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ECE 546 – Jose Schutt‐Aine 1
ECE 546Lecture 11
MOS AmplifiersSpring 2020
Jose E. Schutt-AineElectrical & Computer Engineering
University of Illinoisjesa@illinois.edu
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• Definitions– Used to increase the amplitude of an input signal to a
desired level– This is a fundamental signal processing function– Must be linear (free of distortion) – Shape of signal
preserved
Amplifiers
( ) ( ),o iv t Av t where A is the voltage gain
vi(t) vo(t)AMP
: ov
i
vVoltage Gain Av
( ) :( )
Lp
I
Load Power PPower Gain AInput Power P
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Amplifiers
: p v iNote A A A
: oi
i
iCurrent Gain Ai
o op
I I
v iAv i
20log VVoltage gain in dB A
20log ICurrent gain in dB A
10log PPower gain in dB A
Expressing gain in dB (decibels)
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Amplifiers
1 1 2 2DCP V I V I
100L
DC
P Power EfficiencyP
Since output associated with the signal is larger than the input signal, power must come from DC supply
DC I L dissipatedP P P P
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Biasing of Amp
( ) ( )I QI IV t V v t
Bias will provide quiescent points for input and output about which variations will take place. Bias maintain amplifier in active region.
( ) ( )o QO oV t V v t
( ) ( )o v Iv t A v to
vI at Q
dvAdv
Amplifier characteristics are determined by bias point
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Small-Signal Model
• What is a small-signal incremental model?
– Equivalent circuit that only accounts for signal level fluctuations about the DC bias operating points
– Fluctuations are assumed to be small enough so as not to drive the devices out of the proper range of operation
– Assumed to be linear
– Derives from superposition principle
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• Bias Characteristics– Operation in saturation region– Stable and predictable drain current
Biasing of MOS Transistors
212D n ox GS T
WI C V VL
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Single-Supply MOS Bias
– Choose R1 and R2 to fix VG– Choose RS and R2 to fix VS– VGS determines ID– Choose RD to fix VD
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Common Source MOSFET Amplifier
Bias is to keep MOS in saturation region
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Small-Signal Equivalent Circuit for MOS (device only)
Common Source MOSFET Amplifier
2'12D n GS T
WI k V VL
2
GS GSQ
D Dm
GS effV V
I IgV V
GS T effwhere V V V /mg is proportional to W L
'2 /m n Dg k W L I
Which leads to
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To calculate rds, account for
21 1
2GS GSQ
DSds
D DPV V ox GS T
Vr WI IC V VL
2'12DP n GS T
WI k V VL
rds, accounts for channel width modulation resistance.
MOSFET Output Impedance
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Midband Frequency Gain
out ds DBMB m
in B g ds D
v r RRA gv R R r R
Incremental model for complete amplifier
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For the circuit shown, k=75 A/V2, VT=1 V, =0(a) Find VDQ, VSQ(b) Find the midband gain
2
1 2
20 5 425
GQ DDRV V V
R R
Example
4 2 GSQ GQ SQ DQV V V I
20.075(9 12 4 )DQ DQ DQI I I
2 24 12 9 13.3 6.33 2.25 0DQ DQ DQ DQ DQI I I I I
2 20.075 4 2 1DQ GSQ T DQI K V V I
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26.33 93.167 0.378 or 5.9532
DQI mA mA
20 10 0.378 16.22DQ DD D DQV V R I V
2 0.378 0.756SQ S DQV R I V
Example (Cont’)
16.22DQV V
0.756SQV V
reject since voltage drop across RD willbe too large0.378DQI mA
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0.337 10 3.37MB m DA g R
Example (Cont’)
'2 4 0.075 0.378 0.337m n DQWg k IL
3.37MBA
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Low-Pass Circuit
In frequency domain:1
1i
oVV
j CRj C
11 1
i oo v
i
V VV Aj RC V j RC
2
1 11 1 /vA
j RC jf f
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21 1
2 2f
RC
2 time constantRC
Low-Pass Circuit
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High-Pass Circuit
1 11i i
oV R VV
Rj C j RC
2
1 11 1 /1
2
ov
i
VAV jf fj
fRC
21
2f
RC
18
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Model for general Amplifying Element
Cc1 and Cc2 are coupling capacitors (large) F
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Cin and Cout are parasitic capacitors (small) pF
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Midband Frequencies
- Coupling capacitors are short circuits
- Parasitic capacitors are open circuits
out in LMB
in g in out L
v R RA Av R R R R
20
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Low Frequency Model- Coupling capacitors are present- Parasitic capacitors are open circuits
1
1
1
1 1 ( )in in in c in
abc g in
g inc
v R v j C Rvj C R RR R
j C
1
1
( )1 ( )
c g ininab in
g in c g in
j C R RRv vR R j C R R
21
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22 Dm n ox eff n ox D
eff
W W Ig C V C IL L V
MOSFET High-Frequency Model
2 2mb m mF sb
g g gV
1/ds A DD
r V II
23gs ox ov oxC WLC WL C
gd ov oxC WL C
1
sbosb
SB
o
CCVV
1
dbodb
DB
o
CCVV
22
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CS - Three Frequency Bands
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o m gs gd gsI g V sC V
Unity-Gain Frequency fTfT is defined as the frequency at which the short-circuit current gain of the common source configuration becomes unity
(neglect sCgdVgs since Cgd is small)
o m gsI g V i
gsgs gd
IVs C C
o m
i gs gd
I gI s C C
s jDefine:
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For s=j, magnitude of current gain becomes unity at
2m m
T Tgs gd gs gd
g gfC C C C
fT ~ 100 MHz for 5-m CMOS, fT ~ several GHz for 0.13m CMOS
Calculating fT
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CS - High-Frequency Response
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ii
1GR
DD
1GR
gg
1GR
dsds
1gr
'i D m gdo
' ' 2 'i i g gs gd gd D i g gd m D gd gs D
G R g sCvv G G s C C sC R G G sC g R s C C R
CS – Miller Effect – Exact Analysis
'D D ds
D ds
1R R || rG g
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2 ' 'gd gs D gd m D gs ms C C R sC g R or sC g
We neglect the terms in s2 since
'
i D m gdo' '
i i g gs gd m D gd D i g
G R g sCvv G G s C C 1 g R C R G G
ii
1RG
If we multiply through by
CS – Miller Effect – Exact Analysis
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Miller
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'D m gdo
' 'i i g i gs gd m D gd D s g
R g sCvv 1 R G s R C C 1 g R C R 1 R G
i g
H ' 'i gs gd m D gd D i g
1 R Gf
2 R C C 1 g R C R 1 R G
From which we extract the 3-dB frequency point
CS – Miller Effect – Exact Analysis
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H ' 'i gs gd m D gd D
1f2 R C C 1 g R C R
H 'gd D
1f2 C R
If Gg is negligible
If Ri =0
CS – Miller Effect – Exact Analysis
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1 2
1 2
...( )
...m
H mm
s Z s Z s ZF s a
s P s P s P
Transfer Function Representation
Z1, Z2,…Zm are the zeros of the transfer function
P1, P2,…Pm are the poles of the transfer function
In general, the transfer function of an amplifier can be expressed as
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s is a complex number s = + j
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•Designer is interested in midband operation•However needs to know upper 3‐dB frequency• In many cases some conditions are met:Zeros are infinity or at very high frequenciesOne of the poles (P1) is at much lower frequency than other poles (dominant pole)
• If the conditions are met then FH(s) can be approximated by:
11
1( ) and we have1 /H H P
P
F ss
( ) ( )M HA s A F s3dB Frequency Determination
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If the lowest frequency pole is at least 4 times away from the nearest pole or zero, it is a dominant pole
If there is no dominant pole, the 3‐dB frequency Hcan be approximated by:
2 2 2 21 2 1 2
1 1 1 11/ ... 2 ...HP P Z Z
3dB Frequency Determination
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21 2
21 2
1 ...( )1 ...
nn
H nn
a s a s a sF sb s b s b s
The coefficients a and b are related to the frequencies of the zeros and poles respectively.
Open-Circuit Time Constants
b1 can be obtained by summing the individual time constants of the circuit using the open-circuit time constant method
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11 2
1 1 1...p p pn
b
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Open-Circuit Time Constant Method• The time constant of each capacitor in the circuit is evaluated. It is the product of the capacitance and the resistance seen across its terminals with:All other internal capacitors open circuitedAll independent voltage sources short circuitedAll independent current sources opened
• The value of b1 is computed by summing the individual time constants
11
n
i ioi
b C R
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• An approximation can be made by using the value of b1 to determine the 3dB upper frequency point H
• If the zeros are not dominant and if one of the poles P1 is dominant, then
11
1
P
b
1
1 1H
i ioi
b C R
Assuming that the 3‐dB frequency will be approximately equal to P1
Open-Circuit Time Constant Method
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• The poles of a multistage amplifier are difficult to obtain analytically
• An approximate value for the 3dB upper frequency point 3dB can be obtained by assigning an open circuit time constant io to each capacitor Ci
Bandwidth of Multistage Amplifier
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• The time constant io is the product of the capacitance and the resistance seen across its terminals with:All other internal capacitors open circuitedAll independent voltage sources short circuitedAll independent current sources opened
• The upper 3dB frequency point 3dB is then found by using :
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dBio
Bandwidth of Multistage Amplifier
MOSFET amplifier has Rsig= 100 k, Cgs=Cgd= 1 pF, gm = 4 mA/V and RL’ =3.33 k. Find midband voltage gain and 3-dB frequency.
MOSFET Amp Bandwidth
' 420 4 3.33 10.8420 100
o inM m L
sig in sig
V RA g RV R R
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|| 420 ||100 80.8gs in sigR R R k k k
12 31 10 80.8 10 80.8gs gs gsC R ns
MOSFET Amp AnalysisTo determine the 3‐dB frequency, we first evaluate the time constant associated with Cgs. First, we determine the resistance Rgs seen by Cgs. The capacitance Cgd is removed and Vsig is short‐circuited
The time constant associated with Cgs is
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gs gsx
in sig
V VI
R R '
gs xV I R
'gs x
x m gsL
V VI g V
R
' ' ' 'x
gd L m Lx
VR R R g R RI
The resistance Rgd seen by Cgd is found by setting Cgs = 0 and short‐circuiting Vsig
MOSFET Amp Analysis
' ||in sigR R R
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12 61 10 1.16 10 1160gd gd gdC R ns
9
1 1 806 /80.8 1160 10H
gs gd
krad s
128.32
HHf kHz
MOSFET Amp Analysis
The open‐circuit time constant of Cgd is
The upper 3‐dB frequency H can now be determined from