EE 435. Lecture 4 Spring Fully Differential Single-Stage Amplifier Design

Transcription

1 EE 435 Lecture 4 Spring 019 ully Differential Single-Stage Amplifier Design General Differential Analysis 5T Op Amp from simple quarter circuit Biasing with CMB circuit Common-mode and differential-mode analysis Common Mode Gain Overall Transfer Characteristics 1

2 Review from last lecture: Where we are at: Basic Op Amp Design undamental Amplifier Design Issues Single-Stage Low Gain Op Amps Single-Stage High Gain Op Amps Two-Stage Op Amp Other Basic Gain Enhancement Approaches

3 Review from last lecture: Where we are at: Single-Stage Low-Gain Op Amps Single-ended input Differential Input (Symbol does not distinguish between different amplifier types) 3

4 Review from last lecture: Differential Input Low Gain Op Amps Will Next Show That : Differential input op amps can be readily obtained from single-ended op amps erformance characteristics of differential op amps can be directly determined from those of the single-ended counterparts 4

5 Review from last lecture: Counterpart Networks Definition: The counterpart network of a network is obtained by replacing all n- channel devices with p- channel devices, replacing all p-channel devices with n- channel devices, replacing SS biases with DD biases, and replacing all DD biases with SS biases. 5

6 Review from last lecture: Counterpart Networks Theorem: The parametric expressions for all small-signal characteristics, such as voltage gain, output impedance, and transconductance of a network and its counterpart network are the same. 6

7 Review from last lecture: Synthesis of fully-differential op amps from symmetric networks and counterpart networks Theorem: If is any network with a single input and is its counterpart network, then the following circuits are fully differential circuits --- op amps. BB d DD BB d d 1 1 BB DD BB d I BIAS SS d 1 7

8 Review from last lecture: Synthesis of fully-differential op amps from symmetric networks and counterpart networks Terminology DD BB BB Quarter Circuit Counterpart Circuit d d I BIAS d 1 Half Circuit Symmetric Half Circuit 8

9 Synthesis of fully-differential op amps from symmetric networks and counterpart networks DD BB BB d d SS What do we do with the extra output? 9

10 What do we do with the extra output? DD BB BB d d SS Use it or ignore it!! 10

11 Synthesis of fully-differential op amps from symmetric networks and counterpart networks Terminology DD BB BB Quarter Circuit Counterpart Circuit d 1 d I BIAS - d 1 Half Circuit 11 Symmetric Half Circuit

12 Synthesis of fully-differential op amps from symmetric networks and counterpart networks A fully differential op amp can be derived from any quarter circuit by combining it with its counterpart to obtain a half-circuit, combining two half-circuits to form a differential symmetric circuit and then biasing the symmetric differential circuit with a current source on the axis of symmetry. DD BB BB Quarter Circuit d d I BIAS urther, most of the properties of the operational amplifier can be obtained by inspection, from those of the quarter circuit. 1

13 Synthesis of fully-differential op amps from symmetric networks and counterpart networks A fully differential op amp can be derived from any quarter circuit by combining it with its counterpart to obtain a half-circuit, combining two half-circuits to form a differential symmetric circuit and then biasing the symmetric differential circuit with a current source on the axis of symmetry. urther, most of the properties of the operational amplifier can be obtained by inspection, from those of the quarter circuit. Implications: Much Op Amp design can be reduced to designing much simpler quarter-circuits where it is much easier to get insight into circuit performance Quarter Circuit 13

14 General Differential Analysis 5T Op Amp from simple quarter circuit Biasing with CMB circuit Common-mode and differential-mode analysis Common Mode Gain Overall Transfer Characteristics 14

15 Characterization of Quarter Circuit If the input impedance is infinite, the two-port network only has two characterizing parameters : G M and G I XX I IN IN 1 G M 1 G SS Two-ort Model of Quarter Circuit IN G 1 sc L G M 1 0 A QC ( s) GM sc G L A OQC G G M BW G GB G C M L 15

16 Characterization of Quarter Circuit (or Counterpart Circuit) with input port terminated in short circuit I I XX IN 1 G M 1 G SS If the input port of a two-port has an ac short, then the two-port reduces to a oneport characterized by the conductance G I I XX G 16

17 Determination of op amp characteristics from quarter circuit characteristics DD -- The differential gain -- Small signal differential half-circuit BB G d d O I BIAS d G M1 1 1 G 1 or SS Derivation: from KCL and KL: o 1 L M1 1 1 G G sc + G = 0 d A o GM1 sc G G d L 1 Note: actor of reduction of gain since only half of the differential input is applied to the half-circuit Note: More reduction of gain since denominator increases 17

18 Determination of op amp characteristics from quarter circuit characteristics DD -- The differential gain -- Small signal differential half-circuit d BB d d G G M1 1 1 G 1 O I BIAS or SS A o G M1 sc G G d L 1 A 0 =? BW=? GB=? A O BW GB GM1 G G G C M1 L 1 G1 G C L 18

19 Determination of op amp characteristics from quarter circuit characteristics -- The differential gain -- Small signal Quarter Circuit IN 1 G M 1 G I A QC ( s) GM sc G L Two-ort Model of Quarter Circuit Small signal differential half-circuit (repeated from last slide) G O G G M1 M G G1G I out d G M1 1 1 G 1 d 1 G M 1 G A G M1 o d scl G1 G A QC Two-ort Model of Half Circuit ( s) GM sc G L 19

20 Determination of op amp characteristics from quarter circuit characteristics -- The differential gain -- Small signal Quarter Circuit Small signal differential amplifier DD I XX BB IN d d SS I BIAS or SS A QC ( s) GM sc G L A o G M1 sc G G d L 1 0

21 Determination of op amp characteristics from quarter circuit characteristics -- The differential gain -- Small signal Quarter Circuit Small signal differential amplifier I XX DD IN d BB d A BW OQC G SS G G Note: actor of 4 reduction of gain if G 1 =G Note: actor of increase of BW if G 1 =G M GB G C Note: actor of reduction of GB if G 1 =G M L (this often occurs) (this often occurs) (this often occurs) Remember this is applicable to ANY quarter circuit! I BIAS M1 or SS A 0 d G1 G BW C G GB C L L GM 1 G G 1 1

22 Comparison of Tail oltage and Tail Current Source Structures -- The differential gain -- DD DD BB BB BB BB d d d d I BIAS SS Small signal half-circuits are identical so differential voltage gains, BW, and GB are all the same

23 Biasing Issues for Differential Amplifier Tail voltage bias not suitable for large common-mode (CM) input range but does offer good output swing Tail current bias provides good CM input range but at the expense of a modest reduction in output signal swing 3

24 Differential Output Amplifiers -- The differential gain -- DD DD BB BB OD d d d d I BIAS or SS I BIAS or SS Single-Ended Outputs Differential Output Theorem: or a symmetric circuit with symmetric outputs and differential excitations: Differential oltage Gain Double that of Single-Ended Structure BW is the same GB Doubles for the Differential Output Structure 4

25 General Differential Analysis 5T Op Amp from simple quarter circuit Biasing with CMB circuit Common-mode and differential-mode analysis Common Mode Gain Overall Transfer Characteristics 5

26 Applications of Quarter-Circuit Concept to Op Amp Design consider initially the basic single-ended amplifier DD I DQ Quarter Circuit IN SS 6

27 Single-stage single-input lowgain op amp DD DD I DQ XX M Counterpart Circuit IN IN M 1 SS SS Basic Structure Quarter Circuit ractical Implementation 7

28 Small signal model of half-circuit DD I XX M IN 1 G M 1 G Two-port model of half-circuit IN SS M 1 G G M G 1 G M1 G 8

29 Single-stage low-gain differential op amp -- The differential gain -- DD B1 M 3 M 4 SS Quarter Circuit Single-Ended Output : Differential Input Gain As A 0 BW GB g g C d L o1 o3 o1 o3 o1 o3 m1 gm1 g g L g C L gm1 sc g g IN M 1 M IN B Circuit is ery Sensitive to B1 and B!! Have synthesized fully differential op amp from quarter circuit! Have obtained analysis of fully differential op amp directly from quarter circuit! Still need to determine what happens if input is not differential! M 5 9

30 General Differential Analysis 5T Op Amp from simple quarter circuit Biasing with CMB circuit Common-mode and differential-mode analysis Common Mode Gain Overall Transfer Characteristics 30

31 Single-stage low-gain differential op amp DD -- The differential gain -- DD B1 M 3 M 4 B1 M 3 M 4 IN M 1 M IN CMB Circuit B M 5 IN M 1 M DQ-DES IN CL Need CMB circuit to establish B1 or B!! B M 5 CMB circuit determines average value of the drain voltages Compares the average to the desired quiescent drain voltages Established a feedback signal B1 to set the right Q-point Shown for B1 but could alternately be applied to B Details about CMB circuits will be discussed later 31

32 Single-stage low-gain differential op amp d DD BB -- The differential gain -- Need CMB circuit d DD B1 M 3 M 4 IN M 1 M IN As A 0 BW GB G C 1 1 M1 L d GM 1 G G G G C L L I BIAS GM sc G G 1 or SS B M 5 A(s) sc A O L g g g O1 m1 O1 gm1 GB C Have obtained differential gain of 5T Op Amp by inspection from quarter circuit L gm1 g g O3 O3 3

33 General Differential Analysis 5T Op Amp from simple quarter circuit Biasing with CMB circuit Common-mode and differential-mode analysis Common Mode Gain Overall Transfer Characteristics 33

34 Common-Mode and Differential-Mode Analysis Consider an output voltage for any linear circuit with two inputs Linear Circuit 1 A B By superposition =A 1 1+A where A 1 and A are the gains (transfer functions) from inputs 1 and to the output respectively Define the common-mode and difference-mode inputs by c = = - 1 d 1 These two equations can be solved for 1 and to obtain d 1 = c + d = c - 34

35 Common-Mode and Differential-Mode Analysis Consider an output voltage for any linear circuit with two inputs Linear Circuit 1 A B =A 1 1+A Substituting into the expression for, we obtain d d =A 1 c +A c Rearranging terms we obtain A1-A = A A + c 1 d If we define A c and A d by A1-A A c =A 1+A A d= Can express as = ca c + dad 35

36 Common-Mode and Differential-Mode Analysis Consider any output voltage for any linear circuit with two inputs Linear Circuit 1 A B =A 1 1+A Implication: Can solve a linear two-input circuit by applying superposition with 1 and as inputs or by applying c and d as inputs Implication: In a circuit with A = - A 1, A C =0 we obtain = A d d A c =A 1+A A = ca c + dad 1-A A d= A 1-A = c A1 A + d Analysis of op amps up to this point have assumed differential excitatation 36

37 Common-Mode and Differential-Mode Analysis Depiction of singe-ended inputs and common/difference mode inputs 1 A Linear Circuit B Linear Circuit A B d d c c =A 1 1+A = ca c + dad Applicable to any linear circuit with two inputs and a single output Op amps often have symmetry and this symmetry further simplifies analysis 37

38 Common-Mode and Differential-Mode Analysis Extension to differential outputs and symmetric circuits Linear Circuit E E 1 A B 1 A B Differential Output Symmetric Circuit with Symmetric Differential Output Theorem 1: The differential output for any linear network can be expressed equivalently as =A 1 1+A or as = ca c + dad and superposition can be applied to either 1 and to obtain A 1 and A or to c and d to obtain A c and A d Theorem : The symmetric differential output voltage for any symmetric linear network excited at symmetric nodes can be expressed as =A d d where A d is the differential voltage gain and the voltage d = 1-38

39 Symmetric Circuit with Symmetric Differential Output E E 1 A B Theorem : The symmetric differential output voltage for any symmetric linear network excited at symmetric nodes can be expressed as =A d d where A d is the differential voltage gain and the voltage d = 1-39 Note: Theorem infers that the common mode gain is 0 if the conditions are satisfied

40 Common-Mode and Differential-Mode Analysis roof of Theorem for Symmetric Circuit with Symmetric Differential Output: By superposition, the single-ended outputs can be expressed as + = T + T 0A 1 0B - = T + T 0NA 1 0NB where T 0A, T 0B, T 0NA and T 0NB are the transfer functions from the A and B inputs to the single-ended + and - outputs + E E - 1 A B taking the difference of these two equations we obtain = - = T -T + T -T + - 0A 0NA 1 0B 0NB by symmetry, we have T OA =T ONB and T ONA =T OB thus can be express as or as = T0A -T 0NA 1 - =A d d where A d = T OA -T ONA and where d = 1-40

41 End of Lecture 4 41