Transcript Ch9-12

Fundamentals of Microelectronics II
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
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CH9
CH10
CH11
CH12
Cascode Stages and Current Mirrors
Differential Amplifiers
Frequency Response
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
Rout2  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
Rout

 1

1
 1  g m1 
|| rO 2 || r 1  rO1 
|| rO 2 || r 1
g m2
 g m2


Rout

g m1 
1
rO1 
 1 
 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  rO3 || g m2 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 rO3 (rO 4 || r 3 ) || g m2 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 m 3 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 )

1
W  R2
I1   n Cox 
VDD  VTH 
2
L  R1  R2

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 copy 
I S1
I S , REF
I 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
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Multiple Copies of IREF
I copy, j 
IS, j
I S , REF
I 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
I copy
1
 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 copy1
I REF

15
4 2

I copy2
10 I REF

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 diodeconnected 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 C1  I C 2
I EE

2
V X  VY  VCC
CH 10 Differential Amplifiers
I EE
 RC
2
55
Common-Mode Rejection
 Due to the fixed tail current source, the input commonmode value can vary without changing the output commonmode 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
I C1
IC2
CH 10 Differential Amplifiers
Vin1  Vin 2
I EE exp
VT

Vin1  Vin 2
1  exp
VT
I EE

Vin1  Vin 2
1  exp
VT
63
Input/Output Characteristics
Vout1  Vout2 
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  vout2
  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  vout2
  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 || rO3 || R1 
CH 10 Differential Amplifiers
70
Half Circuit Example II
Av   g m1 rO1 || rO3 || R1 
CH 10 Differential Amplifiers
71
Half Circuit Example III
Av  
CH 10 Differential Amplifiers
RC
1
RE 
gm
72
Half Circuit Example IV
Av  
CH 10 Differential Amplifiers
RC
RE 1

2 gm
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
VGS  VTH equil 
I SS
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 commonmode 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
V P  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
I D1  I D 2
4 I SS
1
W
2
  n Cox Vin1  V in2 
 Vin1  Vin 2 
W
2
L
 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
I D1  I D 2
4 I SS
1
W
W
  n Cox Vin1  Vin 2 
  n Cox I SS Vin1  Vin 2 
W
2
L
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
V P  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

Av   g m1 
|| rO 3 || rO1 
 g 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 rO3  rO1 
CH 10 Differential Amplifiers
90
Bipolar Telescopic Cascode
Av   g m1 g m3 rO3 rO1 || r 3  || g m5 rO5 (rO7 || r 5 )
CH 10 Differential Amplifiers
91
Example: Bipolar Telescopic Parasitic Resistance
R1 
R1


Rop  rO 5 1  g m5  rO 7 || r 5 ||   rO 7 || r 5 ||
2 
2


Av   g m1 g m3 rO 3 (rO1 || r 3 ) || Rop
CH 10 Differential Amplifiers
92
MOS Cascode Differential Pair
Av   g m1rO3 g m3 rO1
CH 10 Differential Amplifiers
93
MOS Telescopic Cascode
Av   g m1  g m3 rO3 rO1  || ( g m5 rO5 rO7 )
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
Vin,CM
RC / 2

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
1
 2[1  g m3 ( R1 || r 3 )]rO 3  R1 || r 3 
g m1
CH 10 Differential Amplifiers
101
CMRR
CMRR 
ADM
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 m N roN (vin1  vin 2 )
RThev  2roN
CH 10 Differential Amplifiers
111
Simplified Differential Pair with Active Load
vout
 g m N (rON || rOP )
vin1  vin 2
CH 10 Differential Amplifiers
112
Proof of VA << Vout
vout
vA  
2 g m PrOP
A
I
vout
I 
 g m4v A
rO 4
CH 10 Differential Amplifiers
113
Chapter 11 Frequency Response









11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
Fundamental Concepts
High-Frequency Models of Transistors
Analysis Procedure
Frequency Response of CE and CS Stages
Frequency Response of CB and CG Stages
Frequency Response of Followers
Frequency Response of Cascode Stage
Frequency Response of Differential Pairs
Additional Examples
CH 10 Differential Amplifiers
114
Chapter Outline
CH 11
10 Frequency
Differential Response
Amplifiers
115
High Frequency Roll-off of Amplifier
 As frequency of operation increases, the gain of amplifier
decreases. This chapter analyzes this problem.
CH 11
10 Frequency
Differential Response
Amplifiers
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 11
10
Frequency
Differential Response
Amplifiers differs from face-to-face conversation.
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.
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
120
Gain Roll-off: Common Source
Vout

1 
  g mVin  RD ||

C
s
L 

 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.
CH 11
10 Frequency
Differential Response
Amplifiers
121
Frequency Response of the CS Stage
Vout
g m RD

Vin
RD2 C L2 2  1
 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 frequency
isFrequency
ω=1/(R
CH 11
10
Differential
Response
Amplifiers
122
DCL), where the gain drops by 3dB.
Example: Figure of Merit
F .O.M . 
1
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.
CH 11
10 Frequency
Differential Response
Amplifiers
123
Example: Relationship between Frequency
Response and Step Response
H  s  j  
1
R12C12 2  1

t 
Vout  t   V0 1  exp
 u t 
R1C1 

 The relationship is such that as R1C1 increases, the
bandwidth drops and the step response becomes slower.
CH 11
10 Frequency
Differential Response
Amplifiers
124
Bode Plot

s 
s 
1 
1 
 
 z1   z 2 

H ( s)  A0



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
ofResponse
-20dB/dec
CH 11
10
Frequency
Differential
Amplifiers
125
Example: Bode Plot
 p1
1

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.
CH 11
10 Frequency
Differential Response
Amplifiers
126
Pole Identification Example I
 p1
 p2 
1

RS Cin
Vout

Vin
CH 11
10 Frequency
Differential Response
Amplifiers
1  
1
RD C L
g m RD
2
 p21 1   2  p2 2 
127
Pole Identification Example II
 p1
1


1 
 RS ||
Cin
gm 

CH 11
10 Frequency
Differential Response
Amplifiers
 p2
1

RD C L
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
ofFrequency
CF isResponse
grounded.
CH 11
10
Differential
Amplifiers
129
Miller’s Theorem
ZF
Z1 
1  Av
ZF
Z2 
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.
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
131
Example: Miller Theorem
1
in 
RS 1  g m RD C F
CH 11
10 Frequency
Differential Response
Amplifiers
out 
1

1 
C F
RD 1 
 g m RD 
132
High-Pass Filter Response
Vout

Vin
R1C1
R12C1212  1
 The voltage division between a resistor and a capacitor can
be configured such that the gain at low frequency is
reduced.
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10
Frequency
Differential Response
Amplifiers
135
Typical Frequency Response
Lower Corner
CH 11
10 Frequency
Differential Response
Amplifiers
Upper Corner
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.
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
138
Example: Capacitance Identification
CH 11
10 Frequency
Differential Response
Amplifiers
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
CH 11
10
Frequency
Differential Response
Amplifiers
source/drain.
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.
CH 11
10 Frequency
Differential Response
Amplifiers
141
Example: Capacitance Identification
CH 11
10 Frequency
Differential Response
Amplifiers
142
Transit Frequency
gm
2f T 
CGS
gm
2f T 
C
 Transit frequency, fT, is defined as the frequency where the
current gain from input to output drops to 1.
CH 11
10 Frequency
Differential Response
Amplifiers
143
Example: Transit Frequency Calculation
2fT 
3 n
VGS  VTH 
2
2L
L  65nm
VGS  VTH  100m V
 n  400cm2 /(V .s )
fT  226GHz
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
146
Frequency Response of CS Stage
with Bypassed Degeneration
Vout
 g m RD RS Cb s  1
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.
CH 11
10 Frequency
Differential Response
Amplifiers
147
Unified Model for CE and CS Stages
CH 11
10 Frequency
Differential Response
Amplifiers
148
Unified Model Using Miller’s Theorem
CH 11
10 Frequency
Differential Response
Amplifiers
149
Example: CE Stage
RS  200
I C  1m A
  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.
CH 11
10 Frequency
Differential Response
Amplifiers
150
Example: Half Width CS Stage
W  2X
 p ,in 
 p ,out
CH 11
10 Frequency
Differential Response
Amplifiers
1
C
 g R C 
RS  in  1  m L  XY 
2  2 
 2 
1

 Cout 
2  C XY 


RL 
 1 

2
g
R
2
m L 



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  CoutC XY  Cin Cout 
 Direct analysis yields different pole locations and an extra
zero.
CH 11
10 Frequency
Differential Response
Amplifiers
152
Example: CE and CS Direct Analysis
 p1
1

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 CoutC XY  Cin Cout 
CH 11
10 Frequency
Differential Response
Amplifiers
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 
Dominant Pole
Miller’s
Exact
 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
CH 11 Frequency Response
154
154
Input Impedance of CE and CS Stages
1
1
Z in 
|| r Z in 
CGS  1  g m RD CGD s
C  1  g m RC C s
CH 11
10 Frequency
Differential Response
Amplifiers
155
Low Frequency Response of CB and CG Stages
Vout
g m RC Ci s
s  
1  g m RS Ci s  g m
Vin
 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).
CH 11
10 Frequency
Differential Response
Amplifiers
156
Frequency Response of CB Stage
 p, X
1


1 
 RS ||
C X
gm 

C X  C
 p ,Y
rO  
CH 11
10 Frequency
Differential Response
Amplifiers
1

RL CY
CY  C  CCS
157
Frequency Response of CG Stage
1
 p , Xr   
O 
1 
 RS ||
C X
gm 

C X  CGS  CSB
 p,Y
rO  
1

RL CY
CY  CGD  C DB
 Similar to a CB stage, the input pole is on the order of fT, so
rarely a speed bottleneck.
CH 11
10 Frequency
Differential Response
Amplifiers
158
Example: CG Stage Pole Identification
 p, X 
1

1 
 RS ||
C SB1  CGD1 
g m1 

CH 11
10 Frequency
Differential Response
Amplifiers
 p ,Y 
1
1
C DB1  CGD1  CGS 2  C DB2 
g m2
159
Example: Frequency Response of CG Stage
RS  200
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
 p , X  2  5.31GHz
 0
 p ,Y  2  442MHz
1
Rd  2 K
CH 10 Differential Amplifiers
CH 11 Frequency Response
160
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 by allowing r to go to infinity.
CH 11
10
Frequency
Differential Response
Amplifiers
161
Direct Analysis of Emitter Follower
Vout
Vin
C
1
s
gm
 2
as  bs  1
CH 11
10 Frequency
Differential Response
Amplifiers
RS
C  C  C  C L  C C L 
a
gm
C  R S
b  RS C  
 1 
gm 
r
 CL

 gm
162
Direct Analysis of Source Follower Stage
Vout
Vin
CGS
1
s
gm
 2
as  bs  1
CH 11
10 Frequency
Differential Response
Amplifiers
RS
CGDCGS  CGDC SB  CGS C SB 
a
gm
CGD  C SB
b  RS CGD 
gm
163
Example: Frequency Response of Source Follower
RS  200
C L  100 fF
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
1
 0
CH
Differential Response
Amplifiers
CH 10
11 Frequency
 p1  2  1.79GHz  j 2.57GHz
 p 2  2  1.79GHz  j 2.57GHz
164
164
Example: Source Follower
Vout
Vin
CGS
1
s
gm
 2
as  bs  1
RS
CGD1CGS1  (CGD1  CGS1 )(C SB1  CGD2  C DB2 )
a
g m1
b  RS CGD1 
CGD1  C SB1 C GD 2 C DB 2
g m1
CH 11
10 Frequency
Differential Response
Amplifiers
165
Input Capacitance of Emitter/Source Follower
rO  
C / CGS
Cin  C / CGD 
1  g m RL
CH 11
10 Frequency
Differential Response
Amplifiers
166
Example: Source Follower Input Capacitance
1
Cin  CGD1 
CGS1
1  g m1 rO1 || rO 2 
CH 11
10 Frequency
Differential Response
Amplifiers
167
Output Impedance of Emitter Follower
VX RS r C s  r  RS

IX
r C s    1
CH 11
10 Frequency
Differential Response
Amplifiers
168
Output Impedance of Source Follower
V X RS CGS s  1

I X CGS s  g m
CH 11
10 Frequency
Differential Response
Amplifiers
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.
CH 11
10 Frequency
Differential Response
Amplifiers
170
Example: Output Impedance
rO  
VX rO1 || rO 2 CGS 3 s  1

IX
CGS 3 s  g m3
CH 11
10 Frequency
Differential Response
Amplifiers
171
Frequency Response of Cascode Stage
Av, XY
 g m1

 1
g m2
C x  2C XY
 For cascode stages, there are three poles and Miller
multiplication is smaller than in the CE/CS stage.
CH 11
10 Frequency
Differential Response
Amplifiers
172
Poles of Bipolar Cascode
 p, X
1

RS || r 1 C 1  2C1 
 p ,out 
CH 11
10 Frequency
Differential Response
Amplifiers
 p ,Y 
1
1
CCS1  C 2  2C1 
g m2
1
RL CCS 2  C  2 
173
Poles of MOS Cascode
 p, X 
1



g m1 
CGD1 
RS CGS1  1 
g m2 



 p ,Y 
CH 11
10 Frequency
Differential Response
Amplifiers
 p ,out 
1
RL C DB 2  CGD2 
1
1
g m2



g m2 
CGD1 
C DB1  CGS 2  1 
g m1 



174
Example: Frequency Response of Cascode
RS  200
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
1
 p , X  2  1.95GHz
 0
 p ,Y  2  1.73GHz
RL  2 K 
 p ,out  2  442MHz
CH
Differential Response
Amplifiers
CH 10
11 Frequency
175
175
MOS Cascode Example
 p, X 
1



g m1 
CGD1 
RS CGS1  1 
g m2 



 p ,Y 
1 
C DB1  CGS 2
CH 11
10 Frequency
Differential Response
Amplifiers
g m2 
 p ,out 
1
RL C DB 2  CGD2 
1


g m2 
CGD1  CGD3  C DB3 
 1 
g m1 


176
I/O Impedance of Bipolar Cascode
1
Z in  r 1 ||
C 1  2C1 s
CH 11
10 Frequency
Differential Response
Amplifiers
Z out
1
 RL ||
C 2  CCS 2 s
177
I/O Impedance of MOS Cascode
1
Z in 


 g m1 
CGD1  s
CGS1  1 
 g m2 


CH 11
10 Frequency
Differential Response
Amplifiers
Z out
1
 RL ||
CGD2  C DB2 s
178
Bipolar Differential Pair Frequency Response
Half Circuit
 Since bipolar differential pair can be analyzed using halfcircuit, its transfer function, I/O impedances, locations of
poles/zeros are the same as that of the half circuit’s.
CH 11
10 Frequency
Differential Response
Amplifiers
179
MOS Differential Pair Frequency Response
Half Circuit
 Since MOS differential pair can be analyzed using halfcircuit, its transfer function, I/O impedances, locations of
poles/zeros are the same as that of the half circuit’s.
CH 11
10 Frequency
Differential Response
Amplifiers
180
Example: MOS Differential Pair
 p, X 
 p ,Y 
 p ,out
CH 11
10 Frequency
Differential Response
Amplifiers
1
RS [CGS1  (1  g m1 / g m 3 )CGD1 ]
1



g m3 
CGD1 
C DB1  CGS 3  1 
g m1 



1

RL C DB3  CGD3 
1
g m3
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
CH 11 Frequency Response
182
182
Tail Node Capacitance Contribution
 Source-Body Capacitance of
M1, M2 and M3
 Gate-Drain Capacitance of M3
CH 11
10 Frequency
Differential Response
Amplifiers
183
Example: Capacitive Coupling
Rin2  RB 2 || r 2    1RE 
L1 
1
 2  542Hz 
r 1 || RB1 C1
CH 10 Differential Amplifiers
CH 11 Frequency Response
L 2
1

   22.9 Hz 
RC  Rin2 C2
184
184
Example: IC Amplifier – Low Frequency Design
Rin2 
CH 11
10 Frequency
Differential Response
Amplifiers
RF
1  Av 2
L1 
g m1RS1  1
 2  42.4MHz
RS1C1
L 2 
1
 2  6.92MHz
RD1  Rin2 C2
185
Example: IC Amplifier – Midband Design
vX
  g m1 RD1 || Rin2   3.77
vin
CH 10 Differential Amplifiers
CH 11 Frequency Response
186
186
Example: IC Amplifier – High Frequency Design
 p1  2  (308 MHz)
 p 2  2  (2.15 GHz)
 p3
1

RL 2 (1.15CGD 2  C DB 2 )
 2  (1.21 GHz)
CH 10 Differential Amplifiers
CH 11 Frequency Response
187
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
A1
Y

X 1  KA1
CH 12 Feedback
190
Feedback Example
Y

X
A1
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
R1
Y
 1
X
R2
CH 12 Feedback
193
Loop Gain
X 0
VN
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
A1 K  1
Y
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
A0
As  
s
1
0
Negative
Feedback
A0
1  KA0
Y
s  
s
X
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
CH 12 Feedback
Rout  RD
202
Example: Closed Loop Voltage Gain
vout

vin
CH 12 Feedback
g m RD
R2
1
g m RD
R1  R2
203
Example: Closed Loop I/O Impedance

R2
1 
1 
Rin 
g m RD 
g m  R1  R2

CH 12 Feedback
Rout
RD

R2
1
g m RD
R1  R2
204
Example: Load Desensitization
W/O Feedback
Large Difference
g m RD  g m RD / 3
CH 12 Feedback
With Feedback
Small Difference
g m RD
g m RD

R2
R2
1
g m RD
3
g m RD
R1  R2
R1  R2
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
CH 12 Feedback
iF
K
 gmF
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 
CH 12 Feedback
I D1 , I D 2 
Vout ,Vx 
Negative Feedback
I D 2 , I D1 
224
Polarity of Feedback Example II
Vin 
CH 12 Feedback
I D1 ,VA 
Vout ,Vx 
Negative Feedback
I D1 ,VA 
225
Polarity of Feedback Example III
I in 
CH 12 Feedback
I D1 ,VX 
Vout , I D 2 
Positive Feedback
I D1 ,VX 
226
Voltage-Voltage Feedback
Vout
A0

Vin 1  KA0
CH 12 Feedback
227
Example: Voltage-Voltage Feedback
Vout

Vin
CH 12 Feedback
g m N (rON || rOP )
R2
1
g m N (rON || rOP )
R1  R2
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
R2
1 
1 

g m RD 
I in g m 
R1  R2

CH 12 Feedback
230
Output Impedance of a V-V Feedback
Rout
VX

I X 1  KA0 
 A better voltage source
CH 12 Feedback
231
Example: V-V Feedback Output Impedance
Rout, closed
CH 12 Feedback

R1  1

 1 
R2  g m N

232
Voltage-Current Feedback
CH 12 Feedback
V out
RO

I in
1  KRO
233
Example: Voltage-Current Feedback
Vout
 g m 2 RD1 RD 2

g m 2 RD1 RD 2
I in
1
RF
CH 12 Feedback
234
Input Impedance of a V-C Feedback
Rin
VX

IX
1 R0 K
 A better current sensor.
CH 12 Feedback
235
Example: V-C Feedback Input Impedance
Rin,closed
CH 12 Feedback
1

.
g m1
1
g m 2 RD1 RD 2
1
RF
236
Output Impedance of a V-C Feedback
Rout
VX

IX
1 R0 K
 A better voltage source.
CH 12 Feedback
237
Example: V-C Feedback Output Impedance
Rout,closed
CH 12 Feedback
RD 2

g m 2 RD1 RD 2
1
RF
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 rO3 || rO5 
|closed 
Vin
1  g m1 g m3 rO3 || rO5 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
Laser
1
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
CH 12 Feedback
I out
AI

I in 1  KAI
245
Input Impedance of C-C Feedback
Rin
VX

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 
CH 12 Feedback
VD1 , I out 
VP , I F 
Negative Feedback
VD1 , I out 
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
CH 12 Feedback
Rout, open  RD ||  R1  R2 
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 m N 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  
CH 12 Feedback
Rout, closed  Rout, open /(1  KAv , open )
258
Breaking the Loop Example IV
Vout
RF RD1
|open 
. g m 2  RD 2 || RF 
1
I in
RF 
g m1
Rin , open
CH 12 Feedback
1

|| RF
g m1
Rout, open  RD 2 || RF
259
Feedback Factor Example IV
K  1 / RF
Vout
Vout
Vout
|closed 
|open /(1  K
|open )
I in
I in
I in
Rin , closed
Rout, closed
CH 12 Feedback
Vout
 Rin , open /(1  K
|open )
I in
Vout
 Rout, open /(1  K
|open )
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
AI , open
Rin , open
 g m 2 rO 2
( R F  RM ) R D

.
1 rO 2  RL  RM || RF
R F  RM 
g m1
1

|| ( 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
Rin , open
1

|| 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
CH 12 Feedback
Stable
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