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Set 3: Single-Stage Amplifiers1SM
EECE488: Analog CMOS Integrated Circuit Design
3. Single-Stage Amplifiers
Shahriar MirabbasiDepartment of Electrical and Computer Engineering
University of British [email protected]
Technical contributions of Pedram Lajevardi in revising the course notes are greatly acknowledged.
Set 3: Single-Stage Amplifiers2SM
Overview
1. Why Amplifiers?2. Amplifier Characteristics3. Amplifier Trade-offs4. Single-stage Amplifiers5. Common Source Amplifiers
1. Resistive Load2. Diode-connected Load3. Current Source Load4. Triode Load5. Source Degeneration
2
Set 3: Single-Stage Amplifiers3SM
Overview
6. Common-Drain (Source-Follower) Amplifiers 1. Resistive Load2. Current Source Load3. Voltage Division in Source Followers
7. Common-Gate Amplifiers
6. Cascode Amplifiers
Set 3: Single-Stage Amplifiers4SM
Reading Assignments
• Reading:Chapter 3 of Razavi’s book
• In this set of slides we will study low-frequency small-signal behavior of single-stage CMOS amplifiers. Although, we assume long-channel MOS models (not a good assumption for deep submicron technologies) the techniques discussed here help us to develop basic circuit intuition and to better understand and predict the behavior of circuits.
Most of the figures in these lecture notes are © Design of Analog CMOS Integrated Circuits, McGraw-Hill, 2001.
3
Set 3: Single-Stage Amplifiers5SM
Why Amplifiers?
• Amplifiers are essential building blocks of both analog and digital systems.
• Amplifiers are needed for variety of reasons including:
– To amplify a weak analog signal for further processing
– To reduce the effects of noise of the next stage
– To provide a proper logical levels (in digital circuits)
• Amplifiers also play a crucial role in feedback systems
• We first look at the low-frequency performance of amplifiers. Therefore, all capacitors in the small-signal model are ignored!
Set 3: Single-Stage Amplifiers6SM
Amplifier Characteristics - 1
• Ideally we would like that the output of an amplifier be a linear function of the input, i.e., the input times a constant gain:
xy 1α=x
y
• In real world the input-output characteristics is typically a nonlinear function:
4
Set 3: Single-Stage Amplifiers7SM
Amplifier Characteristics - 2
• It is more convenient to use a linear approximation of a nonlinear function.
• Use the tangent line to the curve at the given (operating) point.
x
y
• The larger the signal changes about the operating point, the worse the approximation of the curve by its tangent line.
• This is why small-signal analysis is so popular!
Set 3: Single-Stage Amplifiers8SM
Amplifier Characteristics - 3
• Let to get:
nn xxxxxxy )()()( 0
202010 −++−+−+≈ αααα L
)( 010 xxy −+≈ αα
xyxfyy ∆≈−=−=∆ 100 )( αα
• If x-x0=∆x is small, we can ignore the higher-order terms (hence the name small-signal analysis) to get:
• α0 is referred to as the operating (bias) point and α1 is the small-signal gain.
!)( 0
nxf n
n =α
• A well-behaved nonlinear function in the vicinity of a given point can be approximated by its corresponding Taylor series:
nn
xxnxf
xxxf
xxxfxfy )(!
)()(
!2)(''
)()(')( 002
00
000 −⋅++−⋅+−⋅+≈ L
5
Set 3: Single-Stage Amplifiers9SM
• In practice, when designing an amplifier, we need to optimize for some performance parameters. Typically, these parameters trade performance with each other, therefore, we need to choose an acceptable compromise.
Amplifier Trade-offs
Set 3: Single-Stage Amplifiers10SM
Single-Stage Amplifiers
• We will examine the following types of amplifiers:1. Common Source2. Common Drain (Source Follower )3. Common Gate4. Cascode and Folded Cascode
• Each of these amplifiers have some advantages and some disadvantages. Often, designers have to utilize a cascade combination of these amplifiers to meet the design requirements.
6
Set 3: Single-Stage Amplifiers11SM
Common Source Basics - 1
• In common-source amplifiers, the input is (somehow!) connected to the gate and the output is (somehow!) taken from the drain.
• We can divide common source amplifiers into two groups:
1. Without source degeneration (no body effect for the main transistor):
2. With source degeneration (have to take body effect into account for the main transistor):
Set 3: Single-Stage Amplifiers12SM
Common Source Basics - 2
In a simple common source amplifier:
• gate voltage variations times gm gives the drain current variations,
• drain current variations times the load gives the output voltagevariations.
• Therefore, one can expect the small-signal gain to be:
Dmv RgA ⋅=
7
Set 3: Single-Stage Amplifiers13SM
Common Source Basics - 3
• Different types of loads can be used in an amplifier:1. Resistive Load2. Diode-connected Load3. Current Source Load4. Triode Load
• The following parameters of amplifiers are very important:1. Small-signal gain2. Voltage swing
Set 3: Single-Stage Amplifiers14SM
Resistive Load - 1
• Let’s use a resistor as the load.
• The region of operation of M1 depends on its size and the values of Vinand R.
• We are interested in the small-signal gain and the headroom (which determines the maximum voltage swing).
• We will calculate the gain using two different methods
1. Small-signal model
2. Large-signal analysis
8
Set 3: Single-Stage Amplifiers15SM
Resistive Load - 2
Gain – Method 1: Small-Signal Model
• This is assuming that the transistor is in saturation, and channel length modulation is ignored.
• The current through RD:
• Output Voltage:
• Small-signal Gain:
INmD vgi ⋅=
DINmDDOUT RvgRiv ⋅⋅−=⋅−=
Dm
IN
OUTv Rg
vvA ⋅−==
Set 3: Single-Stage Amplifiers16SM
Resistive Load - 3
Gain – Method 2: Large-Signal Analysis• If VIN<VTH, M1 is off, and VOUT= VDD = VDS.
mDTHINoxnD
IN
OUTv
THINoxnDddDDddOUT
gRVVLWCR
VVA
VVLWCRViRVV
⋅−=−⋅⋅⋅⋅−=∂∂
=
−⋅⋅⋅⋅⋅−=⋅−=
)(
)(21 2
µ
µ
• As VIN becomes slightly larger than VTH, M1 turns on and goes into saturation (VDS≈ VDD > VGS- VTH ≈0).
0=∂∂
=
=⋅−=
IN
OUTv
DDDDDDOUT
VVA
ViRVV
• As VIN increases, VDS decreases, and M1 goes into triode when VIN- VTH = VOUT. We can find the value of VIN that makes M1 switch its region of operation.
)()(21 2
THINTHINoxnDddDDddOUT VVVVLWCRViRVV −=−⋅⋅⋅⋅⋅−=⋅−= µ
9
Set 3: Single-Stage Amplifiers17SM
Resistive Load - 4
Gain – Method 2: Large-Signal Analysis (Continued)• As VIN increases, VDS decreases, and M1 goes into triode.
⎥⎥⎦
⎤
⎢⎢⎣
⎡
∂∂⋅−+
∂∂⋅−⋅⋅⋅⋅−=
∂∂
⎥⎥⎦
⎤
⎢⎢⎣
⎡−⋅−⋅⋅⋅⋅−=⋅−=
IN
OUTOUTOUT
IN
OUTTHINoxnD
IN
OUT
OUTOUTTHINoxnDDDDDDDOUT
VVVV
VVVV
LWCR
VV
VVVVLWCRViRVV
)(
2)(
2
µ
µ
• We can find Av from above. It will depend on both VIN and VOUT.
• If VIN increases further, M1 goes into deep triode if VOUT<< 2(VIN- VTH).
DON
ONDD
OND
DD
THINoxnD
DDOUT
OUTTHINoxnDDDDDDDOUT
RRRV
RR
V
VVLWCR
VV
VVVLWCRViRVV
+⋅=
⋅+=
−⋅⋅⋅⋅+=
⋅−⋅⋅⋅⋅−=⋅−=
11)(1
)(
µ
µ
Set 3: Single-Stage Amplifiers18SM
Resistive Load - 5
Example: Sketch the drain current and gm of M1 as a function of VIN.
• gm depends on VIN, so if VIN changes by a large amount the small-signal approximation will not be valid anymore.
• In order to have a linear amplifier, we don’t want gain to depend on parameters like gm which depend on the input signal.
10
Set 3: Single-Stage Amplifiers19SM
Resistive Load - 6
• Gain of common-source amplifier:
• To increase the gain:
1. Increase gm by increasing W or VIN (DC portion or bias). Either way, ID increases (more power) and VRD increases, which limits the voltage swing.
2. Increase RD and keep ID constant (gm and power remain constant). But, VRD increases which limits the voltage swing.
3. Increase RD and reduce ID so VRD remains constant.
If ID is reduced by decreasing W, the gain will not change.
If ID is reduced by decreasing VIN (bias), the gain will increase. Since RD is increased, the bandwidth becomes smaller (why?).
• Notice the trade-offs between gain, bandwidth, and voltage swings.
eff
RD
D
RDoxn
D
RDTHINoxnDmv V
VI
VLWC
IVVV
LWCRgA ⋅−
=⋅−=⋅−−=⋅−=22)( µµ
Set 3: Single-Stage Amplifiers20SM
Resistive Load - 7
• Now let’s consider the simple common-source circuit with channel length modulation taken into account.
• Channel length modulation becomes more important as RD increases (in the next slide we will see why!).
• Again, we will calculate the gain in two different methods
1. Small-signal Model
2. Large Signal Analysis
11
Set 3: Single-Stage Amplifiers21SM
Resistive Load - 8
Gain – Method 1: Small-Signal Model
• This is assuming that the transistor is in saturation.
• The current through RD:
• Output Voltage:
• Small-signal Gain:
INmD vgi ⋅=
( ) ( )oDINmoDDOUT rRvgrRiv ⋅⋅−=⋅−=
( )oDm
IN
OUTv rRg
vvA ⋅−==
Set 3: Single-Stage Amplifiers22SM
Resistive Load - 9
Gain – Method 2: Large-Signal Analysis
( )
( )
( )
( )oDmDo
mDo
oD
mD
DD
mD
THINoxnD
OUTTHINoxnDv
IN
OUTTHINOUTTHINoxnD
IN
OUT
OUTTHINoxnDDDDDDDOUT
rRgRr
gRrr
R
gRIRgR
VVLWCR
VVVLWCR
A
VVVVVVV
LWCR
VV
VVVLWCRVIRVV
⋅−=+
⋅⋅−=
⋅+
⋅−=
λ⋅⋅+⋅−
=λ⋅−⋅⋅⋅µ⋅⋅+
⋅λ+⋅−⋅⋅⋅µ⋅−=
⎥⎦
⎤⎢⎣
⎡
∂∂⋅λ⋅−⋅+⋅λ+⋅−⋅⋅⋅µ⋅−=
∂∂
⋅λ+⋅−⋅⋅⋅µ⋅⋅−=⋅−=
111)(211
1)(
)(211)(
1)(21
2
2
• As VIN becomes slightly larger than VTH, M1 turns on and goes into saturation (VDS≈ VDD > VGS- VTH ≈0).
12
Set 3: Single-Stage Amplifiers23SM
Resistive Load - 10
Example:• Assuming M1 is biased in active region, what is the
small-signal gain of the following circuit?
( )omom
IN
OUTv rgrg
vvA ⋅−=∞⋅−==
• I1 is a current source and ideally has an infinite impedance.
• This is the maximum gain of this amplifier (why?), and is known as the intrinsic gain.
• How can VIN change if I1 is constant?
( )OUTTHINoxnD VVVLWCI ⋅+⋅−⋅⋅⋅⋅= λµ 1)(
21 2
• Here we have to take channel-length modulation into account. As VINchanges, VOUT also changes to keep I1 constant.
Set 3: Single-Stage Amplifiers24SM
Diode Connected Load - 1
• Often, it is difficult to fabricate tightly controlled or reasonable size resistors on chip. So, it is desirable to replace the load resistor with a MOS device.
• Recall the diode connected devices:
Body Effect RX (when λ≠0) RX (when λ=0)
NO
YES
m
oX grR 1
=m
X gR 1
=
mbm
oX ggrR
+=
1mbm
X ggR
+=
1
13
Set 3: Single-Stage Amplifiers25SM
Diode Connected Load - 2
• Now consider the common-source amplifier with two types of diode connected loads:1. PMOS diode connected load:
(No body effect)
2. NMOS diode connected load:(Body effect has to be taken into account)
Set 3: Single-Stage Amplifiers26SM
Diode Connected Load - 3
PMOS Diode Connected Load:• Note that this is a common source configuration
with M2 being the load. We have:
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛⋅−=⋅−== 12
2
111
1oo
m
moXm
IN
OUTv rr
ggrRg
vvA
• Ignoring the channel length modulation (ro1=ro2=∞), we can write:
11
22
222
111
2
1
2
1
22
11
2
1
21
2
2
2
21
THGS
THSG
THSGD
THGSD
m
mv
p
n
Doxp
Doxn
m
m
mmv
VVVV
VVI
VVI
gg
A
LWLW
ILWC
ILWC
gg
ggA
−
−−=
−⋅−⋅
−=−=
⎟⎠⎞
⎜⎝⎛⋅µ
⎟⎠⎞
⎜⎝⎛⋅µ
−=
⋅⎟⎠⎞
⎜⎝⎛⋅⋅µ
⋅⎟⎠⎞
⎜⎝⎛⋅⋅µ
−=−=⎟⎟⎠
⎞⎜⎜⎝
⎛∞∞⋅−=
14
Set 3: Single-Stage Amplifiers27SM
Diode Connected Load - 4
NMOS Diode Connected Load:
• Again, note that this is a common source configurationwith M2 being the load. We have:
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛+
⋅−=⋅−== 12
22
111
1oo
mbm
moXm
IN
OUTv rr
gggrRg
vvA
• Ignoring the channel length modulation (ro1=ro2=∞), we can write:
THGS
THGSv
m
m
mbm
m
mbmmv
VVVV
LWLW
A
gg
ggg
gggA
−−
⋅+
−=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
+−=
+⋅−=
+−=⎟⎟
⎠
⎞⎜⎜⎝
⎛∞∞
+⋅−=
1
2
2
1
2
1
22
1
221
11
11
)1(1
ηη
η
Set 3: Single-Stage Amplifiers28SM
Diode Connected Load - 5
• For a diode connected load we observe that (to the first order approximation):1. The amplifier gain is not a function of the bias current. So,
the change in the input and output levels does not affect the gain, and the amplifier becomes more linear.
2. The amplifier gain is not a function of the input signal (amplifier becomes more linear).
3. The amplifier gain is a weak function (square root) of the transistor sizes. So, we have to change the dimensions by a considerable amount so as to increase the gain.
15
Set 3: Single-Stage Amplifiers29SM
Diode Connected Load - 6
4. The gain of the amplifier is reduced when body effect should be considered.
5. We want M1 to be in saturation, and M2 to be on (M2 cannot be in triode (why?)):
6. The voltage swing is constrained by both the required overdrive voltages and the threshold voltage of the diode connected device.
7. A high amplifier gain leads to a high overdrive voltage for the diode connected device which limits the voltage swing.
2111 :2,:1 THDDOUTeffTHGSOUT VVVMVVVVM −<=−>
Set 3: Single-Stage Amplifiers30SM
Diode Connected Load - 6
Example:• Find the gain of the following circuit if M1 is biased in saturation and
Is=0.75I1.
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛⋅−=⎟⎟
⎠
⎞⎜⎜⎝
⎛∞⋅−=⋅−== 12
2
112
2
111
11oo
m
moo
m
moIsXm
IN
OUTv rr
ggrr
ggrrRg
vvA
• Ignoring the channel length modulation (ro1=ro2=∞) we get:
11
22
2
1
222
111
22
11
2
1
21
42
2
2
2
21
THGS
THSG
p
n
v
THSGD
THGSD
Doxp
Doxn
m
m
mmv
VVVV
LWLW
A
VVI
VVI
ILWC
ILWC
gg
ggA
−
−⋅−=
⎟⎠⎞
⎜⎝⎛⋅
⎟⎠⎞
⎜⎝⎛⋅
⋅−=
−⋅−⋅
−=
⋅⎟⎠⎞
⎜⎝⎛⋅⋅
⋅⎟⎠⎞
⎜⎝⎛⋅⋅
−=−=⎟⎟⎠
⎞⎜⎜⎝
⎛∞∞⋅−=
µ
µ
µ
µ
16
Set 3: Single-Stage Amplifiers31SM
Diode Connected Load - 7
Example (Continued):
• We observe for this example that:
1. For fixed transistor sizes, using the current source increases the gain by a factor of 2.
2. For fixed overdrive voltages, using the current source increasesthe gain by a factor of 4.
3. For a given gain, using the current source allows us to make thediode connected load 4 times smaller.
4. For a given gain, using the current source allows us to make theoverdrive voltage of the diode connected load 4 times smaller. This increases the headroom for voltage swing.
Set 3: Single-Stage Amplifiers32SM
Current Source Load - 1
• Note that current source M2 is the load.•• Recall that the output impedance of M2 seen from Vout:
( ) ( )12111
2
oomoXm
IN
OUTv
o
X
XX
rrgrRgvvA
rivR
⋅−=⋅−==
==
• For large gain at given power, we want large ro and
Increase L and W keeping the aspect ratio constant (so ro increasesand ID remains constant). However, this approach increases thecapacitance of the output node.
• We want M2 to be in saturation so
WL
LW
LI
rD
o2
111
=⋅
∝⋅λ
=
2222 effDDOUTeffTHSGOUTDDSD VVVVVVVVV −<→=−>−=
17
Set 3: Single-Stage Amplifiers33SM
Current Source Load - 2
• We also want M1 to be in saturation:
1111 effOUTeffTHGSOUTDS VVVVVVV >→=−>=
• Thus, we want Veff1 and Veff2 to be small, so that there is more headroom for output voltage swing. For a constant ID, we can increase W1 and W2 to reduce Veff1 and Veff2.
• The intrinsic gain of this amplifier is: • In general, we have:
omv rgA ⋅−=
LAWLr
LWg vom ∝→∝∝
2
,
• But since current in this case is roughly constant:
LWALI
rLWI
LWCg v
D
oDoxnm ∝→∝⋅
=∝⋅⋅⋅=λ
µ 1,2
Set 3: Single-Stage Amplifiers34SM
Triode Load
• We recognize that this is a common source configuration with M2 being the load. Recall that if M2 is in deep triode, i.e., VSD<<2(VSG-|VTH|), it behaves like a resistor.
( )
( ) ( )( )1212 ,11
:2
oONmv
THbddoxpTHSGoxp
ON
THSGSD
rRgAVVV
LWCVV
LWC
R
VVVIf
⋅−=−−⋅⋅⋅
=−⋅⋅⋅
=
−<<
µµ
• Vb should be low enough to make sure that M2 is in deep triode region and usually requires additional complexity to be precisely generated.
• RON2 depends on µp, Cox, and VTH which in turn depend on the technology being used.
• In general, this amplifier with triode load is difficult to design and use!
• However, compared to diode-connected load, triode load consumes less headroom:
DDOUTeffTHGSOUT VVMVVVVM ≈=−> :,: 2111
18
Set 3: Single-Stage Amplifiers35SM
Source Degeneration - 1
• The following circuit shows a common source configuration with a degeneration resistor in the source.
• We will show that this configuration makes the common source amplifier more linear.
• We will use two methods to derive the gain of this circuit:1. Small-signal Model2. Using the following Lemma
Lemma:In linear systems, the voltage gain is equal to –GmRout.
Set 3: Single-Stage Amplifiers36SM
Source Degeneration - 2
Gain – Method 1: Small Signal Model
( )( ) SDSmbmO
DOm
IN
OUTv
DINmO
S
O
DSmbSmOUT
O
SD
OUTOUT
SD
OUTmbS
D
OUTINm
D
OUT
SD
OUTSOUTBSS
D
OUTINSOUTIN
D
OUTOUT
O
SOUTOUTBSmbmOUT
RRRggrRrg
vvA
RvgrR
rRRgRgv
r
RR
vvR
RvgR
Rvvg
Rv
RR
vRivRR
vvRivv
Rvi
rRivvgvgi
++⋅++⋅⋅⋅−
==
⋅⋅−=⎟⎟⎠
⎞⎜⎜⎝
⎛++⋅+⋅+⋅
⋅++⎟⎟
⎠
⎞⎜⎜⎝
⎛⋅⋅+⎟⎟
⎠
⎞⎜⎜⎝
⎛⋅+⋅=
−
⋅=⋅−=⋅+=⋅−=
−=
⋅−+⋅+⋅=
1
1
,
,
1
1
19
Set 3: Single-Stage Amplifiers37SM
Source Degeneration - 3
Gain – Method 2: Lemma
• The Lemma states that in linear systems, the voltage gain is equal to –GmRout. So we need to find Gm and Rout.
1. Gm:Recall that the equivalent transconductance of the aboveCircuit is:
( ) ( ) SSmbm
m
SSmbSm
mm RRggr
rgRRgRgrr
rgviG
O
O
OO
O
IN
OUT
+⋅++⋅
=+⋅+⋅⋅+
⋅==
]1[
Set 3: Single-Stage Amplifiers38SM
Source Degeneration - 4
Gain – Method 2: Lemma (Continued)1. ROUT:
We use the following small signal model to derive the smallsignal output impedance of this amplifier:
( )( ) ( )( )
( ) ( )
( )( )( )( ) DOSmbmS
DOSmbmSDXOUT
OSmbmSOSmbSmSX
XX
OSXmbSXmXSX
OBSmbmXSXX
SXBSSX
RrRggRRrRggRRRR
rRggRrRgRgRivR
rRigRigiRirvgvgiRiv
RivRiv
+⋅⋅+++⋅⋅⋅+++
==
⋅⋅+++=⋅⋅+⋅++==
⋅⋅−⋅−⋅−⋅−+⋅=⋅⋅−⋅−+⋅=
⋅−=⋅−=
)(1)(1
)(11
,
1
1
• Since typically rO>>RS:
( ) ( ) ( ) OSmbmOSmbmOSmbmSX rRggrRggrRggRR ⋅⋅+=⋅⋅++=⋅⋅+++= )(1)(1
20
Set 3: Single-Stage Amplifiers39SM
Source Degeneration - 5
Gain – Method 2: Lemma (Continued)
( )( )( )( )
( ) DOSmbmO
DOm
DSOmbOmO
DSOmbOmO
SSmbSmOO
OmOUTmv
RrRggrRrg
RRrgrgrRRrgrgr
RRgRgrrrgRGA
+⋅⋅+++⋅⋅−
=
+⋅⋅+⋅++⋅⋅⋅+⋅++
⋅+⋅+⋅⋅+
⋅−=⋅−=
)(1
11
( )( )( ) DOSmbmS
DOSmbmSOUT RrRggR
RrRggRR+⋅⋅+++⋅⋅⋅+++
=)(1)(1
( ) SSmbmO
Omm RRggr
rgG+⋅+⋅+
⋅=
)1(
Set 3: Single-Stage Amplifiers40SM
Source Degeneration - 6
• If we ignore body effect and channel-length modulation:
Method 1 – Small-signal Model:
Sm
DmOUTmv Rg
RgRGA⋅+⋅−
=⋅−=1
( )( )( )
( )( )( ) D
Smbm
DSmbm
DSOmbOmO
DSOmbOmOOUT R
RggRRgg
RRrgrgrRRrgrgrR
mbgor
=⋅++⋅⋅++
=+⋅⋅+⋅++⋅⋅⋅+⋅++
=→∞→ 1
111lim
0
( ) ( ) Sm
m
Smbm
m
SSmbSmOO
Omm Rg
gRgg
gRRgRgrr
rgGmbg
or ⋅+=
⋅++=
+⋅+⋅⋅+⋅
=→∞→ 11
lim0
Sm
Dm
IN
OUTv
Sm
INGS
SGSmINGSDGSmOUT
RgRg
vvA
Rgvv
RvgvvRvgv
⋅+⋅−
==→⋅+
⋅=
⋅⋅−=⋅⋅−=
111
,
Method 2 – Taking limits:
21
Set 3: Single-Stage Amplifiers41SM
Source Degeneration - 7
Obtaining Gm and Rout directly assuming λ=γ=0:1. Gm:
Sm
m
IN
Dm
Sm
INGS
SGSmINGSGSmD
Rgg
viG
Rgvv
Rvgvvvgi
⋅+==→
⋅+⋅=
⋅⋅−=⋅=
111,
2. ROUT:
D
X
XOUT
D
XGSm
D
XX
GSSGSmGS
RivR
Rvvg
Rvi
vRvgv
==
=⋅+=
=→⋅⋅−= 0
Sm
DmOUTmv Rg
RgRGA⋅+⋅−
=⋅−=1
Set 3: Single-Stage Amplifiers42SM
Source Degeneration - 8
• If we ignore body effect and channel-length modulation:
Sm
DmvDOUT
Sm
mm Rg
RgARRRg
gG⋅+⋅−
=→=⋅+
=1
,1
• We Notice that as RS increases Gm becomes less dependent on gm:
SSm
mm RRg
gGsRsR
11
limlim =⋅+
=∞→∞→
• That is for large RS: OUTSIN
SIN
OUTm iRv
RviG ⋅≈→≈=
1
• Therefore, the amplifier becomes more linear when RS is large enough. Intuitively, an increase in vIN tend to increase ID, however, the voltage drop across RS also increases. This makes the amplifier less sensitive to input changes, and makes ID smoother!
• The linearization is achieved at the cost of losing gain and voltage headroom.
22
Set 3: Single-Stage Amplifiers43SM
Source Degeneration - 9
• We can manipulate the gain equation so the numerator is the resistance seen at the drain node, and the denominator is the resistance in the source path.
• The following are ID and gm of a transistor without RS.
S
m
D
Sm
Dmv
Rg
RRg
RgA+
−=
⋅+⋅−
= 11
• ID and gm of a transistor considering RS are:• When ID is small such that RS gm<<1, Gm ≈ gm.• When ID is large such that RS gm>>1, Gm ≈1/ RS.
Set 3: Single-Stage Amplifiers44SM
Alternative Method to Find the Output-Resistance of a Degenerated Common-Source Amplifier
23
Set 3: Single-Stage Amplifiers45SM
Why Buffers?
• Common Source amplifiers needed a large load impedance to provide a large gain.
• If the load is small but we need a large gain (can you think of an example?) a buffer is used.
• Source-follower (common-drain) amplifiers can be used as buffers.
Ideal Buffer:1. RIN=∞: the input current is zero; it doesn’t load the previous stage.
2. ROUT=0: No voltage drop at the output; behaves like a voltage source.
1,0, ==∞= VOUTIN ARR
Set 3: Single-Stage Amplifiers46SM
Resistive Load - 1
• We will examine the Source follower amplifier with two different loads:1. Resistive Load2. Current Source Load
– Resistive Load:• As shown below the output (source voltage) will follow the input (gate
voltage). We will analyze the following circuit using large-signal and small-signal analysis.
24
Set 3: Single-Stage Amplifiers47SM
Resistive Load - 2
Large Signal Analysis:• The relationship between VIN and VOUT is:
( ) ( )
( ) ( )SOUTDDTHOUTINoxnOUT
SDSTHGSoxnDSOUT
RVVVVVLWCV
RVVVLWCIRV
⋅⋅−⋅+⋅−−=
⋅⋅+⋅−=⋅=
λλµ
λµ
121
121
2
2
( )
IN
OUT
IN
OUT
SBFIN
OUT
OUT
TH
IN
TH
OUTSBFSBFTHTH
VV
VV
VVV
VV
VV
VVVVV
∂∂⋅=
∂∂⋅
+Φ⋅=
∂∂⋅
∂∂
=∂∂
=Φ⋅−+Φ⋅⋅+=
ηγ
γ
22
,220
( ) ( )
( ) ( )IN
OUTSTHGSoxn
SDS
IN
TH
IN
OUTTHOUTINoxn
IN
OUT
VVRVV
LWC
RVVV
VVVVV
LWC
VV
∂∂⋅−⋅⋅−+
⋅⋅+⋅⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−∂∂
−⋅−−=∂∂
λµ
λµ
2
21
11
• Differentiate with respect to VIN:
• Need to find the derivative of VTH with respect to VIN:
Set 3: Single-Stage Amplifiers48SM
Resistive Load - 3
( )
S
o
mbm
Sm
o
SSmbSm
Sm
IN
OUTV
SmSDSmSm
IN
OUT
Rr
gg
Rg
rRRgRg
RgVVA
RgRIRgRgVV
⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+++
⋅=
+⋅+⋅+
⋅=
∂∂
=
⋅=⋅⋅+⋅⋅+⋅+⋅∂∂
111
1 λη
Large Signal Analysis (Continued):• The small signal gain can be found:
• If channel-length modulation is ignored (ro=∞) we get:
( ) Smbm
Sm
IN
OUTV Rgg
RgVVA
⋅++⋅
=∂∂
=1
25
Set 3: Single-Stage Amplifiers49SM
Resistive Load - 4
Small Signal Analysis:• We get the following small signal model:
( )
( ) ( )( )
( )
( )
11
1
,,
+⎟⎟⎠
⎞⎜⎜⎝
⎛++⋅
⋅=
+⋅+⋅+⋅⋅⋅
=⋅⋅+⋅⋅++
⋅⋅==
⋅⋅⋅=⋅⋅+⋅⋅++⋅+⋅
⋅−⋅+−⋅=
−=−=⋅⋅+⋅=
mbm
O
S
Smv
OOmbOmS
OSm
OSmbOSmOS
OSm
IN
OUTv
INOSmOSmbOSmOSOUT
OS
OSOUTmbOUTINmOUT
OUTBSOUTINGSOSBSmbGSmOUT
ggr
R
RgA
rrgrgRrRg
rRgrRgrRrRg
vvA
vrRgrRgrRgrRvrRrRvgvvgv
vvvvvrRvgvgv
Set 3: Single-Stage Amplifiers50SM
Resistive Load - 5
• Graph of the gain of a source-follower amplifier:
1. M1 never enters the triode region as long as VIN<VDD.
2. Gain is zero if VIN is less than VTH (because gm is 0).
3. As VIN increases, gm increases and the gain becomes:
4. As VOUT increases, η decreases, and therefore, the maximum gain increases.
5. Even if RS=∞, the gain is less than one:
6. Gain depends heavily on the DC level of the input (nonlinear amplifier).
η+=
+≈
11
mbm
mv gg
gA
11 <++
≈
o
mbm
mV
rgg
gA
26
Set 3: Single-Stage Amplifiers51SM
Current Source Load
• In a source follower with a resistive load, the drain current depends on the DC level of VIN, which makes the amplifier highly nonlinear.
• To avoid this problem, we can use a current source as the load.
• The output resistance is:
2
11
11121
11
11
11,11o
mbm
oIMOUToI
mbm
oM rgg
rRRRrRgg
rR ==→==
• If channel length modulation is ignored (ro1=ro2=∞) :
111111
11111mbmmbmmbm
OUT ggggggR
+==∞∞=
• Note that the body effect reduces the output impedance of the source follower amplifiers.
Set 3: Single-Stage Amplifiers52SM
Voltage Division in Source Followers - 1
• When Calculating output resistance seen at the source of M1,i.e., RM1,, we force vIN to zero and find the output impedance:
11
11
11mbm
oM ggrR =
• However, if we were to find the gain of the amplifier, we would not suppress vIN.
• Here, we would like to find an equivalent circuit of M1, from which we can find the gain.
• Consider the small-signal model of M1:
27
Set 3: Single-Stage Amplifiers53SM
Voltage Division in Source Followers - 2
• For small-signal analysis vBS= vDS, so gmbvBS dependant current source can be replaced by a resistor (1/gmb) between source and drain.
• Note that, when looking at the circuit from the source terminal, we can replace the gmvGS dependant current source with a resistor (of value 1/gm) between source and gate.
• Simplified circuit:
Set 3: Single-Stage Amplifiers54SM
Voltage Division in Source Followers - 3
Example:• Find the gain of a source follower amplifier with a resistive
load.• We draw the small signal model of this amplifier as shown
below to get:
mmb
OS
mb
OS
IN
OUTvIN
mmb
OS
mb
OS
OUT
ggrR
grR
vvAv
ggrR
grR
v11
1
11
1
+==→⋅
+=
mOSmbOSOS
mOS
mmbOSOS
OS
mbOSOS
OS
mmb
OS
mb
OSv grRgrRrR
grR
ggrRrRrR
grRrRrR
ggrR
grRA
⋅⋅+⋅⋅++⋅⋅
=+
⋅⋅++⋅
⋅⋅++⋅
=+
++
++= 11
111
111
• We can show that this is equal to what we obtained before:
28
Set 3: Single-Stage Amplifiers55SM
Voltage Division in Source Followers - 4
Example:• Find the gain of a source follower amplifier with a current
source load.• Small-signal model of this amplifier is:
11
12
1
12
11
12
1
12
11
1
11
1
mmb
OO
mb
OO
IN
OUTvIN
mmb
OO
mb
OO
OUT
ggrr
grr
vvAv
ggrr
grr
v+
==→⋅+
=
• If we ignore channel length modulation:
11
1
11
1
11
1
11
1
mmb
mb
IN
OUTvIN
mmb
mbOUT
gg
gvvAv
gg
gv+
==→⋅+
=
Set 3: Single-Stage Amplifiers56SM
Voltage Division in Source Followers - 5
Example:• Find the gain of a source follower amplifier with a resistive
load and biased with a current source.
• Small-signal model of this amplifier is:
11
12
1
12
11
12
1
12
11
1
11
1
mmb
OLO
mb
OLO
IN
OUTvIN
mmb
OLO
mb
OLO
OUT
ggrRr
grRr
vvAv
ggrRr
grRr
v+
==→⋅
+
=
29
Set 3: Single-Stage Amplifiers57SM
Voltage Division in Source Followers - 6
Example:• Find the gain of a source follower amplifier with a resistive
load.
• Small-signal model of this amplifier is:
1221
12
221
12
1221
12
221
12
1111
111
1111
111
mmbmmb
OO
mbmmb
OO
IN
OUTvIN
mmbmmb
OO
mbmmb
OO
OUT
ggggrr
gggrr
vvAv
ggggrr
gggrr
v+
==→⋅+
=
Set 3: Single-Stage Amplifiers58SM
Advantages and Disadvantages - 1
1. Source followers have typically small output impedance.
2. Source followers have large input impedance.
3. Source followers have poor driving capabilities..
4. Source followers are nonlinear. This nonlinearity is caused by:Variable bias current which can be resolved if we use a current source to bias the source follower.Body effect; i.e., dependence of VTH on the source (output) voltage. This can be resolved for PMOS devices, because each PMOS transistor can have a separate n-well. However, because of low mobility, PMOS devices have higher output impedance. (In more advanced technologies, NMOS in a separate p-well, can be implemented that potentially has no body effect)Dependence of ro on VDS in submicron devices.
30
Set 3: Single-Stage Amplifiers59SM
Advantages and Disadvantages - 2
5. Source followers have voltage headroom limitations due to level shift.Consider this circuit (a common source followed by a source follower):
• If we only consider the common source stage, VX>VGS1-VTH1.
• If we only consider the source follower stage, VX>VGS3-VTH3+ VGS2.
• Therefore, adding the source follower will reduce the allowable voltage swing at node X.
• The DC value of VOUT is VGS2 lower than the DC value of VX.
Set 3: Single-Stage Amplifiers60SM
Common-Gate
Av = (gm + gmb )RD = gm(1 + η)RD
31
Set 3: Single-Stage Amplifiers61SM
Common-Gate
Av =(gm + gmb )ro +1
ro + (gm + gmb )roRS + RS + RD
RD
Set 3: Single-Stage Amplifiers62SM
Common-Gate
Av =(gm + gmb )ro +1
ro + (gm + gmb )roRS + RS + RD
RD
)||)((:0 DombmvS RrggAR +≈= for
32
Set 3: Single-Stage Amplifiers63SM
Common-Gate Input Impedance
ombm
o
ombm
D
ombm
oDin rgg
rrgg
Rrgg
rRR)(1)(1)(1 ++
+++
=+++
=
)1||1||()(1 mbm
oombm
Din gg
rrgg
RR +++
=
Set 3: Single-Stage Amplifiers64SM
Common-Gate Input Impedance
• Input impedance of common-gate stage is relatively low only if RD is small
• Example: Find the input impedance of the following circuit.
33
Set 3: Single-Stage Amplifiers65SM
Example
• Calculate the voltage gain of the following circuit:
ombmv rggA )(1 ++=
Set 3: Single-Stage Amplifiers66SM
Common-Gate Output Impedance
DSoSmbmout RRrRggR ||])(1[ +++=
34
Set 3: Single-Stage Amplifiers67SM
Example
• Compare the gain of the following two circuits (λ=γ=0 and 50Ω transmission
lines!)
Set 3: Single-Stage Amplifiers68SM
Cascode Stage
Doombm
Dooombmout
RrrggRrrrggR
||])[( ||])(1[
2122
12122
+≈+++=
AV ≈ gm1[ro1ro2 (gm2 + gmb2 )] || RD ]
• Cascade of a common-source stage and a common-gate stage is called a “cascode” stage.
35
Set 3: Single-Stage Amplifiers69SM
Cascode Stage
AV ≈ gm1[(ro1ro2 gm2 ) || (ro3ro4gm3 )]
Set 3: Single-Stage Amplifiers70SM
Output Impedance Comparison
36
Set 3: Single-Stage Amplifiers71SM
Shielding Property
Set 3: Single-Stage Amplifiers72SM
Board Notes
37
Set 3: Single-Stage Amplifiers73SM
Triple Cascode
• What is the output resistance of this circuit?
• Problem?
Set 3: Single-Stage Amplifiers74SM
Folded Cascode
38
Set 3: Single-Stage Amplifiers75SM
Output Impedance of a Folded Cascode
231222 )||]()(1[ oooombmout rrrrggR +++=