1 fundamentals of microelectronics ii ch9 cascode stages and current mirrors ch10 differential...
TRANSCRIPT
1
Fundamentals of Microelectronics II
CH9 Cascode Stages and Current Mirrors CH10 Differential Amplifiers CH11 Frequency Response CH12 Feedback
2
Chapter 9 Cascode Stages and Current Mirrors
9.1 Cascode Stage
9.2 Current Mirrors
CH 9 Cascode Stages and Current Mirrors 3
Boosted Output Impedances
SOSmout
EOEmout
RrRgR
rRrrRgR
1
||||1
2
1
CH 9 Cascode Stages and Current Mirrors 4
Bipolar Cascode Stage
1211
12112
||
||)]||(1[
rrrgR
rrrrrgR
OOmout
OOOmout
CH 9 Cascode Stages and Current Mirrors 5
Maximum Bipolar Cascode Output Impedance
The maximum output impedance of a bipolar cascode is bounded by the ever-present r between emitter and ground of Q1.
11max,
11max, 1
Oout
Omout
rR
rrgR
CH 9 Cascode Stages and Current Mirrors 6
Example: Output Impedance
Typically r is smaller than rO, so in general it is impossible to double the output impedance by degenerating Q2 with a resistor.
21
122
O
OoutA rr
rrR
CH 9 Cascode Stages and Current Mirrors 7
PNP Cascode Stage
1211
12112
||
||)]||(1[
rrrgR
rrrrrgR
OOmout
OOOmout
CH 9 Cascode Stages and Current Mirrors 8
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 9
False Cascodes
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.
12
12
1
122
1122
1
21
1
||||1
||||1
1
Om
Om
mout
Om
OOm
mout
rg
rg
gR
rrg
rrrg
gR
CH 9 Cascode Stages and Current Mirrors 10
MOS Cascode Stage
211
21211
OOmout
OOOmout
rrgR
rrrgR
CH 9 Cascode Stages and Current Mirrors 11
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 12
PMOS Cascode Stage
211
21211
OOmout
OOOmout
rrgR
rrrgR
CH 9 Cascode Stages and Current Mirrors 13
Example: Parasitic Resistance
RP will lower the output impedance, since its parallel combination with rO1 will always be lower than rO1.
2121 )||)(1( OPOOmout rRrrgR
CH 9 Cascode Stages and Current Mirrors 14
Short-Circuit Transconductance
The short-circuit transconductance of a circuit measures its strength in converting input voltage to output current.
0
outvin
outm v
iG
CH 9 Cascode Stages and Current Mirrors 15
Transconductance Example
1mm gG
CH 9 Cascode Stages and Current Mirrors 16
Derivation of Voltage Gain
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.
outminout
outinmoutoutout
RGvv
RvGRiv
CH 9 Cascode Stages and Current Mirrors 17
Example: Voltage Gain
11 Omv rgA
CH 9 Cascode Stages and Current Mirrors 18
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 19
Voltage Gain of Bipolar Cascode Amplifier
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.
)||( 21111
1
rrgrgA
gG
OmOmv
mm
CH 9 Cascode Stages and Current Mirrors 20
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 21
Practical Cascode Stage
Since no current source can be ideal, the output impedance drops.
)||(|| 21223 rrrgrR OOmOout
CH 9 Cascode Stages and Current Mirrors 22
Improved Cascode Stage
In order to preserve the high output impedance, a cascode PNP current source is used.
)||(||)||( 21223433 rrrgrrrgR OOmOOmout
CH 9 Cascode Stages and Current Mirrors 23
MOS Cascode Amplifier
2211
21221 )1(
OmOmv
OOOmmv
outmv
rgrgA
rrrggA
RGA
CH 9 Cascode Stages and Current Mirrors 24
Improved MOS Cascode Amplifier
Similar to its bipolar counterpart, the output impedance of a MOS cascode amplifier can be improved by using a PMOS cascode current source.
oponout
OOmop
OOmon
RRR
rrgR
rrgR
||
433
122
CH 9 Cascode Stages and Current Mirrors 25
Temperature and Supply Dependence of Bias Current
Since VT, IS, n, and VTH all depend on temperature, I1 for both bipolar and MOS depends on temperature and supply.
2
21
21
1212
2
1
)ln()(
THDDoxn
STCC
VVRR
R
L
WCI
IIVRRVR
CH 9 Cascode Stages and Current Mirrors 26
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 27
Bipolar Current Mirror Circuitry
The diode-connected QREF produces an output voltage V1
that forces Icopy1 = IREF, if Q1 = QREF.
REFREFS
Scopy I
I
II
,
1
CH 9 Cascode Stages and Current Mirrors 28
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 29
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 30
Multiple Copies of IREF
Multiple copies of IREF can be generated at different locations by simply applying the idea of current mirror to more transistors.
REFREFS
jSjcopy I
I
II
,
,,
CH 9 Cascode Stages and Current Mirrors 31
Current Scaling
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.
REFjcopy nII ,
CH 9 Cascode Stages and Current Mirrors 32
Example: Scaled Current
CH 9 Cascode Stages and Current Mirrors 33
Fractional Scaling
A fraction of IREF can be created on Q1 by scaling up the emitter area of QREF.
REFcopy II3
1
CH 9 Cascode Stages and Current Mirrors 34
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 35
Mirroring Error Due to Base Currents
111
n
nII REFcopy
CH 9 Cascode Stages and Current Mirrors 36
Improved Mirroring Accuracy
Because of QF, the base currents of QREF and Q1 are mostly supplied by QF rather than IREF. Mirroring error is reduced times.
111
2
n
nII REFcopy
CH 9 Cascode Stages and Current Mirrors 37
Example: Different Mirroring Ratio Accuracy
2
2
2
1
154
10
154
REFcopy
REFcopy
II
II
CH 9 Cascode Stages and Current Mirrors 38
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 39
Generation of IREF for PNP Current Mirror
CH 9 Cascode Stages and Current Mirrors 40
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 41
MOS Current Mirror
The same concept of current mirror can be applied to MOS transistors as well.
CH 9 Cascode Stages and Current Mirrors 42
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 43
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 44
CMOS Current Mirror
The idea of combining NMOS and PMOS to produce CMOS current mirror is shown above.
45
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
CH 10 Differential Amplifiers 46
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 47
“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 48
Supply Ripple Rejection
Since both node X and Y contain the ripple, their difference will be free of ripple.
invYX
rY
rinvX
vAvv
vv
vvAv
CH 10 Differential Amplifiers 49
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 50
Common Inputs to Differential Amplifier
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.
0
YX
rinvY
rinvX
vv
vvAv
vvAv
CH 10 Differential Amplifiers 51
Differential Inputs to Differential Amplifier
When the inputs are applied differentially, the outputs are 180° out of phase; enhancing each other when sensed differentially.
invYX
rinvY
rinvX
vAvv
vvAv
vvAv
2
CH 10 Differential Amplifiers 52
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 53
Single-ended vs. Differential Signals
CH 10 Differential Amplifiers 54
Differential Pair
With the addition of a tail current, the circuits above operate as an elegant, yet robust differential pair.
CH 10 Differential Amplifiers 55
Common-Mode Response
2
221
21
EECCCYX
EECC
BEBE
IRVVV
III
VV
CH 10 Differential Amplifiers 56
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 57
Differential Response I
CCY
EECCCX
C
EEC
VV
IRVV
I
II
02
1
CH 10 Differential Amplifiers 58
Differential Response II
CCX
EECCCY
C
EEC
VV
IRVV
I
II
01
2
CH 10 Differential Amplifiers 59
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 60
Small-Signal Analysis
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.
II
I
II
I
EEC
EEC
2
2
2
1
CH 10 Differential Amplifiers 61
Virtual Ground
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.
VgI
VgI
V
mC
mC
P
2
1
0
CH 10 Differential Amplifiers 62
Small-Signal Differential Gain
Since the output changes by -2gmVRC and input by 2V, 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.
CmCm
v RgV
VRgA
2
2
CH 10 Differential Amplifiers 63
Large Signal Analysis
T
inin
EEC
T
inin
T
ininEE
C
V
VVI
I
V
VVV
VVI
I
212
21
21
1
exp1
exp1
exp
CH 10 Differential Amplifiers 64
Input/Output Characteristics
T
ininEEC
outout
V
VVIR
VV
2tanh 21
21
CH 10 Differential Amplifiers 65
Linear/Nonlinear Regions
The left column operates in linear region, whereas the right column operates in nonlinear region.
CH 10 Differential Amplifiers 66
Small-Signal Model
CH 10 Differential Amplifiers 67
Half Circuits
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.
Cminin
outout Rgvv
vv
21
21
CH 10 Differential Amplifiers 68
Example: Differential Gain
Ominin
outout rgvv
vv
21
21
CH 10 Differential Amplifiers 69
Extension of Virtual Ground
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.
0XV
CH 10 Differential Amplifiers 70
Half Circuit Example I
1311 |||| RrrgA OOmv
CH 10 Differential Amplifiers 71
Half Circuit Example II
1311 |||| RrrgA OOmv
CH 10 Differential Amplifiers 72
Half Circuit Example III
mE
Cv
gR
RA
1
CH 10 Differential Amplifiers 73
Half Circuit Example IV
m
E
Cv
g
RR
A1
2
CH 10 Differential Amplifiers 74
MOS Differential Pair’s Common-Mode Response
Similar to its bipolar counterpart, MOS differential pair produces zero differential output as VCM changes.
2SS
DDDYX
IRVVV
CH 10 Differential Amplifiers 75
Equilibrium Overdrive Voltage
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.
L
WC
IVV
oxn
SSequilTHGS
CH 10 Differential Amplifiers 76
Minimum Common-mode Output Voltage
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.
THCMSS
DDD VVI
RV 2
CH 10 Differential Amplifiers 77
Differential Response
CH 10 Differential Amplifiers 78
Small-Signal Response
Similar to its bipolar counterpart, the MOS differential pair exhibits the same virtual ground node and small signal gain.
Dmv
P
RgA
V
0
CH 10 Differential Amplifiers 79
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 80
MOS Differential Pair’s Large-Signal Response
221121
4
2
12 inin
oxn
SSinoxnDD VV
L
WC
IVV
L
WCII
in
CH 10 Differential Amplifiers 81
Maximum Differential Input Voltage
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.
equilTHGSinin VVVV 2max21
CH 10 Differential Amplifiers 82
Contrast Between MOS and Bipolar Differential Pairs
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.
MOS Bipolar
CH 10 Differential Amplifiers 83
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 to accommodate this change, thus Vin,max increases by as well. Moreover, since ISS is doubled, the differential output swing will double.
22
CH 10 Differential Amplifiers 84
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 to accommodate this change, thus Vin,max
will be lowered by as well. Moreover, the differential output swing will remain unchanged since neither ISS nor RD has changed
22
CH 10 Differential Amplifiers 85
Small-Signal Analysis of MOS Differential Pair
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.
212121
4
2
1ininSSoxn
oxn
SSininoxnDD VVI
L
WC
L
WC
IVV
L
WCII
CH 10 Differential Amplifiers 86
Virtual Ground and Half Circuit
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.
Cmv
P
RgA
V
0
CH 10 Differential Amplifiers 87
MOS Differential Pair Half Circuit Example I
133
1 ||||1
0
OOm
mv rrg
gA
CH 10 Differential Amplifiers 88
MOS Differential Pair Half Circuit Example II
3
1
0
m
mv g
gA
CH 10 Differential Amplifiers 89
MOS Differential Pair Half Circuit Example III
mSS
DDv gR
RA
12
2
0
CH 10 Differential Amplifiers 90
Bipolar Cascode Differential Pair
133131 || OOOmmv rrrrggA
CH 10 Differential Amplifiers 91
Bipolar Telescopic Cascode
)||(|||| 575531331 rrrgrrrggA OOmOOmmv
CH 10 Differential Amplifiers 92
Example: Bipolar Telescopic Parasitic Resistance
opOOmmv
OOmOop
RrrrggA
Rrr
RrrgrR
||)||(
2||||
2||||1
31331
157
15755
CH 10 Differential Amplifiers 93
MOS Cascode Differential Pair
1331 OmOmv rgrgA
CH 10 Differential Amplifiers 94
MOS Telescopic Cascode
)(|| 7551331 OOmOOmmv rrgrrggA
CH 10 Differential Amplifiers 95
Example: MOS Telescopic Parasitic Resistance
)||(
]1[||
1331
77515
OmOopmv
OOmOop
rgrRgA
rrgRrR
CH 10 Differential Amplifiers 96
Effect of Finite Tail Impedance
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.
mEE
C
CMin
CMout
gR
R
V
V
2/1
2/
,
,
CH 10 Differential Amplifiers 97
Input CM Noise with Ideal Tail Current
CH 10 Differential Amplifiers 98
Input CM Noise with Non-ideal Tail Current
CH 10 Differential Amplifiers 99
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 100
CM to DM Conversion, ACM-DM
If finite tail impedance and asymmetry are both present, then the differential output signal will contain a portion of input common-mode signal.
EEm
D
CM
out
Rg
R
V
V
2/1
CH 10 Differential Amplifiers 101
Example: ACM-DM
3133131
||)]||(1[21
rRrrRgg
R
A
Omm
C
DMCM
CH 10 Differential Amplifiers 102
CMRR
CMRR defines the ratio of wanted amplified differential input signal to unwanted converted input common-mode noise that appears at the output.
DMCM
DM
A
ACMRR
CH 10 Differential Amplifiers 103
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 104
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 105
Better Alternative
This circuit topology performs differential to single-ended conversion with no loss of gain.
CH 10 Differential Amplifiers 106
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 107
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 108
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 109
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 110
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 111
Thevenin Equivalent of the Input Pair
oNThev
ininoNmNThev
rR
vvrgv
2
)( 21
CH 10 Differential Amplifiers 112
Simplified Differential Pair with Active Load
)||(21
OPONmNinin
out rrgvv
v
CH 10 Differential Amplifiers 113
I
A
Proof of VA << Vout
OPmP
outA rg
vv
2
AmO
out vgr
vI 4
4
CH 10 Differential Amplifiers 114
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
114
CH 10 Differential Amplifiers 115
Chapter Outline
CH 11 Frequency Response 115
CH 10 Differential Amplifiers 116CH 11 Frequency Response 116
High Frequency Roll-off of Amplifier
As frequency of operation increases, the gain of amplifier decreases. This chapter analyzes this problem.
CH 10 Differential Amplifiers 117
Example: Human Voice I
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 differs from face-to-face conversation.
CH 11 Frequency Response 117
Natural Voice Telephone System
CH 10 Differential Amplifiers 118
Example: Human Voice II
CH 11 Frequency Response 118
Mouth RecorderAir
Mouth EarAir
Skull
Path traveled by the human voice to the voice recorder
Path traveled by the human voice to the human ear
Since the paths are different, the results will also be different.
CH 10 Differential Amplifiers 119
Example: Video Signal
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 Frequency Response 119
High Bandwidth Low Bandwidth
CH 10 Differential Amplifiers 120
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 Frequency Response 120
CH 10 Differential Amplifiers 121CH 11 Frequency Response 121
Gain Roll-off: Common Source
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.
1||out m in D
L
V g V RC s
CH 10 Differential Amplifiers 122CH 11 Frequency Response 122
Frequency Response of the CS Stage
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 is ω=1/(RDCL), where the gain drops by 3dB.
1222
LD
Dm
in
out
CR
Rg
V
V
CH 10 Differential Amplifiers 123CH 11 Frequency Response 123
Example: Figure of Merit
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.
LCCT CVVMOF
1...
CH 10 Differential Amplifiers 124
Example: Relationship between Frequency Response and Step Response
CH 11 Frequency Response 124
2 2 21 1
1
1H s j
R C
0
1 1
1 expoutt
V t V u tR C
The relationship is such that as R1C1 increases, the bandwidth drops and the step response becomes slower.
CH 10 Differential Amplifiers 125CH 11 Frequency Response 125
Bode Plot
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
21
210
11
11
)(
pp
zz
ss
ss
AsH
CH 10 Differential Amplifiers 126CH 11 Frequency Response 126
Example: Bode Plot
The circuit only has one pole (no zero) at 1/(RDCL), so the slope drops from 0 to -20dB/dec as we pass ωp1.
LDp CR
11
CH 10 Differential Amplifiers 127CH 11 Frequency Response 127
Pole Identification Example I
inSp CR
11
LDp CR
12
22
221
2 11 pp
Dm
in
out Rg
V
V
CH 10 Differential Amplifiers 128CH 11 Frequency Response 128
Pole Identification Example II
inm
S
p
Cg
R
1||
11
LDp CR
12
CH 10 Differential Amplifiers 129CH 11 Frequency Response 129
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.
CH 10 Differential Amplifiers 130CH 11 Frequency Response 130
Miller’s Theorem
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.
v
F
A
ZZ
11v
F
A
ZZ
/112
CH 10 Differential Amplifiers 131CH 11 Frequency Response 131
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 10 Differential Amplifiers 132CH 11 Frequency Response 132
Example: Miller Theorem
FDmSin CRgR
1
1F
DmD
out
CRg
R
1
1
1
CH 10 Differential Amplifiers 133
High-Pass Filter Response
121
21
21
11
CR
CR
V
V
in
out
The voltage division between a resistor and a capacitor can be configured such that the gain at low frequency is reduced.
CH 11 Frequency Response 133
CH 10 Differential Amplifiers 134
Example: Audio Amplifier
nFCi 6.79 nFCL 8.39
In order to successfully pass audio band frequencies (20 Hz-20 KHz), large input and output capacitances are needed.
200/1
100
m
i
g
KR
CH 11 Frequency Response 134
CH 10 Differential Amplifiers 135
Capacitive Coupling vs. 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.
Capacitive Coupling Direct Coupling
CH 11 Frequency Response 135
CH 10 Differential Amplifiers 136
Typical Frequency Response
Lower Corner Upper Corner
CH 11 Frequency Response 136
CH 10 Differential Amplifiers 137CH 11 Frequency Response 137
High-Frequency Bipolar Model
At high frequency, capacitive effects come into play. Cb
represents the base charge, whereas C and Cje are the junction capacitances.
b jeC C C
CH 10 Differential Amplifiers 138CH 11 Frequency Response 138
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 10 Differential Amplifiers 139CH 11 Frequency Response 139
Example: Capacitance Identification
CH 10 Differential Amplifiers 140CH 11 Frequency Response 140
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.
CH 10 Differential Amplifiers 141CH 11 Frequency Response 141
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 10 Differential Amplifiers 142CH 11 Frequency Response 142
Example: Capacitance Identification
CH 10 Differential Amplifiers 143CH 11 Frequency Response 143
Transit Frequency
Transit frequency, fT, is defined as the frequency where the current gain from input to output drops to 1.
C
gf mT 2
GS
mT C
gf 2
CH 10 Differential Amplifiers 144
Example: Transit Frequency Calculation
THGSn
T VVL
f 22
32
GHzf
sVcm
mVVV
nmL
T
n
THGS
226
)./(400
100
65
2
CH 11 Frequency Response 144
CH 10 Differential Amplifiers 145
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 Frequency Response 145
CH 10 Differential Amplifiers 146
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 Frequency Response 146
CH 10 Differential Amplifiers 147
Frequency Response of CS Stagewith Bypassed Degeneration
1
1
SmbS
bSDm
X
out
RgsCR
sCRRgs
V
V
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 Frequency Response 147
CH 10 Differential Amplifiers 148CH 11 Frequency Response 148
Unified Model for CE and CS Stages
CH 10 Differential Amplifiers 149CH 11 Frequency Response 149
Unified Model Using Miller’s Theorem
CH 10 Differential Amplifiers 150
Example: CE Stage
fFC
fFC
fFC
mAI
R
CS
C
S
30
20
100
100
1
200
GHz
MHz
outp
inp
59.12
5162
,
,
The input pole is the bottleneck for speed.
CH 11 Frequency Response 150
CH 10 Differential Amplifiers 151
Example: Half Width CS Stage
XW 2
22
12
1
221
2
1
,
,
XY
Lm
outL
outp
XYLminS
inp
CRg
CR
CRgCR
CH 11 Frequency Response 151
CH 10 Differential Amplifiers 152CH 11 Frequency Response 152
Direct Analysis of CE and CS Stages
Direct analysis yields different pole locations and an extra zero.
outinXYoutXYinLThev
outXYLinThevThevXYLmp
outXYLinThevThevXYLmp
XY
mz
CCCCCCRR
CCRCRRCRg
CCRCRRCRg
C
g
1||
1
1||
||
2
1
CH 10 Differential Amplifiers 153CH 11 Frequency Response 153
Example: CE and CS Direct Analysis
outinXYoutXYinOOS
outXYOOinSSXYOOmp
outXYOOinSSXYOOmp
CCCCCCrrR
CCrrCRRCrrg
CCrrCRRCrrg
21
212112
212111
||
)(||||1
)(||||1
1
CH 10 Differential Amplifiers 154
Example: Comparison Between Different Methods
MHz
MHz
outp
inp
4282
5712
,
,
GHz
MHz
outp
inp
53.42
2642
,
,
GHz
MHz
outp
inp
79.42
2492
,
,
KR
g
fFC
fFC
fFC
R
L
m
DB
GD
GS
S
2
0
150
100
80
250
200
1
Miller’s Exact Dominant Pole
CH 11 Frequency Response 154
CH 10 Differential Amplifiers 155CH 11 Frequency Response 155
Input Impedance of CE and CS Stages
rsCRgC
ZCm
in ||1
1
sCRgC
ZGDDmGS
in
1
1
CH 10 Differential Amplifiers 156
Low Frequency Response of CB and CG Stages
miSm
iCm
in
out
gsCRg
sCRgs
V
V
1
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 Frequency Response 156
CH 10 Differential Amplifiers 157CH 11 Frequency Response 157
Frequency Response of CB Stage
Xm
S
Xp
Cg
R
1||
1,
CCX
YLYp CR
1,
CSY CCC Or
CH 10 Differential Amplifiers 158CH 11 Frequency Response 158
Frequency Response of CG Stage
OrX
mS
Xp
Cg
R
1||
1,
SBGSX CCC
YLYp CR
1,
DBGDY CCC
Similar to a CB stage, the input pole is on the order of fT, so rarely a speed bottleneck.
Or
CH 10 Differential Amplifiers 159CH 11 Frequency Response 159
Example: CG Stage Pole Identification
111
,1
||
1
GDSBm
S
Xp
CCg
R
22112
, 11
DBGSGDDBm
Yp
CCCCg
CH 10 Differential Amplifiers 160
Example: Frequency Response of CG Stage
KR
g
fFC
fFC
fFC
R
d
m
DB
GD
GS
S
2
0
150
100
80
250
200
1
MHz
GHz
Yp
Xp
4422
31.52
,
,
CH 11 Frequency Response 160
CH 10 Differential Amplifiers 161CH 11 Frequency Response 161
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 10 Differential Amplifiers 162CH 11 Frequency Response 162
Direct Analysis of Emitter Follower
1
1
2
bsas
sg
C
V
V m
in
out
m
LS
mS
LLm
S
g
C
r
R
g
CCRb
CCCCCCg
Ra
1
CH 10 Differential Amplifiers 163CH 11 Frequency Response 163
Direct Analysis of Source Follower Stage
1
1
2
bsas
sg
C
V
V m
GS
in
out
m
SBGDGDS
SBGSSBGDGSGDm
S
g
CCCRb
CCCCCCg
Ra
CH 10 Differential Amplifiers 164
Example: Frequency Response of Source Follower
0
150
100
80
250
100
200
1
m
DB
GD
GS
L
S
g
fFC
fFC
fFC
fFC
R
GHzjGHz
GHzjGHz
p
p
57.279.12
57.279.12
2
1
CH 11 Frequency Response 164
CH 10 Differential Amplifiers 165CH 11 Frequency Response 165
Example: Source Follower
1
1
2
bsas
sg
C
V
V m
GS
in
out
1
22111
22111111
))((
m
DBGDSBGDGDS
DBGDSBGSGDGSGDm
S
g
CCCCCRb
CCCCCCCg
Ra
CH 10 Differential Amplifiers 166CH 11 Frequency Response 166
Input Capacitance of Emitter/Source Follower
Lm
GSGDin Rg
CCCCC
1
//
Or
CH 10 Differential Amplifiers 167CH 11 Frequency Response 167
Example: Source Follower Input Capacitance
1211
1 ||1
1GS
OOmGDin C
rrgCC
CH 10 Differential Amplifiers 168CH 11 Frequency Response 168
Output Impedance of Emitter Follower
1
sCr
RrsCrR
I
V SS
X
X
CH 10 Differential Amplifiers 169CH 11 Frequency Response 169
Output Impedance of Source Follower
mGS
GSS
X
X
gsC
sCR
I
V
1
CH 10 Differential Amplifiers 170CH 11 Frequency Response 170
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 10 Differential Amplifiers 171CH 11 Frequency Response 171
Example: Output Impedance
33
321 1||
mGS
GSOO
X
X
gsC
sCrr
I
V
Or
CH 10 Differential Amplifiers 172CH 11 Frequency Response 172
Frequency Response of Cascode Stage
For cascode stages, there are three poles and Miller multiplication is smaller than in the CE/CS stage.
12
1,
m
mXYv g
gA
XYx CC 2
CH 10 Differential Amplifiers 173CH 11 Frequency Response 173
Poles of Bipolar Cascode
111, 2||
1
CCrRS
Xp 121
2
,
21
1
CCC
g CSm
Yp
22,
1
CCR CSL
outp
CH 10 Differential Amplifiers 174CH 11 Frequency Response 174
Poles of MOS Cascode
12
11
,
1
1
GDm
mGSS
Xp
Cg
gCR
11
221
2
,
11
1
GDm
mGSDB
m
Yp
Cg
gCC
g
22,
1
GDDBLoutp CCR
CH 10 Differential Amplifiers 175
Example: Frequency Response of Cascode
KR
g
fFC
fFC
fFC
R
L
m
DB
GD
GS
S
2
0
150
100
80
250
200
1
MHz
GHz
GHz
outp
Yp
Xp
4422
73.12
95.12
,
,
,
CH 11 Frequency Response 175
CH 10 Differential Amplifiers 176CH 11 Frequency Response 176
MOS Cascode Example
12
11
,
1
1
GDm
mGSS
Xp
Cg
gCR
3311
221
2
,
11
1
DBGDGDm
mGSDB
m
Yp
CCCg
gCC
g
22,
1
GDDBLoutp CCR
CH 10 Differential Amplifiers 177CH 11 Frequency Response 177
I/O Impedance of Bipolar Cascode
sCCrZ in
111 2
1||
sCCRZ
CSLout
22
1||
CH 10 Differential Amplifiers 178CH 11 Frequency Response 178
I/O Impedance of MOS Cascode
sCg
gC
Z
GDm
mGS
in
12
11 1
1
sCCRZ
DBGDLout
22
1||
CH 10 Differential Amplifiers 179CH 11 Frequency Response 179
Bipolar Differential Pair Frequency Response
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.
Half Circuit
CH 10 Differential Amplifiers 180CH 11 Frequency Response 180
MOS Differential Pair Frequency Response
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.
Half Circuit
CH 10 Differential Amplifiers 181CH 11 Frequency Response 181
Example: MOS Differential Pair
33,
11
331
3
,
1311,
1
11
1
])/1([
1
GDDBLoutp
GDm
mGSDB
m
Yp
GDmmGSSXp
CCR
Cg
gCC
g
CggCR
CH 10 Differential Amplifiers 182
Common Mode Frequency Response
12
1
SSmSSSS
SSSSDm
CM
out
RgsCR
CRRg
V
V
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 11 Frequency Response 182
CH 10 Differential Amplifiers 183
Tail Node Capacitance Contribution
Source-Body Capacitance of M1, M2 and M3
Gate-Drain Capacitance of M3
CH 11 Frequency Response 183
CH 10 Differential Amplifiers 184
Example: Capacitive Coupling
EBin RrRR 1|| 222
HzCRr B
L 5422||
1
1111
HzCRR inC
L 9.221
222
CH 11 Frequency Response 184
CH 10 Differential Amplifiers 185
MHzCRR inD
L 92.621
2212
MHzCR
Rg
S
SmL 4.422
1
11
111
22 1 v
Fin A
RR
Example: IC Amplifier – Low Frequency Design
CH 11 Frequency Response 185
CH 10 Differential Amplifiers 186
77.3|| 211 inDmin
X RRgv
v
Example: IC Amplifier – Midband Design
CH 11 Frequency Response 186
CH 10 Differential Amplifiers 187
Example: IC Amplifier – High Frequency Design
)21.1(2
)15.1(
1
)15.2(2
)308(2
2223
2
1
GHz
CCR
GHz
MHz
DBGDLp
p
p
CH 11 Frequency Response 187
188
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
CH 12 Feedback 189
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 190
Close-loop Transfer Function
1
1
1 KA
A
X
Y
CH 12 Feedback 191
Feedback Example
A1 is the feedforward network, R1 and R2 provide the sensing and feedback capabilities, and comparison is provided by differential input of A1.
121
2
1
1 ARR
RA
X
Y
CH 12 Feedback 192
Comparison Error
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.
KA
XE
11
E
CH 12 Feedback 193
Comparison Error
2
11R
R
X
Y
CH 12 Feedback 194
Loop Gain
When the input is grounded, and the loop is broken at an arbitrary location, the loop gain is measured to be KA1.
test
N
V
VKA 10X
CH 12 Feedback 195
Example: Alternative Loop Gain Measurement
testN VKAV 1
CH 12 Feedback 196
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 197
Gain Desensitization
A large loop gain is needed to create a precise gain, one that does not depend on A1, which can vary by ±20%.
11 KAKX
Y 1
CH 12 Feedback 198
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 199
Merits of Negative Feedback
1) Bandwidth enhancement
2) Modification of I/O Impedances
3) Linearization
CH 12 Feedback 200
Bandwidth Enhancement
Although negative feedback lowers the gain by (1+KA0), it also extends the bandwidth by the same amount.
0
0
1s
AsA
00
0
0
11
1
KA
sKA
A
sX
Y
Open LoopClosed Loop
Negative
Feedback
CH 12 Feedback 201
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 202
Example: Open Loop Parameters
Dout
min
Dm
RR
gR
RgA
1
0
CH 12 Feedback 203
Example: Closed Loop Voltage Gain
Dm
Dm
in
out
RgRR
RRg
v
v
21
21
CH 12 Feedback 204
Example: Closed Loop I/O Impedance
Dmm
in RgRR
R
gR
21
211
Dm
Dout
RgRR
RR
R
21
21
CH 12 Feedback 205
Example: Load Desensitization
3/DmDm RgRg Dm
Dm
Dm
Dm
RgRR
RRg
RgRR
RRg
21
2
21
2 31
W/O Feedback
Large Difference
With Feedback
Small Difference
CH 12 Feedback 206
Linearization
Before feedback
After feedback
CH 12 Feedback 207
Four Types of Amplifiers
CH 12 Feedback 208
Ideal Models of the Four Amplifier Types
CH 12 Feedback 209
Realistic Models of the Four Amplifier Types
CH 12 Feedback 210
Examples of the Four Amplifier Types
CH 12 Feedback 211
Sensing a Voltage
In order to sense a voltage across two terminals, a voltmeter with ideally infinite impedance is used.
CH 12 Feedback 212
Sensing and Returning a Voltage
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.
21 RR
Feedback
Network
CH 12 Feedback 213
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 214
Sensing and Returning a Current
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.
0SR
Feedback
Network
CH 12 Feedback 215
Addition of Two Voltage Sources
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.
Feedback
Network
CH 12 Feedback 216
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 217
Addition of Two Current Sources
In order to add two current sources, we place them in parallel. So the feedback network is placed in parallel with the input signal.
Feedback
Network
CH 12 Feedback 218
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 219
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 220
Example: Feedback Factor
mFout
F gv
iK
CH 12 Feedback 221
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 222
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 223
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 224
Polarity of Feedback Example I
inV 21 , DD II xout VV , 12 , DD II
Negative Feedback
CH 12 Feedback 225
Polarity of Feedback Example II
inV AD VI ,1 xout VV , AD VI ,1
Negative Feedback
CH 12 Feedback 226
Polarity of Feedback Example III
inI XD VI ,1 2, Dout IV XD VI ,1
Positive Feedback
CH 12 Feedback 227
Voltage-Voltage Feedback
0
0
1 KA
A
V
V
in
out
CH 12 Feedback 228
Example: Voltage-Voltage Feedback
)||(1
)||(
21
2OPONmN
OPONmN
in
out
rrgRR
Rrrg
V
V
CH 12 Feedback 229
Input Impedance of a V-V Feedback
)1( 0KARI
Vin
in
in
A better voltage sensor
CH 12 Feedback 230
Example: V-V Feedback Input Impedance
Dmmin
in RgRR
R
gI
V
21
211
CH 12 Feedback 231
Output Impedance of a V-V Feedback
01 KA
R
I
V out
X
X
A better voltage source
CH 12 Feedback 232
Example: V-V Feedback Output Impedance
mNclosedout gR
RR
11
2
1,
CH 12 Feedback 233
Voltage-Current Feedback
O
O
in
out
KR
R
I
V
1
CH 12 Feedback 234
Example: Voltage-Current Feedback
F
DDm
DDm
in
out
R
RRgRRg
I
V
212
212
1
CH 12 Feedback 235
Input Impedance of a V-C Feedback
KR
R
I
V in
X
X
01
A better current sensor.
CH 12 Feedback 236
Example: V-C Feedback Input Impedance
F
DDmmclosedin
R
RRggR
2121,
1
1.
1
CH 12 Feedback 237
Output Impedance of a V-C Feedback
A better voltage source.
KR
R
I
V out
X
X
01
CH 12 Feedback 238
Example: V-C Feedback Output Impedance
F
DDm
Dclosedout
R
RRgR
R212
2,
1
CH 12 Feedback 239
Current-Voltage Feedback
m
m
in
out
KG
G
V
I
1
CH 12 Feedback 240
Example: Current-Voltage Feedback
MOOmm
OOmmclosed
in
out
Rrrgg
rrgg
V
I
5331
5331
||1
|||
Laser
CH 12 Feedback 241
Input Impedance of a C-V Feedback
A better voltage sensor.
)1( minin
in KGRI
V
CH 12 Feedback 242
Output Impedance of a C-V Feedback
A better current source.
)1( moutX
X KGRI
V
CH 12 Feedback 243
Laser
Example: Current-Voltage Feedback
)1(1
|
)1(1
|
1|
212
211
21
21
MDmmm
closedout
MDmmm
closedin
MDmm
Dmmclosed
in
out
RRggg
R
RRggg
R
RRgg
Rgg
V
I
CH 12 Feedback 244
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 245
Current-Current Feedback
I
I
in
out
KA
A
I
I
1
CH 12 Feedback 246
Input Impedance of C-C Feedback
A better current sensor.
I
in
X
X
KA
R
I
V
1
CH 12 Feedback 247
Output Impedance of C-C Feedback
A better current source.
)1( IoutX
X KARI
V
CH 12 Feedback 248
Example: Test of Negative Feedback
inI outD IV ,1 FP IV , outD IV ,1
Negative Feedback
Laser
CH 12 Feedback 249
Example: C-C Negative Feedback
)]/(1[|
)/(1
1.
1|
)/(1|
22
21
2
2
FDmOclosedout
FMDmmclosedin
FMDm
DmclosedI
RRRgrR
RRRggR
RRRg
RgA
M
Laser
CH 12 Feedback 250
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 251
Rules for Breaking the Loop of Amplifier Types
CH 12 Feedback 252
Intuitive Understanding of these Rules
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.
Voltage-Voltage Feedback
CH 12 Feedback 253
Rules for Calculating Feedback Factor
CH 12 Feedback 254
Intuitive Understanding of these Rules
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.
Voltage-Voltage Feedback
CH 12 Feedback 255
Breaking the Loop Example I
21,
1,
211,
||
/1
||
RRRR
gR
RRRgA
Dopenout
mopenin
Dmopenv
CH 12 Feedback 256
Feedback Factor Example I
)1/(
)1(
)1/(
)/(
,,,
,,,
,,,
212
openvclosedoutclosedout
openvopeninclosedin
openvopenvclosedv
KARR
KARR
KAAA
RRRK
CH 12 Feedback 257
Breaking the Loop Example II
21,
,
21,
||||
||||
RRrrR
R
RRrrgA
OPONopenout
openin
OPONmNopenv
CH 12 Feedback 258
Feedback Factor Example II
)1/(
)1/(
)/(
,,,
,
,,,
212
openvopenoutclosedout
closedin
openvopenvclosedv
KARR
R
KAAA
RRRK
CH 12 Feedback 259
Breaking the Loop Example IV
FDopenout
Fm
openin
FDm
mF
DFopen
in
out
RRR
Rg
R
RRg
gR
RR
I
V
||
||1
||.1
|
2,
1,
22
1
1
CH 12 Feedback 260
Feedback Factor Example IV
)|1/(
)|1/(
)|1/(||
/1
,,
,,
openin
outopenoutclosedout
openin
outopeninclosedin
openin
outopen
in
outclosed
in
out
F
I
VKRR
I
VKRR
I
VK
I
V
I
V
RK
CH 12 Feedback 261
Breaking the Loop Example V
MOopenout
openin
MLO
OmOOmopen
in
out
RrR
R
RRr
rgrrg
V
I
1,
,
1
11533 |||
CH 12 Feedback 262
]|)/(1[
]|)/(1/[)|/()|/(
,,
,
openinoutopenoutclosedout
closedin
openinoutopeninoutclosedinout
M
VIKRR
R
VIKVIVI
RK
Feedback Factor Example V
CH 12 Feedback 263
Breaking the Loop Example VI
Mmopenout
mopenin
mML
Dmopen
in
out
RgR
gR
gRR
Rg
V
I
)/1(
/1
/1|
2,
1,
2
1
CH 12 Feedback 264
Feedback Factor Example VI
]|)/(1[
]|)/(1[
]|)/(1/[)|/()|/(
,,
,,
openinoutopenoutclosedout
openinoutopeninclosedin
openinoutopeninoutclosedinout
M
VIKRR
VIKRR
VIKVIVI
RK
CH 12 Feedback 265
Breaking the Loop Example VII
MFOopenout
MFm
openin
FMLO
Om
mMF
DMFopenI
RRrR
RRg
R
RRRr
rg
gRR
RRRA
||
)(||1
||.
1)(
2,
1,
2
22
1
,
CH 12 Feedback 266
Feedback Factor Example VII
)1(
)1/(
)1/(
)/(
,,,
,,,
,,,
openIopenoutclosedout
openIopeninclosedin
openIopenIclosedI
MFM
KARR
KARR
KAAA
RRRK
CH 12 Feedback 267
Breaking the Loop Example VIII
MFopenout
Fm
openin
MFmmF
DFopen
in
out
RRR
Rg
R
RRggR
RR
I
V
||
||1
)]||([/1
|
,
1,
21
CH 12 Feedback 268
Feedback Factor Example VIII
]|)/(1/[
]|)/(1/[
]|)/(1/[|)/(|)/(
/1
,,
,,
openinoutopenoutclosedout
openinoutopeninclosedin
openinoutopeninoutclosedinout
F
IVKRR
IVKRR
IVKIVIV
RK
CH 12 Feedback 269
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 270
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 271
Instability of a Negative Feedback Loop
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.
)(1
)()(
sKH
sHs
X
Y
CH 12 Feedback 272
“Barkhausen’s Criteria” for Oscillation
180)(
1|)(|
1
1
jKH
jKH
CH 12 Feedback 273
Time Evolution of Instability
CH 12 Feedback 274
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 275
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 276
Condition for Stability
ωPX, (“phase crossover”), is the frequency at which KH=-180o.
ωGX, (“gain crossover”), is the frequency at which |KH|=1.
PXGX
CH 12 Feedback 277
Stability Example I
1
1||
K
H p
CH 12 Feedback 278
Stability Example II
5.0
1||5.0
K
H p
CH 12 Feedback 279
Marginally Stable vs. Stable
Marginally Stable Stable
CH 12 Feedback 280
Phase Margin
Phase Margin = H(ωGX)+180
The larger the phase margin, the more stable the negative feedback becomes
CH 12 Feedback 281
Phase Margin Example
45PM
CH 12 Feedback 282
Frequency Compensation
Phase margin can be improved by moving ωGX closer to origin while maintaining ωPX unchanged.
CH 12 Feedback 283
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 284
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 285
Example: 45o Phase Margin Compensation
2pPM
CH 12 Feedback 286
Miller Compensation
To save chip area, Miller multiplication of a smaller capacitance creates an equivalent effect.
cOOmeq CrrgC )]||(1[ 655