Common-Source (CS) Common-Source (CS) Amplifier With R p

Chapter 3 p

Single-Stage Amplifiers

Analog Design Analog Design

Tradeoffs

N li

Non-linear system

Common-Source (CS) Common-Source (CS) Amplifier With R p

Load

Design Method Constraints and Design Method, Constraints and

Tradeoffs-L5

Simple CS Amplifier Simple CS Amplifier

• Given: k_n’^’,V_TH, λ

• Need: Specific voltage

i A R

gain A_v= -g_mR_D Design Parameters:

V I V W/L R V_DD,I_D,V_G,W/L,R_D Constraints:

S i M d

Saturation Mode

Current-Voltage relationship Swing

Common Source Large Signal Analysis Common Source – Large Signal Analysis

A_v= −g_mR_D

• If VIf V_iin is low (below threshold) transistor is OFFis low (below threshold) transistor is OFF

• For V_in not-too-much-above threshold, transistor is in Saturation and V decreases

is in Saturation, and V_out decreases.

• For large enough input (V_in>V_in1) – Triode Mode.

Common Source (cont ) Common Source (cont.)

A_v = −g_mR_D

V^RD A^v = − 2

μ

Design Tradeoffs Design Tradeoffs

A 2 C

V

A

= − 2 μ

C

I

• Gain is determined by 3 factors: W/L, RGain is determined by 3 factors: W/L, R_DD DCDC voltage and I_D

• If current and Rcu e t a d _DD are kept constant, an increase a e ept co sta t, a c ease of W/L increases the gain, but it also increases the gate capacitance – lower bandwidth

Design Tradeoffs Design Tradeoffs

A 2 C

V

A

= − 2 μ

C

I

• If IIf I_DD and W/L are kept constant, and we increaseand W/L are kept constant, and we increase R_D, then V_DS becomes smaller. Operating Point gets closer to the borderline of Triode Mode.

• It means – less “swing” (room for the amplified signal)

Design Tradeoffs Design Tradeoffs

A 2 C

V

A

= − 2 μ

C

I

• If I_DD decreases and W/L and V_RDRD are kept constant, then p , we must increase R_D

• Large Rg _DD consumes too much space, increases the p

noise level and slows the amplifier down (time constant with input capacitance of next stage).

Common Source Tradeoffs Common Source Tradeoffs

A

= −g

r

|| R

• Larger R_D values increase the influence of channel-length modulation (r term begins to channel-length modulation (r_o term begins to strongly affect the gain)

Common Source Maximum Gain Common Source Maximum Gain

A = −g r ^o

A_v = −g_m r^o || R_D

A _v = g _m r ^o

" intrinsic gain"

CS Amplifier with Diode-Connected Load-L6

• “Diode-Connected” MOSFET is only a name InDiode Connected MOSFET is only a name. In BJT if Base and Collector are short-circuited, then the BJT acts exactly as a diode

then the BJT acts exactly as a diode.

CS Amplifier with Diode-Connected Load

• We want to replace RWe want to replace R_DD with a MOSFET that willwith a MOSFET that will operate like a small-signal resistor.

V

X m o

X X

X g V

r I V

V V

1 1

1 = ⇒ = +

m o

m

d r g

R g 1

1 || ≈

=

⇒

Common Source with Diode-Connected NMOS Load (Body Effect included)

(g^m + g^mb)V^x + V^x

= I^x (g^m + g^mb)V^x +

r^o I^x V^x 1

|| r ≈ ¹ I^x =

g^m + g^{mb ||} r^o ≈

g^m + g^mb

CS with Diode-Connected NMOS Load Voltage Gain Calculation

A^v = −g^{m1 1} = − g^m1 1 A^v = g^m1

g^m2 + g^mb2 =

g^m2 1 +η

− μ

= 1

) / (

) / (

2 _{n OX 1 D1}

v

I L W

A C

η μ ( / ) 1+ 2 _nC_OX W L ₂ I_D2

A

= − (W / L)

(W / L)

1

1 +

(W / L)

1 + η

CS with Diode-Connected NMOS Load Voltage Gain Calculation

A

= − (W / L)

1 A

(W / L)

1 + η

• If variations of η with the output voltage are neglected, the gain is independent of the bias currents and voltages (as long as M stays in Saturation)

(as long as M₁ stays in Saturation)

• The point: Gain does not depend on V_in (as we so in the case of R_D load)

case of R_D load)

Amplifier is relatively linear (even for large signal analysis!)

W

W ⎞ ⎛ ⎞

⎛ 2

2 2

2 1

) (

5 . 0 )

( 5

.

0 _n^C _{in TH1 n}^C V_DD V_out V_TH

L V W

L V W

OX

OX ⎟ − −

⎠

⎜ ⎞

⎝

= ⎛

⎟ −

⎠

⎜ ⎞

⎝

⎛ μ

μ

⎞

⎛

⎞

⎛ ( ) ( ₂)

2 1

TH out

DD TH1

in V V V

L V W

L V

W ⎟ − −

⎠

⎜ ⎞

⎝

= ⎛

⎟ −

⎠

⎜ ⎞

⎝

⎛

Assumptions made along the way:

Assumptions made along the way:

) (

)

( ₂

2 1

TH out

DD TH1

in V V V

L V W

L V

W ⎟ − −

⎠

⎜ ⎞

⎝

= ⎛

⎟ −

⎠

⎜ ⎞

⎝

⎛

• We neglected channel-length modulation effectWe neglected channel length modulation effect

• We neglected V_TH voltage dependent variations

W d th t t t i t t hi i

• We assumed that two transistors are matching in terms of C_OX.

Comment about “cutting off”

Comment about cutting off

• If IIf I₁₁ is made smaller and smaller what happensis made smaller and smaller, what happens to V_out?

• It should be equal to V at the end of the

• It should be equal to V_DD at the end of the current reduction process, or is it?

Comment about “cutting off” (cont’d) Comment about cutting off (cont d)

• If IIf I₁₁ is made smaller and smaller, Vis made smaller and smaller, V_GSGS gets closergets closer and closer to V_TH.

• Very near Ie y ea ₁₁=0, if we neglect sub-threshold 0, e eg ect sub t es o d conduction, we should have V_GS≈V_TH2, and therefore V_out≈V_DD-V_TH2 !

Cutoff conflict resolved:

Cutoff conflict resolved:

• In reality, sub-threshold conduction, at a very low current, y, , y , eventually brings V_out to the value V_DD.

• Output node capacitance slows down this transition.p p

• In high-speed switching, sometimes indeed V_out doesn’t make it to V_DD

Back to large-signal behavior of the CS amplifier with diode-connected NMOS load:

) (

)

( ₂

2 1

TH out

DD TH1

in V V V

L V W

L V

W ⎟ − −

⎠

⎜ ⎞

⎝

= ⎛

⎟ −

⎠

⎜ ⎞

⎝

⎛

Large signal behavior of CS

amplifier with diode-connected load

• When VWhen V_inin<VV_TH1TH1 (but near), we have the above(but near), we have the above

“imperfect cutoff” effect.

• Then we have a, more or less, linear region.e e a e a, o e o ess, ea eg o

• When V_in exceeds V_out+V_TH1 amplifier enters the Triode mode, and becomes nonlinear.,

CS with Diode-Connected PMOS Load Voltage Gain

) /

( W L ₁ ) /

(

) /

(

L W

L A ^v _{= −} μ ⁿ W

) 2

/

( W L

μ p

No body effect!

Numerical Example Numerical Example

• Say that we want the voltage gain to be 10Say that we want the voltage gain to be 10.

Then:

) 10 /

( ₁ = −

−

= W L

A^v μⁿ

) 100 /

(

) 10 /

(

1

2

⇒ W L

L A W

n

p v

μ

μ

) 100 /

( ₂ =

⇒ _p W L

n

μ

Example continued Example continued

) 100 /

(

) /

(

=

L W

L

n

W

μ μ

) /

( W L

μ

• Typically Typically

2

p

n

μ

μ ≈ 2

Example continued Example continued

) /

( W L ) 50 /

(

) /

(

2 1

≈ L

W

L W

• Need a “strong” input device, and a “weak”

load device.

)

(

• Large dimension ratios lead to either a larger input capacitance (if we make input device

id (W/L) 1) t l t t

very wide (W/L)₁>>1) or to a larger output

capacitance (if we make the load device very narrow (W/L)₂<<1)

narrow (W/L)₂<<1).

• Latter option (narrow load) is preferred from bandwidth considerations.

CS with Diode-Connected Load Swing Issues

ID1 ID2

ID1 = ID2, ∴

2

2 ( )

)

( ^{GS1 TH1} W V^GS2 V^TH2 V

W V

⎟⎞

⎜⎛

⎟ ≈

⎜ ⎞

⎛ μ

μ

2 1

) (

)

( ^{GS1 TH1} _p ^{GS2 TH2}

n V V

V L

L V ⎟ −

⎜ ⎠

≈ ⎝

⎟ −

⎜ ⎠

⎝ μ

μ

) (

|

| )

/ (

) /

( ₁ ^{GS2 TH2}

v n

V V

V V

L W

L

A = − W ≈ −

μ μ

) (

) /

( ₂ ^{GS1 TH1}

p W L V −V

μ

This implies substantial voltage swing constraint. Why?

Numerical Example to illustrate the swing problems

A f i t V 3V

• Assume for instance V_DD=3V

• Let’s assume that V_GS1-V_TH1=200mV (arbitrary selection consistent with current selection)

selection, consistent with current selection)

• Assume also that |V_TH2|=0.7V (for PMOS load V_TH2=-0.7V)

V_TH2 0.7V)

• For a gain of 10, we now need |V_GS2|=2.7V (for PMOS load V_GS2GS2<-2.7V)

• Therefore because V_GD2=0: V_DS2=V_GS2=-2.7V).

• Now V_DS1DS1=V_DDDD-V_SD2SD2<3-2.7=0.3V, and recall that V_DS1>V_GS1-V_TH1=0.2V. Not much room left for the amplified signal v_ds1.

DC Q Point of the Amplifier DC Q-Point of the Amplifier

• VV_G1G1 determines the current and the voltage Vdetermines the current and the voltage V_GS2GS2

• If we neglect the effect of the transistors’ λ, the error in predicting the Q point solution may be error in predicting the Q-point solution may be large!

CS with Diode-Connected Load –How do we take into account λ?

)

||

1 ||

2

( o1 o

b2 2

m1

v r r

g g g

A = − +

mb2 m2 g g +

1 || || ) 1

2 ( 1

1 o o

m2

v r r

g g A = − _m

Example as Introduction to Current Source Load

• MM₁₁ is biased to be in Saturation and have ais biased to be in Saturation and have a current of I₁.

• A current source of Icu e t sou ce o _SS=0.75I0 5 ₁₁ is hooked up in s oo ed up

parallel to the load – does this addition ease up the amplifier’s swing problems?

Example as Introduction to Current Source Load

• Now INow I_D2D2=II₁₁/4 Therefore (from the ratio of the two/4. Therefore (from the ratio of the two transconductances with different currents):

) / (

4 W L

2 1

) / (

) / (

4

L W

L A W

p n

v

μ

− μ

≈

)

(

μ

Example: Swing Issues Example: Swing Issues

4I

I_D1 = 4I_D2, ∴ I

2

2 4 ( )

)

( ^{GS1 TH1} W V^GS2 V^TH2 V

W V

⎟⎞

⎜⎛

⎟ ≈

⎜ ⎞

⎛ μ

μ

2 1

) (

4 )

( ^{GS1 TH1} _p ^{GS2 TH2}

n V V

V L

L V ⎟ −

⎜ ⎠

≈ ⎝

⎟ −

⎜ ⎠

⎝ μ

μ

) (

|

|

4 GS1 TH1

TH2 A GS2

V V

V

v

V

−

≈ −

) ( V

V

It sure helps in terms of swing.

CS Amplifier with Current-Source Load-L7

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

How can V

change the current of M

if I

is constant?

1 2

1

1 0.5 (V V ) (1 V ) I

L

I_{D n}^C_OX W ⎟ _in − _TH1 + _out =

⎠

⎜ ⎞

⎝

= μ ⎛ λ

• As V_in increases, V_out must decrease

Simple Implementation: Current Source

obtained from M

in Saturation

CS with Current Source Load CS with Current Source Load

)

||

( r 1 r 2

g

A _v = − g _m ⁽ r _o1 || r _o2 )

A

DC Conditions DC Conditions

) 1

( ) (

5 .

0 ² _{1 2}

1

1 nC in TH1 out D

D V V V I

L

I _OX W ⎟ − + =

⎠

⎜ ⎞

⎝

= μ ⎛ λ

]) [

1 ( ) (

5 .

0 ₂ ² ₂

2

DD out

TH DD

C b

p V V V V V

L W

OX ⎟ − − + −

⎠

⎜ ⎞

⎝

= μ ⎛ λ

• {(W/L){(W/L)₁₁,VV_G1G1} and {(W/L)} and {(W/L)₂₂,VV_bb} need to be more or} need to be more or less consistent, if we wish to avoid too much

dependence on λ values dependence on λ values.

• Need DC feedback to fix better the DC V_out

Swing Considerations Swing Considerations

) 1

( ) (

5 .

0 ² _{1 2}

1

1 nC in TH1 out D

D V V V I

L

I _OX W ⎟ − + =

⎠

⎜ ⎞

⎝

= μ ⎛ λ

]) [

1 ( ) (

5 .

0 ₂ ² ₂

2

DD out

TH DD

C b

p V V V V V

L W

OX ⎟ − − + −

⎠

⎜ ⎞

⎝

= μ ⎛ λ

• We can make |VWe can make |V_DS2DS2|>|V|>|V_GS2GS2-VV_TH2TH2| small (say a| small (say a few hundreds of mV), if we compensate by making W₂ wider

making W₂ wider.

How do we make the gain large?

How do we make the gain large?

)

||

A

g

r

|| r

) A = −

• Recall: λ is inversely proportional to the channel length L.y p p g

• To make λ values smaller (so that r_o be larger) need to increase L. In order to keep the same current, need to p increase W by the same proportion as the L increase.

CS Amplifier with Current-Source Load Gains

T i l i th t h lifi

• Typical gains that such an amplifier can achieve are in the range of -10 to -100.

• To achieve similar gains with a R

load would require much larger V q g

values.

• For low-gain and high-frequency

applications R load may be preferred applications, R

load may be preferred because of its smaller parasitic

capacitance (compared to a MOSFET load)

capacitance (compared to a MOSFET load)

Numerical Example Numerical Example

• Let W/L for both transistors be W/L = 100µm /Let W/L for both transistors be W/L 100µm / 1.6µm

• Let µ C =90µA/V² µ C =30µA/V²

• Let µ_nC_ox=90µA/V², µ_pC_ox=30µA/V²

• Bias current is I_D=100µA

Numerical Example (Cont’d) Numerical Example (Cont d)

• Let rLet r_o11=8000L/I8000L/I_DD and rand r_o22=12000L/I12000L/I_DD where L is inwhere L is in µm and I_D is in mA.

• What is the gain of this stage?

• What is the gain of this stage?

Numerical Example (Cont’d) Numerical Example (Cont d)

128 1

0 / 6 1 8000

/ 06

. 1 )

/ (

2

1

1 1

Ω

=

⋅

=

=

= _{n OX D}

m

K r

V mA I

L W

C

g μ

4 81 )

||

(

192 1

. 0 / 6 . 1 12000

128 1

. 0 / 6 . 1 8000

2 1

Ω

=

⋅

=

Ω

o o

A

K r

K r

4 . 81 )

||

( _{1 2}

1 = −

−

= _{m o o}

V g r r

A

How does L influence the gain?

How does L influence the gain?

A = g r || r A_v = −g_m r_o1 || r_o2

Assuming r_o2 large,

W ⎟⎞ 1

⎜⎛

D D

ox n

o m

v L I

I W C

r g

A 2

μ λ

1

1 ⎟⎟⎟

⎠

⎞

⎜⎜

⎜

⎝

− ⎛

=

−

≈

How does L influence the gain?

How does L influence the gain?

D D

ox n o

m

v L I

I W C

r g A

1 1 1

2 1 μ λ

⎟⎟

⎟

⎠

⎞

⎜⎜

⎜

⎝

− ⎛

=

−

≈

A L i th i i b λ d d

• As L₁ increases the gain increases, because λ₁ depends on L₁ more strongly than g_m1 does!

• As I_DD increases the gain decreases.g

• Increasing L₂ while keeping W₂ constant increases r_o2 and the gain, but |V_DS2| necessary to keep M₂ in

Saturation increases Saturation increases.

CS with Triode Region Load CS with Triode Region Load

2 1 ON m

v g R

A = −

1

) (

1

2 2

2

b TH L DD

ox W p

ON C V V V

R = − +

⎟⎟⎠

⎜⎜ ⎞

⎝

μ

2

How should V be chosen?

How should V

be chosen?

• M₂ must conduct: V_b-V_DD≤V_TH2, or V_b≤V_DD+V_TH2

M must be in Triode Mode: V V ≤V V V or

• M₂ must be in Triode Mode: V_out-V_DD≤V_b-V_DD-V_TH2, or V_b≥V_out+V_TH2

• MM₂₂ must be “deep inside” Triode Mode: 2(Vmust be deep inside Triode Mode: 2(V_bb-VV_DDDD- V_TH2)>>V_out-V_DD, or: V_b>>V_DD/2+V_TH2+V_out/2

• Also V_b and (W/L)₂ determine the desired value of R_on2

It’s not easy at all to determine (and implement) a working value

for V for V

CS Amplifiers with Triode Region CS Amplifiers with Triode Region

load is rarely used

CS Amplifier with Source CS Amplifier with Source

Degeneration-L9 g

The effects of adding a resistor R

The effects of adding a resistor R

between Source and ground.

CS Amplifier with Source Degeneration

m

m g R

G g

≈ + 1

A = −G R_D

S mR + g

1

A_v G_mR_D

D mR A − g

Summary of key formulas, Neglecting

S m

D m

v g R

A g

≈ + 1

Neglecting

Channel-Length Modulation and Body Effect

Linearizing effect of R : Linearizing effect of R

:

A = −g_m R_D

= − R_D A_v =

1+ g_m R_S =

1/ g_m + R_S

Linearizing effect of R : Linearizing effect of R

:

A_v = −g_m R_D

1+ g_m R_S = − R_D

1/ g_m + R_S g_{m S} g_{m S}

• For low current levels 1/g >>RFor low current levels 1/g_m>>R_SS and thereforeand therefore G_m≈g_m.

• For very large V if transistor is still in

• For very large V_in, if transistor is still in Saturation, G_m approaches 1/R_S.

Estimating Gain by Inspection Estimating Gain by Inspection

R R

A_v = −g_m R_D

1+ g_m R_S = − R_D

1/ g_m + R_S

• Denominator: Resistance seen the Source path, “looking up” from

ground towards Source.

• Numerator: Resistance seen at Drain.

Example to demonstrate method:

Example to demonstrate method:

• Note that MNote that M₂₂ is “diode-connected” thus actingis diode connected , thus acting like a resistor 1/g_m2

• A = R /(1/g +1/g )

• A_V=-R_D/(1/g_m1+1/g_m2)

CS Amplifier with Source Degeneration Key Formulas with λ and Body-Effect Included

G_m = gg_m R_S

r +[1+ (g_m + g_mb)R_S] r_o

R = [1 + (g + g )r ]R + r

)

||

( R R G

A

R

= [1 + (g

+ g

)r

]R

+ r

)

||

(

m

v

G R R

A = −

Derivation is similar to that of the simplified case – generalized transconductance

It doesn’t matter what R_D is because we treat II _D as “input”

ro bs

mb m

D

R I I V

g V

g

I = ₁ + +

D D p

o S D X

mb m

R I r

R V I

g V

g ₁ − −

=

D

o S D S

D mb

S D in

m

r g I

r R R I

I g

R I V

g ( − )+ (− )−

=

o S mb m

S

o m in

D

m R g g R r

r g V

G I

] ) (

1

[ + +

= +

=

⇒

R effect on CS Output Resistance R

effect on CS Output Resistance

I V

g V

g

I = + +

ro S

X mb S

X m

ro bs

mb m

X

I R

I g

R I

g

I V

g V

g I

+

−

−

=

+ +

=

) (

1

R I

I R g

g I

r

V_X = r_o (I_X + (g_m + g_mb)R_S I_X ) + I_X R_S V = ( + ( + ) ) +

CS Output Resistance CS Output Resistance

R [1 ( ) ]R

R_OUT = [1 + (g_m + g_mb)r_o]R_S + r_o

R_OUT = r_o' ≈ r_o[1+ (g_m + g_mb)R_S]

R i ifi t i i th

R_S causes a significant increase in the output resistance of the amplifier

CS Amplifier with Source Degeneration Gain Formula with λ and Body-Effect Included

G g_m

G_m = g_m R_S

r +[1+ (g_m + g_mb)R_S]

R

= [1 + (g

+ g

)r

]R

+ r

r_o

)

||

(

m

v

G R R

A = −

Source Follower

Main use: Voltage Buffer Main use: Voltage Buffer

• To achieve a high voltage gain with limited supply voltage, in a CS amplifier, the load impedance must be as large as possible.

• If such a stage is to drive a low impedance load, g then a “buffer” must be placed after the amplifier so as to drive the load with negligible loss of the signal level.

• The source follower (also called the “common-The source follower (also called the common drain” stage) can operate as a voltage buffer.

Buffering Action Buffering Action

• Input resistance of source follower is large.

• CS amplifier connected to a Source CS amplifier, connected to a Source Follower, will see as a load R

of the Source Follower

Source Follower.

• R

of the Source Follower is unaffected by R

of the Source Follower. Variations in R

has no effect on R

of the MOSFET.

has no effect on R

of the MOSFET.

Source Follower with R resistance Source Follower with R

resistance

• Source Follower: Input signal comes into theSource Follower: Input signal comes into the Gate; Output signal comes out of the Source.

• Load connected between Source and ground

• Load connected between Source and ground.

Source Follower with R

resistance: Large Signal Behavior

• If VIf V_iin<V<V_THTH MM₁₁ is offis off.

• As V_in exceed V_TH M₁ is in Saturation.

M i t T i d M d l h V

• M₁ goes into Triode Mode only when V_in exceeds V_DD.

Why V follows V ? Why V

follows V

?

• Source followers exhibit a Body Effect: As ISource followers exhibit a Body Effect: As I_DD

increases, V_S=I_DR_S increases. As V_SB increases, V_TH increases.

Why V follows V ? (Cont’d) Why V

follows V

? (Cont d)

• If V_inin slightly increases, Ig y , _DD slightly increases and therefore g y V_out slightly increases.

• As I_DD increases V_THTH increases due to Body Effect.y

• FACT: V_GS increases but not at the same rate that V_in increases.

Source Follower Gain

Source Follower Gain

Source Follower Gain

Source Follower Gain

Gain Dependence on V Gain Dependence on V

• When VWhen V_GG is slightly above Vis slightly above V_THTH, gg is very small_m is very small, and therefore A_V is small.

• When g becomes large enough (i e g R >>1)

• When g_m becomes large enough (i.e. g_mR_S>>1), then A_V approaches 1/(1+η).

Gain Dependence on V (cont’d) Gain Dependence on V

(cont d)

• Recall η=γ/(2(2ΦRecall η γ/(2(2Φ_FF+V+V_SBSB)) ) γ/(2(2Φ^1/2)= γ/(2(2Φ_FF+V+V_outt)) )^1/2)

• As V_G increases, and V_out increases, η

decreases and the gain may approach 1 decreases, and the gain may approach 1.

• In most practical circuits η remains >0.2.

Source Follower with Current Source Load

• Left: Conceptual diagramLeft: Conceptual diagram

• Right: Actual implementation, using a NMOS operating in Saturation Mode

operating in Saturation Mode.

Example Example

• Let (W/L)₁=20/0.5, I₁=200µA, V_THO=0.6V, 2Φ_F=0.7V, µ_nC_OX=50µA/V², γ=0.4V^1/2

2Φ_F 0.7V, µ_nC_OX 50µA/V , γ 0.4V

• Let V_in=1.2V, what is V_out?

Output Resistance of the Ideal Source Follower with Current Source Load

V V

I

g

V

g

V

0

I − − =

1

⇒ 0

−

= V

V

mb m

out

g g

R = + 1

X 1

mb

m

g

g

Output Resistance of the Ideal Source Follower with Current Source Load becomes smaller with with Current Source Load becomes smaller with

the help of the Body Effect!

• Only in a Source Follower the current source

g_mbV_bs is equivalent to a resistor 1/g_mb in parallel g_mbV_bs is equivalent to a resistor 1/g_mb in parallel to the output.

Gain of Source Follower with Ideal Current Source Load

m V

A = g

mb m

V g + g

Gain Formula: NMOS Source Follower with NMOS Current Source and R_L Loads

||

||

1 ||

2 R

1 1

1 1

||

||

|| _{o o L}

mb v

R r

g r A =

1 2

1 1

) 1

||

||

1 ||

(

m L

o o

mb r r R g

g +

Gain Formula: NMOS Source Follower with PMOS Current Source Load

|| 1

||

1 ||

2 2

2 1

1

1 1

1

||

||

||

mb m

o o

mb v

g r g

g r

A = +

1 2

2 2

1 1

) 1

|| 1

||

1 ||

(

m mb

m o

o

mb r r g g g

g +

+

Sources of Nonlinearities in NMOS Source Followers

B d Eff t i th d i i NMOS t i t

• Body Effect in the driving NMOS transistor causes V_TH to vary with V_in

A ll d t t b t t t i

• Are we allowed to connect substrate to source in the driving NMOS? (to eliminate the body effect).

Answer: No! All NMOS transistors in the entire Answer: No! All NMOS transistors in the entire circuit share the same substrate, so it has to be grounded!

grounded!

• r_o resistors vary with V_DS. Problem becomes more and more aggravated as L becomes more and more aggravated as L becomes smaller and smaller

PMOS Source Follower PMOS Source Follower

• Key idea: PMOS transistors have each a y separate substrate. Each can be powered differentlyy

CMOS fabrication process: All NMOS share p the same substrate, each PMOS has a

separate substrate

separate substrate

PMOS Source Follower Advantage PMOS Source Follower Advantage

• Body Effects eliminated – device is more y linear than NMOS Source Follower