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Differential Amplifiers (Ch. 10)p
김 영 석김 영 석
충북대학교 전자정보대학
2012.9.1.9.
Email: [email protected]
전자정보대학 김영석 10-1
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Contents10.1 General Considerations
10.2 Bipolar Differential Pair
10.3 MOS Differential Pair
10 4 C d Diff ti l A lifi10.4 Cascode Differential Amplifiers
10.5 Common-Mode Rejection
10.6 Differential Pair with Active Load1 .6 Differential Pair with Active Load
전자정보대학 김영석 10-2
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10.1 General ConsiderationsHumming Noise in Audio Amplifier from 60Hz Power Line
CCCCout IRVV −=
10-3전자정보대학 김영석
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10.1 General ConsiderationsHow to suppress the Hum ?
Vout should not to be referenced to GND
Since both node X and Y contain the ripple, their difference will be free of ripple
CCrCCout IRvVV −+= )(
will be free of ripple
But, wasting current in Q2
rinvX vvAv +=
) ( rinvYXout
rY
vNOvAvvVvv
=−==
10-4전자정보대학 김영석
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10.1 General ConsiderationsTo utilize the wasting current in Q2
Apply the inputs differentially
• the outputs are 180° out of phase
h i V t h d diff ti ll• enhancing Vout when sensed differentially
But, Common-Mode (CM) problem: increase VCM => increase gm => increase gaing g
rinvX
AvvAv +=
invYXout
rinvY
vAvvVvvAv
2=−=+−=
10-5전자정보대학 김영석
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10.1 General ConsiderationsTo have the differential-mode gain independent of input CM level
Add a tail current
10-6전자정보대학 김영석
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10.2 Bipolar Differential PairCommon-Mode Response
21 = BEBE VV
CM gain
221
==
==
EE
EECC
IRVVV
III
oft independen2
∂
−==
CM
CCCYX
VV
RVVV
0=∂∂
=∴CM
outCM V
VA
CM i tCM input range
EECCCCM
IRVV −< Region)Activein(Q1
BECM
CCCCM
VVEEV
RVV
+>
< Region)Activein (Q2 1
10-7전자정보대학 김영석
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Differential Pair Characteristics
(a)
EEC
III =0
1EEC II =2
(b)
CCY
EECCCX
C
VVIRVV
I
=−=
= 02
CCX
EECCCY
C
VVIRVV
I
=−=
= 01
(b)( ) (b)(b)
(a)(a)
10-8전자정보대학 김영석
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Small-Signal Analysis
EEEE
mEEEE
C
VgIIII
VgIIII
Δ−=Δ−=
Δ+=Δ+=221
mCCY
mCCX
mC
VgRIRVVgRIRV
VgII
Δ+=Δ+=ΔΔ−=Δ−=Δ
Δ=Δ=222
in
mCYXout
mCCY
VVV
VgRVVVΔ=
Δ−=Δ−Δ=2
2
Cmin
outv Rg
VVA −==
Virtual Ground: 0=Δ PV
10-9전자정보대학 김영석
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10.2.2 Large Signal Analysis
Tin VV 4max, ≈Δ
EECC III 21 =+
inin
SSTinSC
TinSC
VVI
IIVVIIVVII
21
21222
111
exp
, /exp/exp
−==
=
T
inin
TEE
C
VVV
VI
I21
1
exp1
exp
−+
=
T
inin
EEC
VVV
II21
2
exp1 −+
=
T
ininEEC
outout
VVVIR
VV
2tanh 21
21
−−
=−
10-10전자정보대학 김영석
TV2
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Linear/Nonlinear Regions
The left column operates in linear region, whereas the right column operates in nonlinear regionright column operates in nonlinear region.
10-11전자정보대학 김영석
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10.2.3 Small-Signal Model
10-12전자정보대학 김영석
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Half Circuits
mCout
ggvgRvvgRv
=−=−= 111 π
inin
mmmCout
vvvvvv
ggvgRv
−+=+=
==
21
2211
21222
,,
ππ
π
Cminin
outout Rgvvvv
−=− 21
21
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.
10-13전자정보대학 김영석
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Example: Differential Gain
outout rgvv−=
− 21Om
inin
rgvv
=− 21
10-14전자정보대학 김영석
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Half Circuit Example I
0=XV
( )1311 |||| RrrgA OOmv −=
10-15전자정보대학 김영석
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Half Circuit Example II
( )1311 |||| RrrgA OOmv −=
10-16전자정보대학 김영석
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Half Circuit Example III
Cv
RA 1−=
mE g
R 1+
10-17전자정보대학 김영석
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Half Circuit Example IV
CRA −=
m
Ev
gRA
12+
mg
10-18전자정보대학 김영석
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10.3 MOS Differential Pair’s Common-Mode Response
ICM gain = 0
The equilibrium overdrive voltage:
In ut CM Range
2SS
DDDYXIRVVV −==
( )
LWC
IVVVoxn
SSequilTHGS
μ=−=Δ
Input CM Range
VVV R i )S ii(M1
Loxnμ
THCMSS
DDD
THGD
VVIRV
VVV
−>−
−>
2
Region)Saturationin (M1
SSGSCM
THSS
DDDCM
VVVAnd
VIRVV
+>
+−<∴
,2SSGSCM,
10-19전자정보대학 김영석
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Differential Response
(b)(a)
( )(a)
(b)(b)
(a) (b)
CH 10 Differential Amplifiers
10-20전자정보대학 김영석
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Small-Signal Response
VΔ 0
Dmv
P
RgAV
−==Δ 0
Dmv
Similar to its bipolar counterpart, the MOS differential pair exhibits the same virtual ground node and small signal gain.
10-21전자정보대학 김영석
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10.3.2 MOS Differential Pair’s Large-Signal Response
W1
THGSoxnD
THGSoxnD
VVL
WCI
VVL
WCI
−=
−=
222
211
)(21
)(21
μ
μ
( ) ( )241 SS VVIVVWCII
SSDD
THGSoxnD
IIIL
=+ 21
22 2
( ) ( )2211214
21
2 inin
oxn
SSinoxnDD VV
LWC
IVVLWCII
in−−−=−
μμ
( )LWC
IVVVVVoxn
DequilTHGSinin /
2 22max21 μ
=−=Δ=−
10-22전자정보대학 김영석
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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.
And,
Tinin
inin
VVVBJT
VVVMOS
4:
2:
max21
max21
≈−
Δ=−
MOS Bipolar
10-23전자정보대학 김영석
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The effects of Doubling the Tail Current
( )I
VVVV
D
equilTHGSinin
2
2max21
=
−=−
LWCoxn /
μ=
ISS is doubledSS
ΔVin,max increases by
ΔVout,max increases by
22
10-24전자정보대학 김영석
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The effects of Doubling W/L
( )I
VVVV
D
equilTHGSinin
2
2max21
=
−=−
LWCoxn /
μ=
W/L is doubled
Vin max decreases by 2Vin,max decreases by
Vout,max same
10-25전자정보대학 김영석
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Small-Signal Analysis of MOS Differential Pair
( ) ( )241 SS VVIVVWCII ( ) ( )
( ) ( ) ( )
21121
41
2 2 inin
oxn
SSinoxnDD
WIW
VV
LWC
VVL
CIIin
−−−=−μ
μ
( ) ( ) ( )2121214
21 ininmininSSoxn
oxn
SSininoxn VVgVVI
LWC
LWC
IVVLWC −=−=−≈ μ
μμ
When the input differential signal is small compared to 4ISS/μnCox(W/L), the output differential current is linearly proportional to it and small-signal model can be appliedproportional to it, and small signal model can be applied.
10-26전자정보대학 김영석
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Virtual Ground and Half Circuit
PV =Δ 0
Cmv RgA −=
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
dbe used to calculate the gain.
10-27전자정보대학 김영석
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MOS Differential Pair Half Circuit Example I
⎞⎛
≠
1
0λ
⎟⎟⎠
⎞⎜⎜⎝
⎛−= 13
31 ||||1
OOm
mv rrg
gA
10-28전자정보대학 김영석
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MOS Differential Pair Half Circuit Example II
0g
=λ
3
1
m
mv g
gA −=
10-29전자정보대학 김영석
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MOS Differential Pair Half Circuit Example III
DDRA 20=λ
mSS
DDv gR
A12+
−=
10-30전자정보대학 김영석
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10.4 Bipolar Cascode Differential Pair
( )[ ])1(|| 333131 OmOOmv rgrrrgA ++−= π
10-31전자정보대학 김영석
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Bipolar Telescopic Cascode
( )[ ] [ ])||(|||| 575531331 ππ rrrgrrrggA OOmOOmmv −≈
10-32전자정보대학 김영석
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Example: Bipolar Telescopic Parasitic Resistance
RrrgrR |||| 1 ⎟⎞
⎜⎛≈
[ ] opOOmmv
OmOop
RrrrggA
rrgrR
||)||(2
||||
31331
5755
π
π
−=
⎟⎠
⎜⎝
≈
10-33전자정보대학 김영석
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MOS Cascode Differential Pair
1331 OmOmv rgrgA −≈
10-34전자정보대학 김영석
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MOS Telescopic Cascode
( )[ ])(|| 7551331 OOmOOmmv rrgrrggA −≈ ( )[ ])(|| 7551331 OOmOOmmv ggg
10-35전자정보대학 김영석
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Example: MOS Telescopic Parasitic Resistance
)||(
)]||(1[||
1331
155715
OmOopmv
OmOOop
rgrRgA
RrgrRrR
−≈
++=
10-36전자정보대학 김영석
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10.5 CM RejectionEffect of Finite Tail Impedance
Δ
mEE
C
CMin
CMout
gRR
VV
2/12/
,
,
+−=
ΔΔ
If the tail current source is not ideal, then when a input CM voltage is applied, the currents in Q1 and Q2 and g pp 1 2 hence output CM voltage will change.
10-37전자정보대학 김영석
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Input CM Noise with Ideal Tail Current
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Input CM Noise with Non-ideal Tail Current
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CM to DM Conversion, ACM-DM
Dout
RgR
VV
2/1 +Δ
=ΔΔ
EEmCM RgV 2/1 +Δ
If finite tail impedance and asymmetry are both present, then the differential output signal will contain a portion of input common mode signalinput common-mode signal.
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Example: ACM-DM
A DMCM =−
[ ])||)(1(2131333 πrRrgr
R
OmO
C
+++
Δ
[ ])||)(( 313331
πgg OmO
m
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CMRR
DMCM
DM
AACMRR =
DMCMA −
CMRR defines the ratio of wanted amplified differential input signal to unwanted converted input common-mode noise that appears at the output. pp p
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10.6 Diff Pair with Active LoadDifferential to Single-ended Conversion
Many circuits require a differential to single-ended i h th b t l i t dconversion, however, the above topology is not very good.
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Supply Noise Corruption
The most critical drawback of this topology is supply noise corruption, since no common-mode cancellation
h i i t Al l h lf f th i lmechanism exists. Also, we lose half of the signal.
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Better Alternative
This circuit topology performs differential to single-ended conversion with no loss of gain.
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Active LoadWith 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”.and is known as an active load .
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Differential Pair with Active Load
Th i t diff ti l i d th t dThe input differential pair decreases the current drawn from RL by ΔI and the active load pushes an extra ΔI into RL by current mirror action; these effects enhance each other.
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Active Load vs. Static Load
The load on the left responds to the input signal and p p g enhances the single-ended output, whereas the load on the right does not.
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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.
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Asymmetric Differential Pair
Because of the vastly different resistance magnitude atBecause 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.
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Thevenin Equivalent of the Input Pair
ininoNmNThev vvrgv )( 21 −−=
oNThev
ininoNmNThev
rRg
2)( 21
=
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Simplified Differential Pair with Active Load
)||( OPONmNout rrgv
=21 inin vv −
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