1

Single Stage Amplifiers

•Basic Concepts•Common Source Stage•Source Follower•Common Gate Stage•Cascode Stage

Hassan AboushadyUniversity of Paris VI

• B. Razavi, “Design of Analog CMOS Integrated Circuits”,McGraw-Hill, 2001.

References

H. Aboushady University of Paris VI

2

Single Stage Amplifiers

•Basic Concepts•Common Source Stage•Source Follower•Common Gate Stage•Cascode Stage

Hassan AboushadyUniversity of Paris VI

Basic Concepts I

H. Aboushady University of Paris VI

• Amplification is an essential function in most analog circuits !

• Why do we amplify a signal ?

• The signal is too small to drive a load• To overcome the noise of a subsequent stage• Amplification plays a critical role in feedback systems

In this lecture:• Low frequency behavior of single stage CMOS amplifiers:• Common Source, Common Gate, Source Follower, ...

• Large and small signal analysis.• We begin with a simple model and gradually add 2nd order effects

• Understand basic building blocks for more complex systems.

3

Approximation of a nonlinear system

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212

210 )(...)()()( xxxtxtxtxty nn ≤≤++++≈ αααα

)()( 10 txty αα +≈

Input-Output Characteristic of a nonlinear system

In a sufficiently narrow range:

where α0 can be consideredthe operating (bias) point andα1 the small signal gain

Analog Design Octagon

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4

Single Stage Amplifiers

•Basic Concepts•Common Source Stage•Source Follower•Common Gate Stage•Cascode Stage

Hassan AboushadyUniversity of Paris VI

Common Source Stage with Resistive Load

H. Aboushady University of Paris VI

2)(2 THinoxn

DDDout VVLWCRVV −−= µ

211 )(

2 THinoxn

DDDTHin VVLWCRVVV −−=− µ

⎥⎦

⎤⎢⎣

⎡−−−=

2)(

2out

outTHinoxnDDDoutVVVV

LWCRVV µ

DDDDout IRVV −=

M1 in the saturation region:

M1 in limit of saturation:

M1 in thelinear region:

5

Common Source Stage with Resistive Load

H. Aboushady University of Paris VI

)(1 THinDoxn

DD

Don

onDDout

VVRLWC

VRR

RVV−+

=+

=µ

M1 in deeplinear region:

Common Source Stage with Resistive Load

H. Aboushady University of Paris VI

Dm

THinoxnDin

outv

Rg

VVLWCR

VVA

−=

−−=∂∂= )(µ

2)(2 THinoxn

DDDout VVLWCRVV −−= µ

M1 in the saturation region:

Small signal gain:

Same relation can be derivedfrom the small signal equivalentcircuit

Small signal model forthe saturation region

To minimize nonlinearity, the gain equation must be a weakfunction of signal dependent parameters such as gm !

6

Example 1

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Sketch ID and gm of M1 as a function of the Vin:

• M1 in the saturation region:

2)(2 THinoxn

D VVLWCI −= µ

)( THinoxnGS

Dm VV

LWC

VIg −=

∂∂= µ

• M1 in the linear region:

⎥⎦

⎤⎢⎣

⎡−−=

2)(

2

1out

outTHinoxnDVVVV

LWCI µ

outoxnGS

Dm V

LWC

VIg µ=

∂∂=

)(1 THinDoxn

DDout

VVRLWC

VV−+

=µ

Voltage Gain of a Common Source Stage

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D

RDDoxnv IVI

LWCA µ2−=

Dmv RgA −=

How to increase Av ?

Trade-offs:

• Increase W/L Greater device capacitances.

• Increase VRD Limits Vout swing.

• Reduce ID Greater Time Constant.

D

RDoxnv I

VLWCA µ2−=

7

Taking Channel Length Modulation into account

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)1()(2

2outTHin

oxnDDDout VVV

LWCRVV λµ +−−=

in

outTHin

oxnD

outTHinoxnDin

outv

VVVV

LWCR

VVVLWCR

VVA

∂∂−−

+−−=∂∂=

λµ

λµ

2)(2

)1)((

DO

DOmv Rr

RrgA+

−=OD rI /1=λ

vDDmDv AIRgRA λ−−=

DD

mDv IR

gRAλ+

−=1

Calculating Av starting from the Large Signal Equations:

Taking Channel Length Modulation into account

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outDOm VRrVg −=)//(1

inVV =1

)//( DOmin

outv RrgVVA −==

Calculating Av starting from the Small Signal model:

8

Example 2

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Assuming M1 biased in saturation,calculate the small signal voltage gain :

• I1 : Ideal current source Infinite Impedance

Omv rgA −=

• Intrinsic gain of a transistor:This quantity represents the maximum voltagegain that can be achieved using a single device.

12

1 )1()(2

IVVVLWCI outTHin

oxnD =+−= λµ

• Constant Current:As Vin increases, Vout must decrease such that theproduct remains constant

CS Stage with Current-Source Load

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)//( 21 OOmv rrgA −=• Both transistors operate in the saturation region:

• The output impedance and the minimumrequired VDS of M2 are less strongly coupled thanthe value and voltage drop of a resistor.

22min,2 THGSDS VVV −=

• This value can be reduced to a few hundred millivolts bysimply increasing the width of M2.•If rO2 is not sufficiently high, the length and width of M2 can beincreased to achieve a smaller λ while maintaining the sameoverdrive voltage.•The penalty is the large capacitance introduced by M2 at theoutput node.•Increasing L2 while keeping W2 constant increases rO2 andhence the voltage gain, but at the cost of higher |VDS2|required to maintain M2 in saturation

9

CS with Source Degeneration

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in

GS

GS

D

in

Dm V

VVI

VIG

∂∂

∂∂=

∂∂=

SDinGS RIVV −=

Sin

D

in

GS RVI

VV

∂∂−=

∂∂ 1

( )mSmm

in

DS

GS

Dm

GRgGVIR

VIG

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂−

∂∂=

1

1

GS

Dm V

Ig∂∂=

Sm

mm Rg

gG+

=1 Dmv RGA −=

Sm

Dmv Rg

RgA+

−=1

Large Signal model: Small Signal model:

Sm

m

in

Dm RVgV

VgVIG

11

1

+==

Sm

mm Rg

gG+

=1

CS with Source Degeneration

H. Aboushady University of Paris VI

SmSm

mm RgRg

gG+

=+

=/1

11 mS gRfor /1>> Sm RG /1≈

ID is linearized at the cost of lower gain.

Small Signal model including body effectand channel length modulation:

O

SoutSoutmbSoutinm

O

XXmbmout

rRIRIgRIVg

rVVgVgI

−−+−=

−−=

)()(

1

OSmbmS

Om

in

outm rRggR

rgVIG

])(1[ +++==

10

With and Without Source Degeneration

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OSmbm

Omm rRgg

rgG])(1[1 +++

=

0=SR 0≠SR

Estimating Gain by Inspection

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Sm

D

Sm

Dmv Rg

RRgRgA

+−=

+−=

/11

Example:

PathSourcein the ResistanceTotalDrain at theseen Resistance−=Gain

21 /1/1 mm

Dv gg

RA+

−=

11

Output Resistance of Degenerated CS

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SX RIV −=1

XSmbmX

mbmX

IRggIVggI

)()( 1

++=+−

The current flowing in rO :

SXXSmbmXOX RIIRggIrV +++= ])([

SSmbmOX

Xout RRggr

IVR +++== ])(1[

OSOmbmout rRrggR +++= ])(1[

OSOmbmout rRrggR ++≈ )( OSmbmout rRggR ])(1[ ++=

Voltage Gain of Degenerated CS

H. Aboushady University of Paris VI

)( 1 BSmbmD

outr VgVg

RVI

O+−−=

])([D

Soutmb

D

Soutinm

D

outr R

RVgRRVVg

RVI

O++−−=

The current through RS must equal that through RD:

D

outRR R

VIISD

==VS

D

SoutS RRVV −=

The current through rO :

SD

outOrout R

RVrIV

O−=

D

SoutO

D

Soutmb

D

SoutinmO

D

outout R

RVrRRVg

RRVVgr

RVV −++−−= ])([

OSmbmOSD

DOm

in

out

rRggrRRRrg

VV

)( ++++−=

12

Voltage Gain of Degenerated CS

H. Aboushady University of Paris VI

OSmbmOSD

DOm

in

out

rRggrRRRrg

VV

)( ++++−=

OSmbmOSD

OSmbmOSD

OSmbmS

Om

in

out

rRggrRRrRggrRR

rRggRrg

VV

)(])([

])(1[ +++++++

+++−=

The output resistanceof a degenerated CS stage:

OSmbmout rRggR ])(1[ ++=

The Transconductanceof a degenerated CS stage:

( )Doutmin

out RRGVV //=

OSmbmS

Om

in

outm rRggR

rgVIG

])(1[ +++==

General expression to calculate Av by inspection

H. Aboushady University of Paris VI

Gm : the transconductance ofthe circuit when the output isshorted to grounded.

outmRGAv −=

Lemma:

Rout : the output resistanceof the circuit when the inputvoltage is set to zero.

• For high voltage gain the output resistance must be high!A “buffer” is needed to drive a low-impedance load.The source follower can operate as a voltage buffer.

13

Single Stage Amplifiers

•Basic Concepts•Common Source Stage•Source Follower•Common Gate Stage•Cascode Stage

Hassan AboushadyUniversity of Paris VI

Source Follower (Common Drain)

H. Aboushady University of Paris VI

STHoutinoxn

out RVVVLWCV 2)(

2−−= µ

Sin

TH

in

outTHoutinoxn

in

out RVV

VVVVV

LWC

VV )1)((

∂∂−

∂∂−−−=

∂∂ µ

M1 turns on in saturation:

SDout RIV =

To calculate gm :

in

out

in

SB

SBFin

SB

SB

TH

in

TH

VV

VV

VVV

VV

VV

∂∂η

∂∂γ

∂∂

∂∂

∂∂

=

+Φ==

22

( ) 220 FSBFTHTH VVV Φ−+Φ+= γSince,

Large Signal Behavior

14

Source Follower Voltage Gain

H. Aboushady University of Paris VI

)1()(1

)(

ηµ

µ

+−−+

−−=

∂∂

STHoutinoxn

STHoutinoxn

in

out

RVVVLWC

RVVVLWC

VV

)( THoutinoxnm VVVLWCg −−= µ

Smbm

Sm

RggRgAv

)(1 ++=

Sin

TH

in

outTHoutinoxn

in

out RVV

VVVVV

LWC

VV )1)((

∂∂−

∂∂−−−=

∂∂ µ

Sin

out

in

outTHoutinoxn

in

out RVV

VVVVV

LWC

VV )1)((

∂∂−

∂∂−−−=

∂∂ ηµ

We also have,

Source Follower Voltage Gain

H. Aboushady University of Paris VI

Smbm

Sm

in

out

RggRg

VVAv

)(1 ++==

Small Signal Equivalent Circuit

[ ][ ] Soutmboutinm

SBSmbmout

RVgVVgRVgVgV

−−=+=

)(1

mmb gg η=Since:

)1(1η+

≈Av

And for : 1>>SmRg

15

Source Follower Output Resistance

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0=−− XmbXmX VgVgImbmX

Xout ggI

VR+

== 1

Rout : the output resistance when the input voltage is set to zero.

Body Effect decreases the output resistance of source followers.

H. Aboushady

↑⇒

↓↑⇒↓

D

THGSX

IVandVV

XBS VVV −==1

Source Follower body effect

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mbmmbmout ggggR

+== 11//1

Small Signal Model Simplification

Note that the value of the current source gmbVbs is linearlyproportional to the voltage across it.

Rout : the output resistance when the input voltage is set to zero.

16

Source Follower Thévenin Equivalent

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mbm

m

mbm

mbv gg

g

gg

gA+

=+

= 11

1

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Channel Length Modulation in M1 and M2

mLOO

mb

LOOmb

gRrr

g

RrrgAv 1//////1

//////1

21

21

+=

17

Source Follower Characteristics

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PMOS source follower with VSB=0

- Nonlinearity

SBTH VV α

outnoutp

mnmpnp

RRgg

>⇒

<⇒< µµ

H. Aboushady

- Limited voltage swing

+ High input impedance and Moderate output impedance

Example:

11 THGSX VVV −>Without the source follower stage:

With the source follower stage:)( 332 THGSGSX VVVV −+>

Low Load Impedance: CS vs SF

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mL

LvSF gR

RA/1+

≈ LmvCS RgA −≈

2/1≈vSFA 1−≈vCSA

Source Followers are not necessarily efficient drivers.

Source Follower Amplifier Common Source Amplifier

Assuming RL=1/gm

18

Single Stage Amplifiers

•Basic Concepts•Common Source Stage•Source Follower•Common Gate Stage•Cascode Stage

Hassan AboushadyUniversity of Paris VI

Common Gate Stage

H. Aboushady University of Paris VI

DTHinboxn

DDout RVVVLWCVV 2)(

2−−−= µ

Din

THTHinboxn

in

out RVVVVV

LWC

VV

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂−−−−−=

∂∂ 1)(µ

( ) DTHinboxnin

out RVVVLWC

VV ηµ +−−=

∂∂ 1)(

Large Signal Behavior

Assuming M1 in saturation:

DDDDout RIVV −=

Dmv RgA )1( η+=

19

Common Gate Stage Input Resistance

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Body Effect:• increases Av• decreases Rin

Same as Output Resistance of Source Follower:

mbmin ggR

+= 1

Common Gate Gain

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01 =+− inSD

out VRRVV

outinSD

outmbm

D

outO VVR

RVVgVg

RVr =+−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−−

11

Small Signal Signal Equivalent Circuit

The current through RS is equal to -Vout / RD :

The current through rO is equal to -Vout / RD - gmV1 - gmbV1 :

outinSD

outin

D

Soutmbm

D

outO VVR

RVV

RRVgg

RVr =+−⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+−− )(

20

Common Gate Gain

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DSOmbmOSD

OmbmvCG R

RrggrRRrggA

)(1)(

++++++=

DSOmbmOSD

OmvCS R

RrggrRRrgA

)( ++++−=

Common Gate Amplifier:

Degenerated Common Source Amplifier:

Common Gate Stage Input Resistance

H. Aboushady University of Paris VI

Since V1 = -VX :

[ ]XmbmXOXDX VggIrIRV )( +−+=

)(1

)(

)(1

mbmOmbm

Din

Ombm

OD

X

X

ggrggRR

rggrR

IV

++

+≈

+++=

• Assume RD = 0 :

)(/11

mbmOin ggrR

++=

• Replace RD by ideal current source:

∞=inR

Rin of a common gate stage is low only if RD is small.

21

Common Gate Stage Output Impedance

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( ) DOSOmbmout RrRrggR //])(1[ +++=

Similar to Output Impedance of aDegenerated Common Source Stage

Single Stage Amplifiers

•Basic Concepts•Common Source Stage•Source Follower•Common Gate Stage•Cascode Stage

Hassan AboushadyUniversity of Paris VI

22

Biasing of a Cascode Stage

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The cascade of CS stage and a CG stage is called “cascode”.

M1 : the input deviceM2 : the cascode device

Biasing conditions:• M1 in saturation:

2GSbX VVV −=

12 THinGSb VVVV −≥−

2THXbXout VVVVV −−≥−

12 THGSinb VVVV −+≥

221 THGSTHinout VVVVV −+−≥

• M2 in saturation:

Cascode Stage Characteristics

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Large signal behavior:As Vin goes from zero to VDDFor Vin < VTH M1 and M2 are OFF

Vout =VDD

21222 ])(1[ OOOmbmout rrrggR +++=

Output Resistance:• Same common source stage witha degeneration resistor equal to rO1

1222 )( OOmbmout rrggR +≈

• M2 boosts the output impedance of M1by a factor of gmr02

• Triple cascode difficult biasing at low supply voltage.

↑↑outR

23

Cascode Stage Voltage Gain

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outmRGAv −=

1mm gG ≈

1222 )( OOmbmout rrggR +≈

11222 )( OmOmbmv rgrggA +≈

Ideal Current Source:

Cascode Current Source:

433122 // OOmOOmout rrgrrgR ≈

( )4331221 // OOmOOmmv rrgrrggA ≈

Shielding Property

H. Aboushady University of Paris VI

Assume VX is higher than VY by ∆V.Calculate the resulting difference between ID1 and ID2 (with λ ≠ 0 ).

)()(2 21

221 DSDSTHb

oxnDD VVVV

LWCII λλµ −−=−

)()(2

221 DSTHb

oxnDD VVV

LWCII ∆−=− λµ

33331333

1

)(])(1[ OmbmOOOmbm

OPQ rgg

Vrrrgg

rVV+∆≈

+++∆=∆

333

221 )(

)(2 Ombm

THboxn

DD rggVVV

LWCII

+∆−=− λµ

24

Folded Cascode

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Simple Folded Cascode

Folded Cascodewith biasing

Folded Cascodewith NMOS input

Large Signal Characteristics:

Output Resistance of Folded Cascode

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231222 )//]()(1[ OOOOmbmout rrrrggR +++=

1111 ])(1[ OSOmbmout rRrggR +++=

Degenerated Common Source Stage:

Folded Cascode Stage:

M1 M2

RS rO1 // rO3