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Fundamentals of Microelectronics II

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					Fundamentals of Microelectronics II




   CH9    Cascode Stages and Current Mirrors
   CH10   Differential Amplifiers
   CH11   Frequency Response
   CH12   Feedback




                                                1
Chapter 9 Cascode Stages and Current Mirrors




               9.1 Cascode Stage

               9.2 Current Mirrors




                                          2
                      Boosted Output Impedances




                     Rout1  1  g m  RE || r rO  RE || r
                     Rout 2  1  g m RS rO  RS

CH 9 Cascode Stages and Current Mirrors                            3
                            Bipolar Cascode Stage




                  Rout  [1  g m (rO 2 || r 1 )]rO1  rO 2 || r 1
                  Rout  g m1rO1 rO 2 || r 1 
CH 9 Cascode Stages and Current Mirrors                                4
     Maximum Bipolar Cascode Output Impedance




                                          Rout , max  g m1rO1r 1
                                          Rout , max  1rO1




 The maximum output impedance of a bipolar cascode is
  bounded by the ever-present r between emitter and ground
  of Q1.
CH 9 Cascode Stages and Current Mirrors                              5
                      Example: Output Impedance




                                                     2rO 2 r 1
                                          RoutA   
                                                    r 1  rO 2

 Typically r is smaller than rO, so in general it is impossible
  to double the output impedance by degenerating Q2 with a
  resistor.
CH 9 Cascode Stages and Current Mirrors                             6
                               PNP Cascode Stage




                  Rout  [1  g m (rO 2 || r 1 )]rO1  rO 2 || r 1
                  Rout  g m1rO1 rO 2 || r 1 
CH 9 Cascode Stages and Current Mirrors                                7
         Another Interpretation of Bipolar Cascode




 Instead of treating cascode as Q2 degenerating Q1, we can
  also think of it as Q1 stacking on top of Q2 (current source)
  to boost Q2’s output impedance.
CH 9 Cascode Stages and Current Mirrors                           8
                                   False Cascodes




                          1                         1
      Rout                 g || rO 2 || r 1  rO1  g || rO 2 || r 1
               1  g m1                    
                          m2                        m2

                    g m1        1
      Rout     1 
                         rO1 
                                      2rO1
                    g m2       g m2
 When the emitter of Q1 is connected to the emitter of Q2, it’s
  no longer a cascode since Q2 becomes a diode-connected
  device instead of a current source.
CH 9 Cascode Stages and Current Mirrors                                    9
                              MOS Cascode Stage




                           Rout  1  g m1rO 2 rO1  rO 2
                           Rout  g m1rO1rO 2
CH 9 Cascode Stages and Current Mirrors                       10
            Another Interpretation of MOS Cascode




 Similar to its bipolar counterpart, MOS cascode can be
  thought of as stacking a transistor on top of a current
  source.
 Unlike bipolar cascode, the output impedance is not limited
  by .
CH 9 Cascode Stages and Current Mirrors                     11
                             PMOS Cascode Stage




                         Rout  1  g m1rO 2 rO1  rO 2
                         Rout  g m1rO1rO 2
CH 9 Cascode Stages and Current Mirrors                     12
                    Example: Parasitic Resistance




                                          Rout  (1  g m1rO 2 )( rO1 || RP )  rO 2




 RP will lower the output impedance, since its parallel
  combination with rO1 will always be lower than rO1.
CH 9 Cascode Stages and Current Mirrors                                        13
                   Short-Circuit Transconductance




                                               iout
                                          Gm 
                                               vin     vout  0




 The short-circuit transconductance of a circuit measures its
  strength in converting input voltage to output current.
CH 9 Cascode Stages and Current Mirrors                      14
                       Transconductance Example




                                      Gm  g m1
CH 9 Cascode Stages and Current Mirrors           15
                        Derivation of Voltage Gain




                                          vout  iout Rout  Gm vin Rout
                                          vout vin  Gm Rout



 By representing a linear circuit with its Norton equivalent,
  the relationship between Vout and Vin can be expressed by
  the product of Gm and Rout.
CH 9 Cascode Stages and Current Mirrors                             16
                           Example: Voltage Gain




                                      Av   g m1rO1


CH 9 Cascode Stages and Current Mirrors                17
Comparison between Bipolar Cascode and CE Stage




 Since the output impedance of bipolar cascode is higher
  than that of the CE stage, we would expect its voltage gain
  to be higher as well.
CH 9 Cascode Stages and Current Mirrors                         18
         Voltage Gain of Bipolar Cascode Amplifier




                       Gm  g m1
                       Av   g m1rO1 g m1 (rO1 || r 2 )

 Since rO is much larger than 1/gm, most of IC,Q1 flows into the
  diode-connected Q2. Using Rout as before, AV is easily
  calculated.
CH 9 Cascode Stages and Current Mirrors                        19
               Alternate View of Cascode Amplifier




 A bipolar cascode amplifier is also a CE stage in series with
  a CB stage.
CH 9 Cascode Stages and Current Mirrors                      20
                          Practical Cascode Stage




                       Rout  rO 3 || g m 2 rO 2 (rO1 || r 2 )
 Since no current source can be ideal, the output impedance
  drops.
CH 9 Cascode Stages and Current Mirrors                           21
                          Improved Cascode Stage




                                    Rout  g m3 rO 3 (rO 4 || r 3 ) || g m 2 rO 2 (rO1 || r 2 )




 In order to preserve the high output impedance, a cascode
  PNP current source is used.

CH 9 Cascode Stages and Current Mirrors                                                  22
                           MOS Cascode Amplifier




                                          Av  Gm Rout
                                          Av   g m1 (1  g m 2 rO 2 )rO1  rO 2 
                                          Av   g m1rO1 g m 2 rO 2




CH 9 Cascode Stages and Current Mirrors                                      23
                 Improved MOS Cascode Amplifier




                                          Ron  g m 2 rO 2 rO1
                                          Rop  g m3 rO 3 rO 4
                                          Rout  Ron || Rop



 Similar to its bipolar counterpart, the output impedance of a
  MOS cascode amplifier can be improved by using a PMOS
  cascode current source.
CH 9 Cascode Stages and Current Mirrors                          24
      Temperature and Supply Dependence of Bias
                       Current




                          R2V CC ( R1  R2 )  VT ln(I1 I S )
                                                                2
                              1       W  R2              
                          I1   n Cox  R  R VDD  VTH 
                                                          
                              2       L 1     2          

 Since VT, IS, n, and VTH all depend on temperature, I1 for
  both bipolar and MOS depends on temperature and supply.
CH 9 Cascode Stages and Current Mirrors                             25
                         Concept of Current Mirror




 The motivation behind a current mirror is to sense the
  current from a “golden current source” and duplicate this
  “golden current” to other locations.
CH 9 Cascode Stages and Current Mirrors                       26
                    Bipolar Current Mirror Circuitry




                                            I S1
                             I copy                  I REF
                                          I S , REF

 The diode-connected QREF produces an output voltage V1
  that forces Icopy1 = IREF, if Q1 = QREF.
CH 9 Cascode Stages and Current Mirrors                       27
                      Bad Current Mirror Example I




 Without shorting the collector and base of QREF together,
  there will not be a path for the base currents to flow,
  therefore, Icopy is zero.
CH 9 Cascode Stages and Current Mirrors                       28
                     Bad Current Mirror Example II




 Although a path for base currents exists, this technique of
  biasing is no better than resistive divider.

CH 9 Cascode Stages and Current Mirrors                         29
                             Multiple Copies of IREF




                                            IS, j
                            I copy , j                I REF
                                           I S , REF
 Multiple copies of IREF can be generated at different
  locations by simply applying the idea of current mirror to
  more transistors.
CH 9 Cascode Stages and Current Mirrors                        30
                                   Current Scaling




                                    I copy, j  nI REF

 By scaling the emitter area of Qj n times with respect to
  QREF, Icopy,j is also n times larger than IREF. This is equivalent
  to placing n unit-size transistors in parallel.
CH 9 Cascode Stages and Current Mirrors                           31
                         Example: Scaled Current




CH 9 Cascode Stages and Current Mirrors            32
                                Fractional Scaling




                                                    1
                                          I copy    I REF
                                                    3
 A fraction of IREF can be created on Q1 by scaling up the
  emitter area of QREF.
CH 9 Cascode Stages and Current Mirrors                       33
                Example: Different Mirroring Ratio




 Using the idea of current scaling and fractional scaling,
  Icopy2 is 0.5mA and Icopy1 is 0.05mA respectively. All coming
  from a source of 0.2mA.
CH 9 Cascode Stages and Current Mirrors                           34
              Mirroring Error Due to Base Currents




                                                 nI REF
                                   I copy   
                                                 1
                                              1  n  1
                                                 
CH 9 Cascode Stages and Current Mirrors                     35
                     Improved Mirroring Accuracy




                                                        nI REF
                                          I copy   
                                                         1
                                                     1  2 n  1
                                                        



 Because of QF, the base currents of QREF and Q1 are mostly
  supplied by QF rather than IREF. Mirroring error is reduced 
  times.
CH 9 Cascode Stages and Current Mirrors                       36
      Example: Different Mirroring Ratio Accuracy




                                                        I REF
                                          I copy1    
                                                           15
                                                       4 2
                                                           
                                                       10 I REF
                                          I copy 2   
                                                            15
                                                       4 2
                                                           


CH 9 Cascode Stages and Current Mirrors                    37
                               PNP Current Mirror




 PNP current mirror is used as a current source load to an
  NPN amplifier stage.

CH 9 Cascode Stages and Current Mirrors                       38
          Generation of IREF for PNP Current Mirror




CH 9 Cascode Stages and Current Mirrors               39
    Example: Current Mirror with Discrete Devices




 Let QREF and Q1 be discrete NPN devices. IREF and Icopy1 can
  vary in large magnitude due to IS mismatch.

CH 9 Cascode Stages and Current Mirrors                          40
                               MOS Current Mirror




 The same concept of current mirror can be applied to MOS
  transistors as well.

CH 9 Cascode Stages and Current Mirrors                  41
                 Bad MOS Current Mirror Example




 This is not a current mirror since the relationship between
  VX and IREF is not clearly defined.
 The only way to clearly define VX with IREF is to use a diode-
  connected MOS since it provides square-law I-V
  relationship.
CH 9 Cascode Stages and Current Mirrors                        42
                        Example: Current Scaling




 Similar to their bipolar counterpart, MOS current mirrors
  can also scale IREF up or down (I1 = 0.2mA, I2 = 0.5mA).
CH 9 Cascode Stages and Current Mirrors                       43
                             CMOS Current Mirror




 The idea of combining NMOS and PMOS to produce CMOS
  current mirror is shown above.
CH 9 Cascode Stages and Current Mirrors             44
Chapter 10 Differential Amplifiers


  10.1 General Considerations

  10.2 Bipolar Differential Pair

  10.3 MOS Differential Pair

  10.4 Cascode Differential Amplifiers

  10.5 Common-Mode Rejection

  10.6 Differential Pair with Active Load


                                             45
                                Audio Amplifier Example




 An audio amplifier is constructed above that takes on a
  rectified AC voltage as its supply and amplifies an audio
  signal from a microphone.
CH 10 Differential Amplifiers                                 46
      “Humming” Noise in Audio Amplifier Example




 However, VCC contains a ripple from rectification that leaks
  to the output and is perceived as a “humming” noise by the
  user.
CH 10 Differential Amplifiers                                47
                                Supply Ripple Rejection




                                                      v X  Av vin  vr
                                                      vY  vr
                                                      v X  vY  Av vin



 Since both node X and Y contain the ripple, their difference
  will be free of ripple.

CH 10 Differential Amplifiers                                      48
                       Ripple-Free Differential Output




 Since the signal is taken as a difference between two
  nodes, an amplifier that senses differential signals is
  needed.
CH 10 Differential Amplifiers                               49
             Common Inputs to Differential Amplifier



                                                 v X  Av vin  vr
                                                 vY  Av vin  vr
                                                 v X  vY  0



 Signals cannot be applied in phase to the inputs of a
  differential amplifier, since the outputs will also be in
  phase, producing zero differential output.
CH 10 Differential Amplifiers                                 50
           Differential Inputs to Differential Amplifier



                                             v X  Av vin  vr
                                             vY   Av vin  vr
                                             v X  vY  2 Av vin




 When the inputs are applied differentially, the outputs are
  180° out of phase; enhancing each other when sensed
  differentially.
CH 10 Differential Amplifiers                                   51
                                Differential Signals




 A pair of differential signals can be generated, among other
  ways, by a transformer.
 Differential signals have the property that they share the
  same average value to ground and are equal in magnitude
  but opposite in phase.
CH 10 Differential Amplifiers                                52
                 Single-ended vs. Differential Signals




CH 10 Differential Amplifiers                            53
                                Differential Pair




 With the addition of a tail current, the circuits above operate
  as an elegant, yet robust differential pair.

CH 10 Differential Amplifiers                                  54
                            Common-Mode Response




                                          VBE1  VBE 2
                                                           I EE
                                          I C1  I C 2   
                                                             2
                                                                   I EE
                                          V X  VY  VCC       RC
                                                                     2




CH 10 Differential Amplifiers                                     55
                                Common-Mode Rejection




 Due to the fixed tail current source, the input common-
  mode value can vary without changing the output common-
  mode value.

CH 10 Differential Amplifiers                           56
                                Differential Response I




                                                      I C1  I EE
                                                      IC2  0
                                                      V X  VCC  RC I EE
                                                      VY  VCC
CH 10 Differential Amplifiers                                          57
                                Differential Response II




                                                       I C 2  I EE
                                                       I C1  0
                                                       VY  VCC  RC I EE
                                                       V X  VCC
CH 10 Differential Amplifiers                                          58
                      Differential Pair Characteristics




 None-zero differential input produces variations in output
  currents and voltages, whereas common-mode input
  produces no variations.
CH 10 Differential Amplifiers                                  59
                                Small-Signal Analysis



                                                               I EE
                                                        I C1        I
                                                                 2
                                                               I EE
                                                        IC2         I
                                                                 2




 Since the input to Q1 and Q2 rises and falls by the same
  amount, and their bases are tied together, the rise in IC1 has
  the same magnitude as the fall in IC2.
CH 10 Differential Amplifiers                                       60
                                Virtual Ground




                                                 VP  0
                                                 I C1  g m V
                                                 I C 2   g m V

 For small changes at inputs, the gm’s are the same, and the
  respective increase and decrease of IC1 and IC2 are the
  same, node P must stay constant to accommodate these
  changes. Therefore, node P can be viewed as AC ground.
CH 10 Differential Amplifiers                                  61
                        Small-Signal Differential Gain




                                                2 g m VRC
                                          Av                 g m RC
                                                   2V
 Since the output changes by -2gmVRC and input by 2V,
  the small signal gain is –gmRC, similar to that of the CE
  stage. However, to obtain same gain as the CE stage,
  power dissipation is doubled.
CH 10 Differential Amplifiers                                      62
                                Large Signal Analysis




                                                                   Vin1  Vin 2
                                                          I EE exp
                                                                       VT
                                                 I C1   
                                                                  Vin1  Vin 2
                                                          1  exp
                                                                      VT
                                                                 I EE
                                                 IC2    
                                                                  Vin1  Vin 2
                                                          1  exp
                                                                      VT



CH 10 Differential Amplifiers                                              63
                         Input/Output Characteristics




                                                Vout1  Vout 2 
                                                               Vin1  Vin 2
                                                 RC I EE tanh
                                                                  2VT




CH 10 Differential Amplifiers                                       64
                                Linear/Nonlinear Regions




 The left column operates in linear region, whereas the right
  column operates in nonlinear region.

CH 10 Differential Amplifiers                                65
                                Small-Signal Model




CH 10 Differential Amplifiers                        66
                                Half Circuits




                                                vout1  vout 2
                                                                 g m RC
                                                 vin1  vin 2


 Since VP is grounded, we can treat the differential pair as
  two CE “half circuits”, with its gain equal to one half
  circuit’s single-ended gain.
CH 10 Differential Amplifiers                                      67
                            Example: Differential Gain




                                vout1  vout 2
                                                 g m rO
                                 vin1  vin 2

CH 10 Differential Amplifiers                               68
                          Extension of Virtual Ground




                                  VX  0


 It can be shown that if R1 = R2, and points A and B go up
  and down by the same amount respectively, VX does not
  move.
CH 10 Differential Amplifiers                                 69
                                Half Circuit Example I




                                Av   g m1 rO1 || rO 3 || R1 
CH 10 Differential Amplifiers                                      70
                                Half Circuit Example II




                                Av   g m1 rO1 || rO 3 || R1 
CH 10 Differential Amplifiers                                      71
                                Half Circuit Example III




                                               RC
                                    Av  
                                                  1
                                             RE 
                                                  gm
CH 10 Differential Amplifiers                              72
                                Half Circuit Example IV




                                               RC
                                    Av  
                                             RE   1
                                                
                                              2 gm
CH 10 Differential Amplifiers                             73
  MOS Differential Pair’s Common-Mode Response




                                                                   I SS
                                        V X  VY  VDD         RD
                                                                    2




 Similar to its bipolar counterpart, MOS differential pair
  produces zero differential output as VCM changes.

CH 10 Differential Amplifiers                                    74
                        Equilibrium Overdrive Voltage




                                                                   I SS
                                          VGS  VTH equil 
                                                                        W
                                                                 n Cox
                                                                        L




 The equilibrium overdrive voltage is defined as the
  overdrive voltage seen by M1 and M2 when both of them
  carry a current of ISS/2.
CH 10 Differential Amplifiers                                      75
            Minimum Common-mode Output Voltage




                                               I SS
                                    VDD    RD       VCM  VTH
                                                2




 In order to maintain M1 and M2 in saturation, the common-
  mode output voltage cannot fall below the value above.
 This value usually limits voltage gain.
CH 10 Differential Amplifiers                                 76
                                Differential Response




CH 10 Differential Amplifiers                           77
                                Small-Signal Response



                                                    VP  0
                                                        Av   g m RD




 Similar to its bipolar counterpart, the MOS differential pair
  exhibits the same virtual ground node and small signal
  gain.
CH 10 Differential Amplifiers                                       78
                                Power and Gain Tradeoff




 In order to obtain the source gain as a CS stage, a MOS
  differential pair must dissipate twice the amount of current.
  This power and gain tradeoff is also echoed in its bipolar
  counterpart.
CH 10 Differential Amplifiers                                 79
     MOS Differential Pair’s Large-Signal Response




                    n Cox Vin1  V in 2 
                   1       W                    4 I SS
I D1  I D 2                                              Vin1  Vin 2 
                                                                          2

                   2       L                           W
                                              n Cox
                                                       L
CH 10 Differential Amplifiers                                            80
                 Maximum Differential Input Voltage




                 Vin1  Vin 2   max
                                         2 VGS  VTH equil

 There exists a finite differential input voltage that
  completely steers the tail current from one transistor to the
  other. This value is known as the maximum differential
  input voltage.
CH 10 Differential Amplifiers                                   81
Contrast Between MOS and Bipolar Differential Pairs


                       MOS                      Bipolar




 In a MOS differential pair, there exists a finite differential
  input voltage to completely switch the current from one
  transistor to the other, whereas, in a bipolar pair that
  voltage is infinite.

CH 10 Differential Amplifiers                                      82
             The effects of Doubling the Tail Current




 Since ISS is doubled and W/L is unchanged, the equilibrium
  overdrive voltage for each transistor must increase by 2
  to accommodate this change, thus Vin,max increases by 2
  as well. Moreover, since ISS is doubled, the differential
  output swing will double.
CH 10 Differential Amplifiers                              83
                          The effects of Doubling W/L




 Since W/L is doubled and the tail current remains
  unchanged, the equilibrium overdrive voltage will be
  lowered by 2 to accommodate this change, thus Vin,max
  will be lowered by 2 as well. Moreover, the differential
  output swing will remain unchanged since neither ISS nor RD
  has changed
CH 10 Differential Amplifiers                              84
           Small-Signal Analysis of MOS Differential Pair




                1       W                   4 I SS            W
I D1  I D 2     n Cox Vin1  Vin 2                n Cox I SS Vin1  Vin 2 
                2       L                          W          L
                                          n Cox
                                                   L
       When the input differential signal is small compared to
        4ISS/nCox(W/L), the output differential current is linearly
        proportional to it, and small-signal model can be applied.

      CH 10 Differential Amplifiers                                         85
                       Virtual Ground and Half Circuit




                                VP  0
                                Av   g m RC

 Applying the same analysis as the bipolar case, we will
  arrive at the same conclusion that node P will not move for
  small input signals and the concept of half circuit can be
  used to calculate the gain.
CH 10 Differential Amplifiers                               86
         MOS Differential Pair Half Circuit Example I




                                0
                                             1                
                                             g || rO 3 || rO1 
                                Av   g m1                   
                                             m3               
CH 10 Differential Amplifiers                                      87
         MOS Differential Pair Half Circuit Example II




                                 0
                                       g m1
                                Av  
                                       g m3
CH 10 Differential Amplifiers                            88
        MOS Differential Pair Half Circuit Example III




                                 0
                                          RDD 2
                                Av  
                                       RSS 2  1 g m
CH 10 Differential Amplifiers                            89
                     Bipolar Cascode Differential Pair




                Av   g m1 g m3 rO1 || r 3 rO 3  rO1 
CH 10 Differential Amplifiers                                  90
                          Bipolar Telescopic Cascode




  Av   g m1 g m3 rO 3 rO1 || r 3  || g m5 rO 5 (rO 7 || r 5 )
CH 10 Differential Amplifiers                                             91
  Example: Bipolar Telescopic Parasitic Resistance




                                                 R1                R1
          Rop  rO 5 1  g m 5  rO 7 || r 5 ||   rO 7 || r 5 ||
                                                  2                2
          Av   g m1 g m 3 rO 3 (rO1 || r 3 ) || Rop
CH 10 Differential Amplifiers                                               92
                       MOS Cascode Differential Pair




                                Av   g m1rO 3 g m3 rO1
CH 10 Differential Amplifiers                              93
                                MOS Telescopic Cascode




      Av   g m1  g m3 rO 3 rO1  || ( g m5 rO 5 rO 7 )
CH 10 Differential Amplifiers                                 94
     Example: MOS Telescopic Parasitic Resistance




                                Rop  rO 5 || [ R1 1  g m5 rO 7   rO 7 ]
                                Av   g m1 ( Rop || rO 3 g m3 rO1 )
CH 10 Differential Amplifiers                                                  95
                       Effect of Finite Tail Impedance




                                Vout ,CM         RC / 2
                                            
                                Vin,CM        REE  1 / 2 g m
 If the tail current source is not ideal, then when a input CM
  voltage is applied, the currents in Q1 and Q2 and hence
  output CM voltage will change.
CH 10 Differential Amplifiers                                     96
                      Input CM Noise with Ideal Tail Current




CH 10 Differential Amplifiers                                  97
          Input CM Noise with Non-ideal Tail Current




CH 10 Differential Amplifiers                          98
                                Comparison




 As it can be seen, the differential output voltages for both
  cases are the same. So for small input CM noise, the
  differential pair is not affected.
CH 10 Differential Amplifiers                                    99
                        CM to DM Conversion, ACM-DM




                                             Vout      RD
                                                   
                                             VCM 1 / g m  2 REE




 If finite tail impedance and asymmetry are both present,
  then the differential output signal will contain a portion of
  input common-mode signal.
CH 10 Differential Amplifiers                                     100
                                Example: ACM-DM




                                   ACM  DM 
                                                         R C
                                         2[1  g m3 ( R1 || r 3 )]rO 3  R1 || r 3 
                                    1
                                   g m1
CH 10 Differential Amplifiers                                                  101
                                CMRR




                                                            ADM
                                            CMRR 
                                                       ACM  DM




 CMRR defines the ratio of wanted amplified differential
  input signal to unwanted converted input common-mode
  noise that appears at the output.
CH 10 Differential Amplifiers                               102
             Differential to Single-ended Conversion




 Many circuits require a differential to single-ended
  conversion, however, the above topology is not very good.
CH 10 Differential Amplifiers                             103
                                Supply Noise Corruption




 The most critical drawback of this topology is supply noise
  corruption, since no common-mode cancellation
  mechanism exists. Also, we lose half of the signal.
CH 10 Differential Amplifiers                              104
                                Better Alternative




 This circuit topology performs differential to single-ended
  conversion with no loss of gain.
CH 10 Differential Amplifiers                                   105
                                Active Load




 With current mirror used as the load, the signal current
  produced by the Q1 can be replicated onto Q4.
 This type of load is different from the conventional “static
  load” and is known as an “active load”.

CH 10 Differential Amplifiers                                    106
                     Differential Pair with Active Load




 The input differential pair decreases the current drawn from
  RL by I and the active load pushes an extra I into RL by
  current mirror action; these effects enhance each other.
CH 10 Differential Amplifiers                               107
                           Active Load vs. Static Load




 The load on the left responds to the input signal and
  enhances the single-ended output, whereas the load on the
  right does not.
CH 10 Differential Amplifiers                            108
               MOS Differential Pair with Active Load




 Similar to its bipolar counterpart, MOS differential pair can
  also use active load to enhance its single-ended output.

CH 10 Differential Amplifiers                                 109
                          Asymmetric Differential Pair




 Because of the vastly different resistance magnitude at the
  drains of M1 and M2, the voltage swings at these two nodes
  are different and therefore node P cannot be viewed as a
  virtual ground.
CH 10 Differential Amplifiers                               110
                Thevenin Equivalent of the Input Pair




                       vThev   g mN roN (vin1  vin 2 )
                       RThev  2roN
CH 10 Differential Amplifiers                               111
         Simplified Differential Pair with Active Load




                       vout
                                 g mN (rON || rOP )
                   vin1  vin 2
CH 10 Differential Amplifiers                            112
                                  Proof of VA << Vout




          vout                     A
vA  
       2 g mP rOP
                                         I


                                                        vout
                                                   I        g m4v A
                                                        rO 4


  CH 10 Differential Amplifiers                                    113
                    Chapter 11 Frequency Response




              11.1      Fundamental Concepts
              11.2      High-Frequency Models of Transistors
              11.3      Analysis Procedure
              11.4      Frequency Response of CE and CS Stages
              11.5      Frequency Response of CB and CG Stages
              11.6      Frequency Response of Followers
              11.7      Frequency Response of Cascode Stage
              11.8      Frequency Response of Differential Pairs
              11.9      Additional Examples


CH 10 Differential Amplifiers                                       114
                                Chapter Outline




   11 Differential Amplifiers
CH 10 Frequency Response                          115
                 High Frequency Roll-off of Amplifier




 As frequency of operation increases, the gain of amplifier
  decreases. This chapter analyzes this problem.
   11 Differential Amplifiers
CH 10 Frequency Response                                       116
                 Example: Human Voice I



          Natural Voice                            Telephone System




 Natural human voice spans a frequency range from 20Hz to
    20KHz, however conventional telephone system passes
    frequencies from 400Hz to 3.5KHz. Therefore phone
    conversation
CH 10 Frequency Response differs from face-to-face conversation.
   11 Differential Amplifiers                                    117
                            Example: Human Voice II


               Path traveled by the human voice to the voice recorder


                  Mouth                   Air                  Recorder



                  Path traveled by the human voice to the human ear


                 Mouth                    Air                    Ear




                                         Skull


 Since the paths are different, the results will also be
  different.
   11 Differential Amplifiers
CH 10 Frequency Response                                                  118
                                Example: Video Signal




                          High Bandwidth        Low Bandwidth



 Video signals without sufficient bandwidth become fuzzy as
  they fail to abruptly change the contrast of pictures from
  complete white into complete black.

   11 Differential Amplifiers
CH 10 Frequency Response                                        119
               Gain Roll-off: Simple Low-pass Filter




 In this simple example, as frequency increases the
  impedance of C1 decreases and the voltage divider consists
  of C1 and R1 attenuates Vin to a greater extent at the output.
   11 Differential Amplifiers
CH 10 Frequency Response                                      120
             Gain Roll-off: Common Source




                                           1 
                 Vout     g mVin  RD ||      
                                          CL s 



 The capacitive load, CL, is the culprit for gain roll-off since
    at high frequency, it will “steal” away some signal current
    and shunt it to ground.
   11 Differential Amplifiers
CH 10 Frequency Response                                         121
         Frequency Response of the CS Stage




                                           Vout          g m RD
                                                
                                           Vin        RD C L  2  1
                                                       2   2




 At low frequency, the capacitor is effectively open and the
    gain is flat. As frequency increases, the capacitor tends to
    a short and the gain starts to decrease. A special
CH 10 Frequency Responseω=1/(RDCL), where the gain drops by 3dB.
    frequency is
   11 Differential Amplifiers                                    122
                                 Example: Figure of Merit




                                                                    1
                                                   F .O.M . 
                                                                VT VCC C L



 This metric quantifies a circuit’s gain, bandwidth, and
  power dissipation. In the bipolar case, low temperature,
  supply, and load capacitance mark a superior figure of
  merit.
    11 Differential Amplifiers
 CH 10 Frequency Response                                               123
            Example: Relationship between Frequency
                 Response and Step Response




                                   1                                  t 
H  s  j                                Vout  t   V0  1  exp       u t 
                            R1 C1  2  1
                             2 2
                                                                     R1C1 


   The relationship is such that as R1C1 increases, the
    bandwidth drops and the step response becomes slower.

      11 Differential Amplifiers
   CH 10 Frequency Response                                                124
                        Bode Plot



                           s       s 
                       1 
                         1    
                                        
           H ( s)  A0      z1     z2 

                           s       s 
                       1      1      
                           
                            p1      p2 




 When we hit a zero, ωzj, the Bode magnitude rises with a
    slope of +20dB/dec.
 When we hit a pole, ωpj, the Bode magnitude falls with a
    slope of -20dB/dec
   11 Differential Amplifiers
CH 10 Frequency Response                                     125
                                Example: Bode Plot




                                            1
                                  p1   
                                          RD C L

 The circuit only has one pole (no zero) at 1/(RDCL), so the
  slope drops from 0 to -20dB/dec as we pass ωp1.
   11 Differential Amplifiers
CH 10 Frequency Response                                        126
                                Pole Identification Example I




                                                                                         1
 p1   
           1                                                                   p2 
         RS Cin                                                                        RD C L


                               Vout                   g m RD
                                    
                               Vin      1     2
                                                      p1 1   2  p 2 
                                                       2              2


          11 Differential Amplifiers
       CH 10 Frequency Response                                                    127
                           Pole Identification Example II




                 1
 p1                                                             1
                 1                                    p2   
          RS ||
                   Cin                                        RD C L
                gm 
                    
     11 Differential Amplifiers
  CH 10 Frequency Response                                               128
             Circuit with Floating Capacitor




 The pole of a circuit is computed by finding the effective
    resistance and capacitance from a node to GROUND.
 The circuit above creates a problem since neither terminal
    of CF is grounded.
   11 Differential Amplifiers
CH 10 Frequency Response                                     129
                                Miller’s Theorem




                             ZF                      ZF
                      Z1                    Z2 
                           1  Av                 1  1 / Av
 If Av is the gain from node 1 to 2, then a floating impedance
  ZF can be converted to two grounded impedances Z1 and Z2.

   11 Differential Amplifiers
CH 10 Frequency Response                                       130
                                Miller Multiplication




 With Miller’s theorem, we can separate the floating
  capacitor. However, the input capacitor is larger than the
  original floating capacitor. We call this Miller multiplication.
   11 Differential Amplifiers
CH 10 Frequency Response                                       131
                                 Example: Miller Theorem




                                                              1
               1                               out 
in                                                             1 
      RS 1  g m RD C F                                    g R C F
                                                        R D 1      
                                                                m D 


    11 Differential Amplifiers
 CH 10 Frequency Response                                          132
                           High-Pass Filter Response




                                 Vout       R1C1
                                      
                                 Vin      R12C1212  1

 The voltage division between a resistor and a capacitor can
  be configured such that the gain at low frequency is
  reduced.

   11 Differential Amplifiers
CH 10 Frequency Response                                    133
                               Example: Audio Amplifier




Ci  79 .6nF                                              CL  39 .8nF
                                      Ri  100K
                                      g m  1 / 200
      In order to successfully pass audio band frequencies (20
       Hz-20 KHz), large input and output capacitances are
       needed.
      11 Differential Amplifiers
   CH 10 Frequency Response                                      134
      Capacitive Coupling vs. Direct Coupling




     Capacitive Coupling                   Direct Coupling




 Capacitive coupling, also known as AC coupling, passes
    AC signals from Y to X while blocking DC contents.
 This technique allows independent bias conditions between
    stages. Direct coupling does not.
   11 Differential Amplifiers
CH 10 Frequency Response                                 135
                         Typical Frequency Response




                          Lower Corner            Upper Corner




   11 Differential Amplifiers
CH 10 Frequency Response                                         136
                       High-Frequency Bipolar Model




                                 C  Cb  C je


 At high frequency, capacitive effects come into play. Cb
  represents the base charge, whereas C and Cje are the
  junction capacitances.
   11 Differential Amplifiers
CH 10 Frequency Response                                     137
     High-Frequency Model of Integrated Bipolar
                    Transistor




 Since an integrated bipolar circuit is fabricated on top of a
    substrate, another junction capacitance exists between the
    collector and substrate, namely CCS.
   11 Differential Amplifiers
CH 10 Frequency Response                                       138
                 Example: Capacitance Identification




   11 Differential Amplifiers
CH 10 Frequency Response                               139
              MOS Intrinsic Capacitances




 For a MOS, there exist oxide capacitance from gate to
    channel, junction capacitances from source/drain to
    substrate, and overlap capacitance from gate to
    source/drain.
   11 Differential Amplifiers
CH 10 Frequency Response                                  140
   Gate Oxide Capacitance Partition and Full Model




 The gate oxide capacitance is often partitioned between
  source and drain. In saturation, C2 ~ Cgate, and C1 ~ 0. They
  are in parallel with the overlap capacitance to form CGS and
  CGD.
   11 Differential Amplifiers
CH 10 Frequency Response                                     141
                 Example: Capacitance Identification




   11 Differential Amplifiers
CH 10 Frequency Response                               142
                                Transit Frequency




               gm                                           gm
       2f T                                       2f T 
               CGS                                          C


 Transit frequency, fT, is defined as the frequency where the
  current gain from input to output drops to 1.
   11 Differential Amplifiers
CH 10 Frequency Response                                         143
            Example: Transit Frequency Calculation




                                                               3 n
                                                        2fT     2
                                                                    VGS  VTH 
                                                               2L




                                L  65nm
                                VGS  VTH  100mV
                                 n  400cm 2 /(V .s)
                                fT  226GHz
   11 Differential Amplifiers
CH 10 Frequency Response                                                   144
                                Analysis Summary



  The frequency response refers to the magnitude of the
   transfer function.
  Bode’s approximation simplifies the plotting of the
   frequency response if poles and zeros are known.
  In general, it is possible to associate a pole with each node
   in the signal path.
  Miller’s theorem helps to decompose floating capacitors
   into grounded elements.
  Bipolar and MOS devices exhibit various capacitances that
   limit the speed of circuits.




   11 Differential Amplifiers
CH 10 Frequency Response                                      145
         High Frequency Circuit Analysis Procedure



 Determine which capacitor impact the low-frequency region
  of the response and calculate the low-frequency pole
  (neglect transistor capacitance).
 Calculate the midband gain by replacing the capacitors with
  short circuits (neglect transistor capacitance).
 Include transistor capacitances.
 Merge capacitors connected to AC grounds and omit those
  that play no role in the circuit.
 Determine the high-frequency poles and zeros.
 Plot the frequency response using Bode’s rules or exact
  analysis.




   11 Differential Amplifiers
CH 10 Frequency Response                                   146
                          Frequency Response of CS Stage
                            with Bypassed Degeneration




                                             g m RD RS Cb s  1
                                Vout
                                     s  
                                VX           RS Cb s  g m RS  1
 In order to increase the midband gain, a capacitor Cb is
  placed in parallel with Rs.
 The pole frequency must be well below the lowest signal
  frequency to avoid the effect of degeneration.
   11 Differential Amplifiers
CH 10 Frequency Response                                             147
                 Unified Model for CE and CS Stages




   11 Differential Amplifiers
CH 10 Frequency Response                              148
               Unified Model Using Miller’s Theorem




   11 Differential Amplifiers
CH 10 Frequency Response                              149
                                Example: CE Stage


                                                              RS  200
                                                              I C  1mA
                                                                100
                                                              C  100 fF
                                                              C   20 fF
                                                              CCS  30 fF




                                  p ,in  2  516MHz
                                  p ,out  2  1.59GHz 

 The input pole is the bottleneck for speed.

   11 Differential Amplifiers
CH 10 Frequency Response                                                    150
                       Example: Half Width CS Stage




                                                                   W  2X



                                                         1
                                 p ,in 
                                                C      g R C 
                                             RS  in  1  m L  XY 
                                                 2          2  2 
                                                            1
                                 p ,out   
                                                  Cout        2  C XY 
                                              RL         g R  2 
                                                        1       
                                                   2         m L      
   11 Differential Amplifiers
CH 10 Frequency Response                                                     151
                  Direct Analysis of CE and CS Stages

         gm
|  z |
         C XY
                                     1
|  p1 |
          1  g m RL C XY RThev  RThevCin  RL C XY  Cout 
           1  g m RL C XY RThev  RThevCin  RL C XY  Cout 
|  p 2 |
                  RThev RL Cin C XY  Cout C XY  Cin Cout 



   Direct analysis yields different pole locations and an extra
    zero.
     11 Differential Amplifiers
  CH 10 Frequency Response                                     152
                   Example: CE and CS Direct Analysis




                                             1
 p1   
         1  g m1 rO1 || rO 2 C XY RS  RS Cin  rO1 || rO 2 (C XY  Cout )
       1  g m1 rO1 || rO 2 C XY RS  RS Cin  rO1 || rO 2 (C XY  Cout )
 p2 
                  RS rO1 || rO 2 Cin C XY Cout C XY  Cin Cout 
      11 Differential Amplifiers
   CH 10 Frequency Response                                                 153
       Example: Comparison Between Different Methods




RS  200
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
               1


 0
RL  2 K
          Miller’s                             Exact                   Dominant Pole

 p ,in  2  571MHz               p ,in  2  264MHz       p ,in  2  249MHz
 p ,out  2  428MHz              p ,out  2  4.53GHz     p ,out  2  4.79GHz 
     CH 10 Differential Amplifiers                                                154
     CH 11 Frequency Response                                                     154
                    Input Impedance of CE and CS Stages




                  1                                      1
Z in                           || r Z in 
       C  1  g m RC C s              CGS  1  g m RD CGD s
        11 Differential Amplifiers
     CH 10 Frequency Response                                  155
    Low Frequency Response of CB and CG Stages




                                Vout
                                     s         g m RC Ci s
                                Vin         1  g m RS Ci s  g m
 As with CE and CS stages, the use of capacitive coupling
  leads to low-frequency roll-off in CB and CG stages
  (although a CB stage is shown above, a CG stage is
  similar).
   11 Differential Amplifiers
CH 10 Frequency Response                                              156
                   Frequency Response of CB Stage


                                                            1
                                          p, X   
                                                            1 
                                                     RS ||
                                                              C X
                                                           gm 
                                                               
                                                  C X  C

                                                      1
                                          p ,Y   
                                                    RL CY
                                rO  
                                                  CY  C  CCS

   11 Differential Amplifiers
CH 10 Frequency Response                                          157
                   Frequency Response of CG Stage

                                                            1
                                          p , Xr   
                                                 O          1 
                                                     RS ||
                                                              C X
                                                           gm 
                                                               
                                            C X  CGS  C SB

                                                      1
                                          p ,Y   
                                                    RL CY
                                rO  
                                            CY  CGD  C DB


 Similar to a CB stage, the input pole is on the order of fT, so
  rarely a speed bottleneck.
   11 Differential Amplifiers
CH 10 Frequency Response                                              158
                   Example: CG Stage Pole Identification




                               1                                       1
 p, X                                     p ,Y 
                   1                                 1
                                                           C DB1  CGD1  CGS 2  C DB 2 
            RS ||
                       C SB1  CGD1 
                  g m1 
                        
                                                      g m2

        11 Differential Amplifiers
     CH 10 Frequency Response                                                     159
           Example: Frequency Response of CG Stage




RS  200
CGS  250 f F
CGD  80 f F
C DB  100 f F
g m  150 
                  1
                                    p , X  2  5.31GHz 
 0
                                    p ,Y  2  442MHz
Rd  2 K
   CH 10 Differential Amplifiers                               160
   CH 11 Frequency Response                                    160
             Emitter and Source Followers




 The following will discuss the frequency response of
    emitter and source followers using direct analysis.
 Emitter follower is treated first and source follower is
    derived easily
CH 10 Frequency Response by allowing r to go to infinity.
   11 Differential Amplifiers                                161
                  Direct Analysis of Emitter Follower




                C                a
                                     RS
                                        C C  C C L  C C L 
             1     s                gm
 Vout           gm
            2                                 C  RS        CL
            as  bs  1           b  RS C      1       
 Vin                                           gm     r   g
                                                              m
   11 Differential Amplifiers
CH 10 Frequency Response                                      162
                  Direct Analysis of Source Follower Stage




                                          RS
          1
             CGS
                 s                     a    CGD CGS  CGD C SB  CGS C SB 
                                          gm
Vout         gm
         2                                         CGD  C SB
Vin      as  bs  1                   b  RS CGD 
                                                       gm
          11 Differential Amplifiers
       CH 10 Frequency Response                                        163
Example: Frequency Response of Source Follower




      RS  200
      C L  100 fF
      CGS  250 fF
      CGD  80 fF
      C DB  100 fF              p1  2  1.79GHz  j 2.57GHz 
      g m  150 
                     1

                                 p 2  2  1.79GHz  j 2.57GHz 
       0
CH 10 Frequency Response
CH 11 Differential Amplifiers                                           164
                                                                        164
                           Example: Source Follower




                                                        CGS
                                                     1     s
                                            Vout        gm
                                                    2
                                            Vin     as  bs  1


   RS
a      CGD1CGS1  (CGD1  CGS1 )(C SB1  CGD 2  C DB 2 )
   g m1
              CGD1  C SB1 C GD 2 C DB 2
b  RS CGD1 
                         g m1
   11 Differential Amplifiers
CH 10 Frequency Response                                       165
       Input Capacitance of Emitter/Source Follower




                                rO  

                                        C / CGS
                      Cin  C  / CGD 
                                        1  g m RL
   11 Differential Amplifiers
CH 10 Frequency Response                              166
      Example: Source Follower Input Capacitance




                                         1
                  Cin  CGD1                          CGS1
                               1  g m1 rO1 || rO 2 
   11 Differential Amplifiers
CH 10 Frequency Response                                      167
               Output Impedance of Emitter Follower




                            V X RS r C s  r  RS
                               
                            IX     r C s    1
   11 Differential Amplifiers
CH 10 Frequency Response                               168
               Output Impedance of Source Follower




                                V X RS CGS s  1
                                   
                                I X CGS s  g m
   11 Differential Amplifiers
CH 10 Frequency Response                             169
                                Active Inductor




 The plot above shows the output impedance of emitter and
  source followers. Since a follower’s primary duty is to
  lower the driving impedance (RS>1/gm), the “active
  inductor” characteristic on the right is usually observed.
   11 Differential Amplifiers
CH 10 Frequency Response                                   170
                         Example: Output Impedance



                                     rO  




                        V X rO1 || rO 2 CGS 3 s  1
                           
                        IX      CGS 3 s  g m3
   11 Differential Amplifiers
CH 10 Frequency Response                                171
             Frequency Response of Cascode Stage




                           g m1
          Av , XY                1     C x  2C XY
                           g m2
 For cascode stages, there are three poles and Miller
  multiplication is smaller than in the CE/CS stage.
   11 Differential Amplifiers
CH 10 Frequency Response                                 172
                                         Poles of Bipolar Cascode

                                                                                     1
 p, X    
                          1                                     p ,Y 
            RS || r 1 C 1  2C 1                                    1
                                                                               CCS1  C 2  2C1 
                                                                          g m2




                                                               1
                                           p ,out 
         CH 10 Frequency Response
            11 Differential Amplifiers                 RL CCS 2  C  2                  173
                                      Poles of MOS Cascode

                                  1
 p, X                                                                               1
                         g m1                              p ,out 
           RS CGS1  1 
                               CGD1                                     RL C DB 2  CGD 2 
                         g m2 
                                     




                                                            1
                              p ,Y 
                                         1                         g m2        
                                               C DB1  CGS 2  1  g
                                                                           CGD1 
                                                                            
                                        g m2                         m1        
         11 Differential Amplifiers
      CH 10 Frequency Response                                                            174
          Example: Frequency Response of Cascode




RS  200
CGS  250 f F
CGD  80 f F
C DB  100 f F
                                  p , X  2  1.95GHz 
g m  150 
                 1


 0                              p ,Y  2  1.73GHz 
RL  2 K                         p ,out  2  442MHz
 CH 10 Frequency Response
 CH 11 Differential Amplifiers                               175
                                                             175
                            MOS Cascode Example

                          1
 p, X                                                                         1
                         g m1                          p ,out 
           RS CGS1  1 
                               CGD1                                 RL C DB 2  CGD 2 
                         g m2 
                                     




                                                 1
     p ,Y 
                    1                          g m2                         
                          C DB  CGS 2
     CH 10 Frequency Response 1
                                           1 
                                                       CGD1  CGD 3  C DB 3 
                                                        
                  g m2                                                      
        11 Differential Amplifiers                                                  176
                                                 g m1
                        I/O Impedance of Bipolar Cascode




                      1                                             1
Z in  r 1 ||                                        RL ||
               C 1  2C1 s                               C 2  CCS 2 s
                                             Z out

        11 Differential Amplifiers
     CH 10 Frequency Response                                         177
                          I/O Impedance of MOS Cascode




                   1
Z in                                       Z out    RL ||
                                                                   1
               g m1                                     CGD 2  C DB 2 s
       CGS1  1  g CGD1  s
                       
                   m2    
       11 Differential Amplifiers
    CH 10 Frequency Response                                          178
       Bipolar Differential Pair Frequency Response




                                              Half Circuit



 Since bipolar differential pair can be analyzed using half-
  circuit, its transfer function, I/O impedances, locations of
  poles/zeros are the same as that of the half circuit’s.
   11 Differential Amplifiers
CH 10 Frequency Response                                         179
          MOS Differential Pair Frequency Response




                                              Half Circuit



 Since MOS differential pair can be analyzed using half-
  circuit, its transfer function, I/O impedances, locations of
  poles/zeros are the same as that of the half circuit’s.
   11 Differential Amplifiers
CH 10 Frequency Response                                         180
                      Example: MOS Differential Pair

                                                            1
                                  p, X 
                                             RS [CGS1  (1  g m1 / g m 3 )CGD1 ]
                                                                1
                                  p ,Y 
                                              1                       g m3      
                                                  C DB1  CGS 3  1 
                                                                            CGD1 
                                             g m3                     g m1 
                                                                                  
                                                       1
                                  p ,out   
                                              RL C DB 3  CGD 3 




   11 Differential Amplifiers
CH 10 Frequency Response                                                     181
                Common Mode Frequency Response




                                Vout   g R R C  1
                                       m D SS SS
                                VCM   RSS CSS s  2 g m RSS  1

 Css will lower the total impedance between point P to
  ground at high frequency, leading to higher CM gain which
  degrades the CM rejection ratio.
CH 10 Differential Amplifiers                                      182
CH 11 Frequency Response                                           182
                 Tail Node Capacitance Contribution




                                   Source-Body Capacitance of
                                    M1, M2 and M3
                                   Gate-Drain Capacitance of M3




   11 Differential Amplifiers
CH 10 Frequency Response                                     183
                           Example: Capacitive Coupling




                                    Rin 2  RB 2 || r 2    1RE 


                           2  542 Hz 
                  1
                                                                                 22 .9 Hz 
                                                                    1
 L1                                                L 2 
         r 1 || RB1 C1                                     RC  Rin 2 C2
    CH 10 Differential Amplifiers                                                      184
    CH 11 Frequency Response                                                           184
    Example: IC Amplifier – Low Frequency Design




                                                         RF
                                             Rin 2 
                                                       1  Av 2

                                         g m1 RS 1  1
                                 L1                   2  42 .4 MHz 
                                            RS 1C1


                                                           2  6.92 MHz 
                                               1
                                L 2 
                                         RD1  Rin 2 C2




   11 Differential Amplifiers
CH 10 Frequency Response                                            185
            Example: IC Amplifier – Midband Design




                                       g m1 RD1 || Rin 2   3.77
                                 vX
                                 vin




CH 10 Differential Amplifiers                                      186
CH 11 Frequency Response                                           186
    Example: IC Amplifier – High Frequency Design




                                  p1  2  (308 MHz)
                                  p 2  2  (2.15 GHz)
                                                     1
                                  p3 
                                          RL 2 (1.15CGD2  CDB 2 )
                                      2  (1.21 GHz)

CH 10 Differential Amplifiers                             187
CH 11 Frequency Response                                  187
     Chapter 12 Feedback


 12.1 General Considerations

 12.2 Types of Amplifiers

 12.3 Sense and Return Techniques

 12.4 Polarity of Feedback

 12.5 Feedback Topologies

 12.6 Effect of Finite I/O Impedances

 12.7 Stability in Feedback Systems

                                         188
                 Negative Feedback System




 A negative feedback system consists of four components:
  1) feedforward system, 2) sense mechanism, 3) feedback
  network, and 4) comparison mechanism.
CH 12 Feedback                                          189
                 Close-loop Transfer Function




                        Y    A1
                          
                        X 1  KA1
CH 12 Feedback                                  190
                  Feedback Example




                                Y            A1
                                  
                                X           R2
                                       1         A1
                                          R1  R2


 A1 is the feedforward network, R1 and R2 provide the
  sensing and feedback capabilities, and comparison is
  provided by differential input of A1.
CH 12 Feedback                                           191
                     Comparison Error



                 E

                                               X
                                          E
                                             1 A1 K



 As A1K increases, the error between the input and fed back
  signal decreases. Or the fed back signal approaches a
  good replica of the input.
CH 12 Feedback                                            192
                 Comparison Error




                                    Y      R1
                                       1
                                    X      R2


CH 12 Feedback                            193
                        Loop Gain




                                                      VN
X 0                                          KA1  
                                                      Vtest




 When the input is grounded, and the loop is broken at an
  arbitrary location, the loop gain is measured to be KA1.
CH 12 Feedback                                               194
    Example: Alternative Loop Gain Measurement




                 VN   KA1Vtest
CH 12 Feedback                                   195
                 Incorrect Calculation of Loop Gain




 Signal naturally flows from the input to the output of a
  feedforward/feedback system. If we apply the input the
  other way around, the “output” signal we get is not a result
  of the loop gain, but due to poor isolation.
CH 12 Feedback                                              196
                     Gain Desensitization




                                     Y   1
                 A1 K  1             
                                     X K
 A large loop gain is needed to create a precise gain, one
  that does not depend on A1, which can vary by ±20%.

CH 12 Feedback                                                197
                    Ratio of Resistors




 When two resistors are composed of the same unit resistor,
  their ratio is very accurate. Since when they vary, they will
  vary together and maintain a constant ratio.

CH 12 Feedback                                               198
                 Merits of Negative Feedback




                  1) Bandwidth
                   enhancement

                  2) Modification of I/O
                   Impedances

                  3) Linearization




CH 12 Feedback                                 199
                         Bandwidth Enhancement


                                                 Closed Loop
   Open Loop
                            Negative
                            Feedback                   A0
               A0
As                                  Y            1  KA0
          1
                    s                    s  
                                       X                  s
                    0                          1
                                                   1  KA0 0

    Although negative feedback lowers the gain by (1+KA0), it
     also extends the bandwidth by the same amount.

   CH 12 Feedback                                                200
                 Bandwidth Extension Example




 As the loop gain increases, we can see the decrease of the
  overall gain and the extension of the bandwidth.

CH 12 Feedback                                             201
                 Example: Open Loop Parameters




                         A0  g m RD
                               1
                         Rin 
                               gm
                         Rout  RD
CH 12 Feedback                                   202
             Example: Closed Loop Voltage Gain




                  vout          g m RD
                       
                  vin           R2
                           1          g m RD
                              R1  R2
CH 12 Feedback                                   203
              Example: Closed Loop I/O Impedance




                                                  RD
      1        R2               Rout   
Rin     1 
          R  R g m RD 
      gm                                 1
                                                R2
                                                      g m RD
              1    2                         R1  R2
  CH 12 Feedback                                          204
                   Example: Load Desensitization




 W/O Feedback                         With Feedback
Large Difference                     Small Difference
                                g m RD               g m RD
g m RD  g m RD / 3                           
                                R2                   R2
                           1          g m RD   3          g m RD
                              R1  R2              R1  R2
  CH 12 Feedback                                            205
                         Linearization




             Before feedback

                          After feedback




CH 12 Feedback                             206
                 Four Types of Amplifiers




CH 12 Feedback                              207
          Ideal Models of the Four Amplifier Types




CH 12 Feedback                                       208
      Realistic Models of the Four Amplifier Types




CH 12 Feedback                                       209
            Examples of the Four Amplifier Types




CH 12 Feedback                                     210
                    Sensing a Voltage




 In order to sense a voltage across two terminals, a
  voltmeter with ideally infinite impedance is used.

CH 12 Feedback                                          211
                 Sensing and Returning a Voltage




    Feedback
     Network




                           R1  R2  
 Similarly, for a feedback network to correctly sense the
  output voltage, its input impedance needs to be large.
 R1 and R2 also provide a mean to return the voltage.
CH 12 Feedback                                               212
                   Sensing a Current




 A current is measured by inserting a current meter with
  ideally zero impedance in series with the conduction path.
 The current meter is composed of a small resistance r in
  parallel with a voltmeter.

CH 12 Feedback                                             213
                 Sensing and Returning a Current




  Feedback
   Network



                             RS  0
 Similarly for a feedback network to correctly sense the
  current, its input impedance has to be small.
 RS has to be small so that its voltage drop will not change
  Iout.

CH 12 Feedback                                                  214
                 Addition of Two Voltage Sources




                                            Feedback
                                             Network




 In order to add or substrate two voltage sources, we place
  them in series. So the feedback network is placed in series
  with the input source.

CH 12 Feedback                                             215
 Practical Circuits to Subtract Two Voltage Sources




 Although not directly in series, Vin and VF are being
  subtracted since the resultant currents, differential and
  single-ended, are proportional to the difference of Vin and
  VF.
CH 12 Feedback                                                  216
                 Addition of Two Current Sources




                                                  Feedback
                                                   Network




 In order to add two current sources, we place them in
  parallel. So the feedback network is placed in parallel with
  the input signal.
CH 12 Feedback                                               217
 Practical Circuits to Subtract Two Current Sources




 Since M1 and RF are in parallel with the input current source,
  their respective currents are being subtracted. Note, RF has
  to be large enough to approximate a current source.


CH 12 Feedback                                                218
                 Example: Sense and Return




 R1 and R2 sense and return the output voltage to
  feedforward network consisting of M1- M4.
 M1 and M2 also act as a voltage subtractor.

CH 12 Feedback                                       219
                 Example: Feedback Factor




                         iF
                     K       g mF
CH 12 Feedback
                        vout                220
    Input Impedance of an Ideal Feedback Network




 To sense a voltage, the input impedance of an ideal
  feedback network must be infinite.
 To sense a current, the input impedance of an ideal
  feedback network must be zero.

CH 12 Feedback                                          221
  Output Impedance of an Ideal Feedback Network




 To return a voltage, the output impedance of an ideal
  feedback network must be zero.
 To return a current, the output impedance of an ideal
  feedback network must be infinite.

CH 12 Feedback                                            222
             Determining the Polarity of Feedback



                   1) Assume the input goes
                    either up or down.

                   2) Follow the signal through
                    the loop.

                   3) Determine whether the
                    returned quantity enhances or
                    opposes the original change.




CH 12 Feedback                                      223
                     Polarity of Feedback Example I




Vin               I D1 , I D 2    Vout ,Vx       I D 2 , I D1 

  CH 12 Feedback
                           Negative Feedback                  224
                    Polarity of Feedback Example II




Vin               I D1  , V A     Vout ,Vx       I D1  , V A 

  CH 12 Feedback
                           Negative Feedback                  225
                     Polarity of Feedback Example III




I in               I D1 , V X      Vout , I D 2    I D1 , V X 

   CH 12 Feedback
                            Positive Feedback                   226
                 Voltage-Voltage Feedback




                      Vout   A0
                           
                      Vin 1  KA0
CH 12 Feedback                              227
             Example: Voltage-Voltage Feedback




                 Vout          g mN (rON || rOP )
                      
                 Vin           R2
                          1          g mN (rON || rOP )
                             R1  R2
CH 12 Feedback                                             228
                 Input Impedance of a V-V Feedback




                        Vin
                              Rin (1  A0 K )
                        I in

 A better voltage sensor
CH 12 Feedback                                       229
         Example: V-V Feedback Input Impedance




                 Vin   1        R2           
                         1 
                                      g m RD 
                                              
                 I in g m     R1  R2        
CH 12 Feedback                                    230
             Output Impedance of a V-V Feedback




                      VX     Rout
                         
                      I X 1  KA0 
 A better voltage source
CH 12 Feedback                                    231
       Example: V-V Feedback Output Impedance




                                       R1  1
                 Rout , closed    1 
                                          
                                       R2  g mN
                                           
CH 12 Feedback                                      232
                 Voltage-Current Feedback




                     V out      RO
                           
                      I in   1  KRO
CH 12 Feedback                              233
             Example: Voltage-Current Feedback




                    Vout      g m 2 RD1 RD 2
                          
                     I in      g m 2 RD1 RD 2
                            1
                                      RF
CH 12 Feedback                                   234
                 Input Impedance of a V-C Feedback




                       VX      Rin
                          
                       IX   1  R0 K

 A better current sensor.
CH 12 Feedback                                       235
         Example: V-C Feedback Input Impedance




                                  1               1
                 Rin, closed         .
                                 g m1        g m 2 RD1 RD 2
                                          1
                                                   RF
CH 12 Feedback                                                236
            Output Impedance of a V-C Feedback




                    VX     Rout
                       
                    IX   1 R0 K

 A better voltage source.

CH 12 Feedback                                   237
       Example: V-C Feedback Output Impedance




                                         RD 2
                 Rout , closed   
                                      g m 2 RD1 RD 2
                                   1
                                            RF
CH 12 Feedback                                         238
                 Current-Voltage Feedback




                     I out   Gm
                           
                     Vin 1  KGm
CH 12 Feedback                              239
             Example: Current-Voltage Feedback




                                                  Laser




                 I out               g m1 g m3 rO 3 || rO 5 
                       |closed 
                 Vin             1  g m1 g m3 rO 3 || rO 5 RM
CH 12 Feedback                                                     240
                 Input Impedance of a C-V Feedback




                       V in
                             Rin (1  KGm )
                       I in

 A better voltage sensor.

CH 12 Feedback                                       241
            Output Impedance of a C-V Feedback




                 VX
                     Rout (1  KGm )
                 IX

 A better current source.
CH 12 Feedback                                   242
             Example: Current-Voltage Feedback




                          I out               g m1 g m 2 RD
                                |closed 
                          Vin             1  g m1 g m 2 RD RM
                                         1
                 Laser    Rin |closed       (1  g m1 g m 2 RD RM )
                                        g m1
                                            1
                          Rout   |closed       (1  g m1 g m 2 RD RM )
                                           g m2



CH 12 Feedback                                                   243
Wrong Technique for Measuring Output Impedance




 If we want to measure the output impedance of a C-V
  closed-loop feedback topology directly, we have to place VX
  in series with K and Rout. Otherwise, the feedback will be
  disturbed.
CH 12 Feedback                                             244
                 Current-Current Feedback




                      I out    AI
                            
CH 12 Feedback
                       I in 1  KAI         245
                 Input Impedance of C-C Feedback




                          VX     Rin
                             
                          I X 1  KAI
 A better current sensor.

CH 12 Feedback                                     246
                 Output Impedance of C-C Feedback




                      VX
                          Rout (1  KAI )
                      IX
 A better current source.

CH 12 Feedback                                      247
               Example: Test of Negative Feedback




                                          Laser




I in              VD1 , I out     VP , I F    VD1 , I out 
  CH 12 Feedback
                          Negative Feedback                248
                 Example: C-C Negative Feedback




                                           Laser




                                      g m 2 RD
                 AI |closed 
                              1  g m 2 R D ( RM / R F )
                                1               1
                 Rin |closed      .
                               g m1 1  g m 2 RD ( RM / RF )
                 Rout |closed  rO 2 [1  g m 2 RD ( R M / RF )]
CH 12 Feedback                                                     249
                 How to Break a Loop




 The correct way of breaking a loop is such that the loop
  does not know it has been broken. Therefore, we need to
  present the feedback network to both the input and the
  output of the feedforward amplifier.

CH 12 Feedback                                               250
    Rules for Breaking the Loop of Amplifier Types




CH 12 Feedback                                       251
           Intuitive Understanding of these Rules

                   Voltage-Voltage Feedback




 Since ideally, the input of the feedback network sees zero
  impedance (Zout of an ideal voltage source), the return
  replicate needs to be grounded. Similarly, the output of the
  feedback network sees an infinite impedance (Zin of an ideal
  voltage sensor), the sense replicate needs to be open.
 Similar ideas apply to the other types.
CH 12 Feedback                                              252
            Rules for Calculating Feedback Factor




CH 12 Feedback                                      253
           Intuitive Understanding of these Rules

                   Voltage-Voltage Feedback




 Since the feedback senses voltage, the input of the
  feedback is a voltage source. Moreover, since the return
  quantity is also voltage, the output of the feedback is left
  open (a short means the output is always zero).
 Similar ideas apply to the other types.
CH 12 Feedback                                                   254
                     Breaking the Loop Example I




                 Av , open  g m1 RD ||  R1  R2 
                 Rin, open  1 / g m1
                 Rout , open  RD ||  R1  R2 
CH 12 Feedback                                          255
                      Feedback Factor Example I




                 K  R2 /( R1  R2 )
                 Av , closed  Av , open /(1  KAv , open )
                 Rin, closed  Rin, open (1  KAv , open )
                 Rout , closed  Rout , closed /(1  KAv , open )
CH 12 Feedback                                                      256
                 Breaking the Loop Example II




           Av ,open  g mN rON || rOP ||  R1  R2 
          Rin,open  
          Rout ,open  rON || rOP ||  R1  R2 
CH 12 Feedback                                           257
                 Feedback Factor Example II




                 K  R2 /( R1  R2 )
                 Av ,closed  Av ,open /(1  KAv ,open )
                 Rin,closed  
                 Rout ,closed  Rout ,open /(1  KAv ,open )
CH 12 Feedback                                                 258
                   Breaking the Loop Example IV




                 Vout           RF RD1
                       |open            . g m 2  RD 2 || RF 
                  I in               1
                               RF 
                                    g m1
                                1
                 Rin, open         || RF
                               g m1
                 Rout , open  RD 2 || RF
CH 12 Feedback                                                       259
                 Feedback Factor Example IV




                 K  1 / RF
                 Vout            Vout                Vout
                       |closed        |open /(1  K       |open )
                  I in            I in                I in
                                                     Vout
                 Rin, closed      Rin, open /(1  K       |open )
                                                      I in
                                                        Vout
                 Rout , closed     Rout , open /(1  K       |open )
CH 12 Feedback                                           I in           260
                 Breaking the Loop Example V




                 I out         g m 3 rO 3 || rO 5 g m1rO1
                       |open 
                 Vin               rO1  RL  RM
                 Rin,open  
                 Rout ,open  rO1  RM
CH 12 Feedback                                                261
                   Feedback Factor Example V




     K  RM
     ( I out / Vin |closed )  ( I out / Vin |open ) /[1  K ( I out / Vin ) |open ]
     Rin, closed  
     Rout , closed  Rout , open [1  K ( I out / Vin ) |open ]
CH 12 Feedback                                                                   262
                 Breaking the Loop Example VI




                   I out               g m1 RD
                         |open 
                   Vin           R L  RM  1 / g m 2
                   Rin,open  1 / g m1
CH 12 Feedback
                   Rout ,open  (1 / g m 2 )  RM       263
                   Feedback Factor Example VI




      K  RM
      ( I out / Vin |closed )  ( I out / Vin |open ) /[1  K ( I out / Vin ) |open ]
      Rin, closed  Rin, open [1  K ( I out / Vin ) |open ]
      Rout , closed  Rout , open [1  K ( I out / Vin ) |open ]
CH 12 Feedback                                                                      264
                  Breaking the Loop Example VII




                    ( R F  RM ) R D        g m 2 rO 2
      AI , open                      .
                                  1 rO 2  RL  RM || RF
                    R F  RM 
                                 g m1
                     1
     Rin, open          || ( RF  RM )
                    g m1
     Rout , open  rO 2  RF || RM
CH 12 Feedback                                             265
                 Feedback Factor Example VII




                   K   RM /( RF  RM )
                   AI ,closed  AI ,open /(1  KAI ,open )
                   Rin,closed  Rin,open /(1  KAI ,open )

CH 12 Feedback
                   Rout ,closed  Rout ,open (1  KAI ,open )   266
                 Breaking the Loop Example VIII




           Vout            RF RD
                 |open                [ g m 2 ( RF || RM )]
            I in         RF  1 / g m1
                          1
           Rin, open         || RF
                         g m1
           Rout ,open  RF || RM
CH 12 Feedback                                                  267
                  Feedback Factor Example VIII




K  1 / RF
(Vout / I in ) |closed  (Vout / I in ) |open /[1  K (Vout / I in ) |open ]
Rin,closed  Rin,open /[1  K (Vout / I in ) |open ]
Rout ,closed  Rout ,open /[1  K (Vout / I in ) |open ]
 CH 12 Feedback                                                         268
                 Example: Phase Response




 As it can be seen, the phase of H(jω) starts to drop at 1/10
  of the pole, hits -45o at the pole, and approaches -90o at 10
  times the pole.
CH 12 Feedback                                                269
                 Example: Three-Pole System




 For a three-pole system, a finite frequency produces a
  phase of -180o, which means an input signal that operates
  at this frequency will have its output inverted.

CH 12 Feedback                                                270
          Instability of a Negative Feedback Loop




                  Y           H ( s)
                    ( s) 
                  X        1  KH ( s)
 Substitute jω for s. If for a certain ω1, KH(jω1) reaches
  -1, the closed loop gain becomes infinite. This implies for a
  very small input signal at ω1, the output can be very large.
  Thus the system becomes unstable.
CH 12 Feedback                                               271
           “Barkhausen’s Criteria” for Oscillation




                 | KH ( j1 ) | 1
                 KH ( j1 )  180

CH 12 Feedback                                       272
                 Time Evolution of Instability




CH 12 Feedback                                   273
                   Oscillation Example




 This system oscillates, since there’s a finite frequency at
  which the phase is -180o and the gain is greater than unity.
  In fact, this system exceeds the minimum oscillation
  requirement.
CH 12 Feedback                                               274
                 Condition for Oscillation




 Although for both systems above, the frequencies at which
  |KH|=1 and KH=-180o are different, the system on the left
  is still unstable because at KH=-180o, |KH|>1. Whereas
  the system on the right is stable because at KH=-180o,
  |KH|<1.
CH 12 Feedback                                            275
                 Condition for Stability




                        GX   PX
 ωPX, (“phase crossover”), is the frequency at which
  KH=-180o.
 ωGX, (“gain crossover”), is the frequency at which |KH|=1.
CH 12 Feedback                                                 276
                 Stability Example I




                                       | H p | 1
                                       K 1



CH 12 Feedback                                      277
                 Stability Example II




                                        0.5 | H p | 1
                                        K  0.5



CH 12 Feedback                                      278
                 Marginally Stable vs. Stable




         Marginally Stable                      Stable
CH 12 Feedback                                           279
                       Phase Margin




                  Phase Margin =
                   H(ωGX)+180

                  The larger the phase
                   margin, the more stable
                   the negative feedback
                   becomes




CH 12 Feedback                               280
                 Phase Margin Example




                     PM  45 
CH 12 Feedback                          281
                 Frequency Compensation




 Phase margin can be improved by moving ωGX closer to
  origin while maintaining ωPX unchanged.
CH 12 Feedback                                           282
                 Frequency Compensation Example




 Ccomp is added to lower the dominant pole so that ωGX
  occurs at a lower frequency than before, which means
  phase margin increases.
CH 12 Feedback                                            283
             Frequency Compensation Procedure




 1) We identify a PM, then -180o+PM gives us the new ωGX, or
  ωPM.
 2) On the magnitude plot at ωPM, we extrapolate up with a
  slope of +20dB/dec until we hit the low frequency gain then
  we look “down” and the frequency we see is our new
  dominant pole, ωP’.
CH 12 Feedback                                             284
       Example: 45o Phase Margin Compensation




                   PM   p 2
CH 12 Feedback                                  285
                      Miller Compensation




                 Ceq  [1  g m5 (rO5 || rO6 )]Cc
 To save chip area, Miller multiplication of a smaller
  capacitance creates an equivalent effect.

CH 12 Feedback                                            286

				
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