here replace Rin by Zin and RD with ZL=1/sCL
91. 91. CG with current source load Input Impedance - Discussion Zin=[1/sCL/((gm+gmb)ro)]+[1/(gm+gmb)]. At high frequencies, or if CL is large, the effect of g q L g CL at input becomes negligible compared to the effect of Cin (as ZinÆ 1/(gm+gmb) )
92. 92. CG and CS Bandwidth Comparison CG and CS Bandwidth Comparison If R i l h b d idth i • If RS is large enough, bandwidth is determined by the input time constant • CG: τin=(CGS+CSB)[RS||(1/(gm+gmb))] • CS: τi =[CGS+(1+g RD)CGD]RS CS: τin [CGS+(1+gmRD)CGD]RS • Typically -3db frequency of CG amplifier is by an order of magnitude larger than that by an order of magnitude larger than that of a CS amplifier. • If RS is small, τout dominates. It is the same in both amplifiers.
93. 93. High Frequency Response- High Frequency Response- L27 Cascode Amplifier Cascode Amplifier
94. 94. Key Ideas: Key Ideas: • Cascode amplifier has the same input resistance and voltage gain as those of a g g CS amplifier. • Cascode amplifier has a much larger • Cascode amplifier has a much larger bandwidth than CS amplifier. • Cascode = CSÆCG. CS has a much reduced Miller Effect because CS gain is reduced Miller Effect because CS gain is low (near -1).
95. 95. Cascode Amplifier Capacitances Cascode Amplifier Capacitances
96. 96. C Insignificant Miller Effect CGD1 Insignificant Miller Effect • M1 is CS with a load of 1/(g 2+g b2) (the input M1 is CS with a load of 1/(gm2+gmb2) (the input resistance of the CG stage). • A = g /(g +g ) which is a fraction • AV= -gm1 /(gm2+gmb2) which is a fraction. • Effect of CGD1 at input is at most 2CGD1
97. 97. Cascode Input Pole (if dominant) Cascode Input Pole (if dominant) 1 f = ] ) 1 ( [ 2 1 2 2 1 1 , GD mb m m GS S A p C g g g C R f + + + = π 2 2 mb m
98. 98. Cascode Mid-section Pole (if dominant) Total resistance seen at Node X is M2 input resistance. By Miller: CGD1 By Miller: CGD1 contribution is at most CGD1 g g + CGD1 ( ) 2 2 1 1 2 2 , 2 GS SB DB GD mb m X p C C C C g g f + + + + ≈ π
99. 99. Cascode Output Pole (if dominant) Cascode Output Pole (if dominant) 1 f ( ) 2 2 , 2 GD L DB D Y p C C C R f + + = π
100. 100. Which of the Cascode’s three poles is dominant? ⎥ ⎤ ⎢ ⎡ ⎟ ⎟ ⎞ ⎜ ⎜ ⎛ + + = 1 1 1 , 1 2 1 m GD GS S A p g C C R f π Either fp A or ⎥ ⎦ ⎢ ⎣ ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ + + + 2 2 1 1 1 2 mb m GD GS S g g C C R π Either fp,A or fp,Y dominates. fp X is typically ( ) 2 2 , 2 mb m X p C C C C g g f + + + + = π fp,X is typically a higher frequency. ( ) 2 2 1 1 2 GS SB DB GD C C C C + + + π frequency. ( ) 2 2 , 2 1 GD L DB D Y p C C C R f + + = π ( ) 2 2 GD L DB D
101. 101. Cascode with current source load Cascode with current source load • This is done whenever we want larger voltage gains gains. • However we pay by having a lower bandwidth:
102. 102. Lower Bandwidth of Cascode with current source load • Key problem: If a CG amplifier feeds a current Key problem: If a CG amplifier feeds a current source load, its Rin may be quite large. • It causes an aggravation of Miller effect at input. t causes a agg a at o o e e ect at put • Also: The impact of CX (mid-section capacitance) may no longer be negligible. y g g g
103. 103. High Frequency Response- High Frequency Response- L28 Differential Amplifiers Differential Amplifiers
104. 104. Analysis Strategy Analysis Strategy Do separate Differential Mode and Common Do separate Differential-Mode and Common- Mode high-frequency response analysis
105. 105. Differential Mode Half Circuit Analysis Differential-Mode – Half-Circuit Analysis Differential Mode HF response identical to that Differential-Mode HF response identical to that of a CS amplifier.
106. 106. Differential Mode HF response Analysis Differential-Mode HF response Analysis Relatively low ADM bandwidth due to CGD Miller Relatively low ADM bandwidth due to CGD Miller effect. Remedy: Cascode Differential Amplifier. y p
107. 107. Common Mode HF Response Analysis Common-Mode HF Response Analysis CP depends on CGD3, CDB3, CSB1 and CSB2 How large is it? Can be substantial if (W/L)’s How large is it? Can be substantial if (W/L) s large
108. 108. Recall: Mismatches in W/L, VTH and other transistor parameters all translate to mismatches transistor parameters all translate to mismatches in gm Y X g g V V 2 1 − − D SS m m m m CM in Y X DM CM V R R g g g g V V V A 1 ) ( 2 1 2 1 , , + + − = = −
109. 109. Common-Mode HF response if M1 and M2 mismatch (solution approach) g g V V − − D SS m m m m CM in Y X DM CM V R R g g g g V V V A 1 ) ( 2 1 2 1 , , + + − − = − = − Replace: R with r ||(1/C s) R with Replace: RSS with ro3||(1/CPs) , RD with RD||(1/CL s)
110. 110. Common-Mode HF response if M1 and M2 mismatch - Result ] 1 || )[ ( − R g g 1 ] 1 || )[ ( ] || )[ ( ) ( 2 1 , = − − = − = − L D m m CM i Y X DM CM V s C R g g V V V s A ) 1 ( ) ( 1 ] || )[ ( 3 2 1 , + + P o m m CM in C R s C r g g V ) 1 )( 1 ( ) 1 ( 1 ) ( ) ( 3 3 3 2 1 2 1 + + + ⋅ + + − − = P o L D P o o m m D m m C r s C sR C sr r g g R g g ) 1 ) ( 1 )( 1 ( 3 2 1 + + + + o m m L D r g g s C sR Zero dominates! AV,CM-DM begins to rise at f 1/(2 C ) fz=1/(2πro3CP)
111. 111. Overall Differential Amplifier’s Bandwidth At t i hi h f f f • At a certain high frequency f=fP,DM differential gain ADM begins to fall. • At another high-frequency f=fZ,CM common-mode gain ACM begins to rise. g CM g • Whichever of the above 3db frequencies is lower determines the amplifier’s bandwidth lower determines the amplifier s bandwidth – we are really interested in the frequency at which the amplifier’s CMRR begins to at which the amplifier s CMRR begins to fall.
112. 112. CMRR vs Bandwidth Tradeoff CMRR vs Bandwidth Tradeoff • In order to reduce the voltage drop IDRD and still obtain a large enough ADM we typically want obtain a large enough ADM we typically want gate widths to be relatively large (for W/L to be large enough) large enough) • The larger W’s the smaller is bandwidth. I i th ll V t • Issue is more severe the smaller VDD gets.
113. 113. Frequency Response of Differential Pairs With High-Impedance Load Differential t t output Here CL include CGD and CDB of PMOS loads
114. 114. Differential Mode Analysis: G Node Differential Mode Analysis: G Node • For differential outputs, CGD3 and CGD4 conduct l d it t t d G equal and opposite currents to node G. • Therefore node G is ground for small-signal analysis. • In practice: We hook up by-pass capacitor y between G and ground.
115. 115. Differential Mode Half Circuit Analysis [ ] 1 ) ( ) 1 ( ) ( ) ( ) ( ) ( 2 + + + + + + + + − = DB GD D GS S GD D m S DB GD SB GS GD GS D S D m GD in out C C R C R C R g R s C C C C C C R R s R g sC s V s V Above (obtained earlier) “exact” transfer function ( ) applies, if we replace RD by ro1||ro3
116. 116. Differential Mode Half Circuit Analysis – final result • Here, because CL is quiet large and because the Here, because CL is quiet large and because the output time constant involves ro1||ro3 which is large, it is the output time constant that dominates. • fh≈1/2πCL(ro1||ro3 )
117. 117. Common Mode HF Analysis Common-Mode HF Analysis Result identical to that of a differential amplifier with RD load. Recall dominant zero due to C (source) Recall dominant zero due to CP (source)
118. 118. Differential Pair with Active Current Mirror Load Single-ended output There are two signal paths, each one has its own transfer function. own transfer function. Overall transfer function is the sum of the t th ’ t f f ti two paths’ transfer functions
119. 119. The Mirror Pole The Mirror Pole C i f C C C C d CE arises from CGS3,CGS4,CDB3,CDB1 and Miller effect of CGD1 and CGD4. Pole is at s=- /C gm3/CE
120. 120. Simplified diagram showing only the largest capacitances: Solution approach: Replace Vin, M1 and M2 by Th i i l t a Thevenin equivalent
121. 121. Thevenin equivalent of bottom circuit (for diff. mode analysis) • We can show that: V g r V g r V • VX=gm1ro1Vin=gmNroNVin • RX=2ro1=2roN
122. 122. Differential Mode analysis Differential Mode analysis • We assume that 1/gmP<<roP V (V V )[(1/(C s+g )] / • VE=(Vout-VX)[(1/(CEs+gmP)] / [RX+(1/(CEs+gmP)] voltage division
123. 123. Differential Mode analysis (cont’d) Differential Mode analysis (cont d) • Current of M4 is gm4VE g V I V (C s+r 1) • - gm4VE-IX=Vout(CLs+roP -1)
124. 124. Differential Mode Analysis Result Differential Mode Analysis - Result s C g r g V ) 2 ( + oP oN mP L E oN oP E mP oN mN in out r r g as s C C r r s C g r g V V ) ( 2 2 ) 2 ( 2 + + + + = L oN mP oP E oP oN C r g r C r r a ) 2 1 ( ) 2 ( + + + =
125. 125. Differential Mode Analysis – Result Interpretation E mP oN mN out s C g r g V ) 2 ( 2 + = L oN mP oP E oP oN oP oN mP L E oN oP in C r g r C r r a r r g as s C C r r V ) 2 1 ( ) 2 ( ) ( 2 2 2 + + + = + + + L oN mP oP E oP oN g ) ( ) ( This type of circuits typically produce a This type of circuits typically produce a “dominant pole”, as the output pole often dominates the mirror pole dominates the mirror pole. In such a case we can use “a” (above) to estimate the dominant pole location.
126. 126. Differential Mode Analysis – Dominant and Secondary Poles ) ( 2 L oN mP oP E oP oN oP oN mP h C r g r C r r r r g f ) 2 1 ( ) 2 ( ) ( 2 ≈ + + + + ≈ L oN mP oP oP oN mP L oN mP oP oP oN mP C r g r r r g C r g r r r g 2 ) ( 2 ) 2 1 ( ) ( 2 = + ≈ + + ≈ L oN oP C r r ) || ( 1 = By substituting fh into the exact transfer f ti fi d th hi h f l function we can find the higher frequency pole: fp2=gmP/CE
127. 127. Single-ended differential amplifier summary of results fp1 ≈ 1 2π(r N || r P )CL 2π(roN || roP )CL g fp2 = gmP 2πCE fZ = 2 fp 2 = 2gmP 2 C fZ fp 2 2πCE
128. 128. In summary In summary Differential amplifiers with current mirror load and differential output have a superior p p high frequency response over differential amplifiers with current mirror load and amplifiers with current mirror load and single-ended output.