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(1)

Chapter 6

Frequency Response of Frequency Response of

Amplifiers

Amplifiers

(2)

MOSFET Capacitances-L22

MOSFET Capacitances L22

(3)

Device Capacitances Device Capacitances

There is some capacitance between every There is some capacitance between every

pair of MOSFET terminals.

Only exception: We neglect the

capacitance between Source and Drain.

(4)

Gate to Substrate capacitance Gate to Substrate capacitance

C1 is the oxide capacitance (between Gate and channel)

C1 = WLCOX

C2 is the depletion capacitance between channel and Substrate

C2 = WL(qεsiNsub/(4ΦF))1/2 = Cd

(5)

Gate-Drain and Gate-Source Overlap Capacitance

C and C are due to overlap between the gate C3 and C4 are due to overlap between the gate poly-silicon and the Source and Drain regions.

No simple formulas for C3 and C4: It is incorrect to write C3=C4=WLDCOX because of fringing electric field lines.

We denote the overlap capacitance per unit width We denote the overlap capacitance per unit width as Cov

(6)

Source-Substrate and Drain-Substrate Junction Capacitances

A i i l f l f C d C

Again, no simple formulas for C5 and C6:

See right figure: Each capacitance should be decomposed into two components:

Bottom-plate capacitance, denoted as Cjj, which depends on the junction area.

Side-wall capacitance Cp jswjsw which depends on the p perimeter of the junction.

(7)

Source-Substrate and Drain-Substrate Junction Capacitances

I MOSFET d l C d C ll i

In MOSFET models Cj and Cjsw are usually given as capacitance-per-unit-area, and capacitance-per-unit- length respectively

length, respectively.

Cj = Cj0 / [1+VRB]m where VR is the junction’s reverse voltage Φ is the junction built in potential and m

voltage, ΦB is the junction built-in potential, and m typically ranges from 0.3 to 0.4

(8)

Examples (both transistors have the same C

j

and C

jsw

parameters)

Calculate C

DB

and C

SB

for each structure

Calculate C

DB

and C

SB

for each structure

(9)

Left structure Left structure

C =C = WEC + 2(W+E)C CDB=CSB = WECj + 2(W+E)Cjsw

(10)

Layout for Low Capacitance (see folded structure on the right)

Two parallel transistors with one Two parallel transistors with one common Drain

(11)

Layout for Low Capacitance (see folded structure on the right)

CDB=(W/2)ECj+2((W/2)+E)Cjsw

C =2((W/2)EC +2((W/2)+E)C = CSB=2((W/2)ECj+2((W/2)+E)Cjsw=

= WECjj +2(W+2E)Cjswjsw

Compare to the left structure:

CDB=CSB = WECj + 2(W+E)Cjsw

CSB slightly larger, CDB a lot smaller. Both transistors have same W/L

(12)

C C at Cutoff C

GS

,C

GD

at Cutoff

C

GD

=C

GS

=C

ov

W

(13)

C at Cutoff C

GB

at Cutoff

C1 = WLCOX,C2= WL(qεsiNsub/(4ΦF))1/2 =Cd CGB=C1C2/(C1+C2)= WLCOXCd/(WLCOX+Cd ) L is the effective length of the channel.

(14)

C and C at all modes C

DB

and C

SB

at all modes

Both depend on the reverse voltages Drain to Substrate and Source to Substrate

Recall Cj = Cj0 / [1+VRB]m

(15)

C C at Triode Mode C

GS

,C

GD

at Triode Mode

If VDS is small enough, we can assume

VGS≈VGD Æ Channel uniform Æ Gate-Channel VGS VGD Æ Channel uniform Æ Gate Channel capacitance WLCOX is distributed uniformly

C

GD

=C

GS

≈ (

WLCOX)/2+

C

ov

W

(16)

C C at Saturation Mode C

GS

,C

GD

at Saturation Mode

C

GD

≈C

ov

W because there isn’t much of a channel near Drain

channel near Drain.

(It can be proved):

(It can be proved):

C

GS

=2WL

eff

C

OX

/3+WC

ov

(17)

C

GBG

at Triode and Saturation Modes

CGB is negligible, because channel acts as a

“shield” between Gate and Substrate. That is, if VG varies, change of charge comes from Source and Drain, and not from the Substrate!

(18)

MOS Small Signal Model with

Capacitance

(19)

C V of NMOS C-V of NMOS

I “W k I i ” C i i

In “Weak Inversion” CGS is a series

combination of COX-capacitance and Cdep-

it Æ L it k

capacitance Æ Low capacitance peak near VGS≈VTH

(20)

High-Frequency Response of High-Frequency Response of

Amplifiers-L23 p

Basic Concepts

Basic Concepts

(21)

Miller’s Theorem Miller s Theorem

Replacement of a bridging component (typically a capacitor) with equivalent input and output capac to ) t equ a e t put a d output impedances, for the sake of facilitating high-

frequency response analysis.

(22)

Miller’s Theorem Statement Miller s Theorem Statement

• Let ALet AVV = V VYY/V/VXX be knownbe known.

• Then (a) and (b) above are equivalent if:

Z Z/(1 A ) d Z Z/(1 A 1)

• Z1=Z/(1-AV) and Z2=Z/(1-AV-1)

• Equivalence in VX,VY and currents to Z’s.

(23)

Proof of Miller’s Theorem Proof of Miller s Theorem

• Current equivalence: (VCurrent equivalence: (VXX-VVYY)/Z=V)/Z VXX/Z/Z11

• That is: Z1=Z/(1-VY/VX) etc.

Si il l (V V )/Z V /Z Æ Z Z/ (1 V /V )

• Similarly: (VY –VX)/Z=VY/Z2 Æ Z2=Z/ (1-VX/VY)

• A simple-looking result, but what does it mean?

(24)

The most common application The most common application

• Mid-band gain of amplifier is known to be –AMid band gain of amplifier is known to be A.

• We want to know (at least approximately) how does the bridging capacitor C influence the

does the bridging capacitor CF influence the amplifier’s bandwidth.

(25)

Exact vs Approximate Analysis Exact vs. Approximate Analysis

• For exact analysis, replace Cy , p FF with its impedance p

ZC=1/sCF and find transfer function VY(s)/VX(s) of circuit.

• Approximate analysis: Use Miller’s theorem to break

capacitor into t o capacitances and then st d the inp t capacitor into two capacitances, and then study the input circuit’s time constant and the output circuit’s time

constant.

(26)

Miller Effect input capacitance Miller Effect – input capacitance

• Input circuit’s added capacitance due to CInput circuit s added capacitance due to CFF::

Since Z1 is smaller than Z by (1-AV), and since CFF appears in denominator of Z, it means that input capacitance C1=CF(1-AV). If |AV| is large, C1 may become pretty large.

(27)

Miller Effect output capacitance Miller Effect – output capacitance

• Output circuit’s added capacitance due to COutput circuit s added capacitance due to CFF:: Since Z2 is smaller than Z by (1-AV-1), and since CFF appears in denominator of Z, it means that

output capacitance C2=CF(1-AV-1). If |AV| is large, C2 will be almost equal to CF .

(28)

Bridging component restriction Bridging component restriction

• Both sides of the bridging component must be fully floating, for Miller’s theorem to be valid.

• If voltage, on either side, if fixed then there is no Miller effect.

• Later we’ll use Miller’s theorem to deal with CGD. Therefore in CG and Source Follower amplifiers, Therefore in CG and Source Follower amplifiers, in which one side of CGD is ground, there will be no Miller effect.

no Miller effect.

• Main impact of Miller is on CS amplifiers.

(29)

Meaning of “approximated analysis”

Meaning of approximated analysis

Th i littl f th l t hi h

• There is little use for the complete high- frequency transfer function (all its poles and zeros).

• We are interested in the amplifier only at p y mid-band frequencies (at which all

MOSFET capacitances are considered OS capac ta ces a e co s de ed open circuit), and may be a little bit

beyond, near high-frequency cutoff (-3db)

beyond, near high frequency cutoff ( 3db)

frequency.

(30)

Approximate Analysis Meaning Approximate Analysis Meaning

• We calculate input circuit’s and output circuit’sWe calculate input circuit s and output circuit s time constant for the sake of determining the amplifier’s bandwidth.

• We use the known mid-band voltage gain in Miller’s theorem.

(31)

Validity of Miller’s analysis at HF Validity of Miller s analysis at HF

• We use the known mid-band voltage gain inWe use the known mid band voltage gain in Miller’s theorem.

• At high frequencies, much higher than the -3db t g eque c es, uc g e t a t e 3db frequency, gain is much lower Æ Miller’s

theorem, using mid-band gain, is not valid!

(32)

Bandwidth Calculation Bandwidth Calculation

• Typically, the two time constants are rarely of the same order of magnitude.g

• The larger time constant determines the amplifier’s -3db frequency

amplifier s 3db frequency.

• The smaller time constant (if much smaller than other) should be discarded as invalid

other) should be discarded as invalid.

(33)

Poles and Zeros Poles and Zeros

• A time constant τ = 1/RC means that the transferA time constant τ 1/RC means that the transfer function has a pole at s=-1/τ.

• Typically every capacitor contributes one pole to

• Typically, every capacitor contributes one pole to the transfer function.

(34)

Example below: All poles are real;

No zeros

)

(s A A

V

) 1

)(

1 )(

1 ( )

( ) (

2 1

2 1

P N

in S in

out

C sR C

sR C

sR

A A s

V

s V

+ +

= +

(35)

Complicated example Complicated example

• There are three poles due to the threeThere are three poles due to the three

capacitors. It’s hard to determine poles location by inspection (some poles may be complex)

• There may be some zeros.

• Difficult due to Vinin,V, outout interaction (via R( 33,C, 33) )

(36)

Example: Feedback capacitor over finite gain

• VVoutt(s)/V(s)/VXX(s)= -A(s) A

• (Vin(s)-VX(s))/RS=sCF(VX(s)-Vout(s))

E t R lt V ( )/V ( ) A/[1 R C (1 A)]

• Exact Result: Vout(s)/Vin(s)=-A/[1+sRSCF(1+A)]

(37)

Example: CG Amplifier Example: CG Amplifier

• CCSS=CCGS1GS1+C+CSB1SB1 Capacitance “seen” from SourceCapacitance seen from Source to ground.

• C =C +C Capacitance “seen” from Drain to

• CD=CDG+CDB Capacitance seen from Drain to ground.

(38)

Example: CG Amplifier Example: CG Amplifier

• Need to determine (by inspection) the two polesNeed to determine (by inspection) the two poles contributed, one by CS and another by CD.

• This is easy to do only if we neglect rs s easy to do o y e eg ect 0101.

• Strategy: Find equivalent resistance “seen” by each capacitor, when sources are nulled.p ,

(39)

Example: CG Amplifier Example: CG Amplifier

• ττSS=CCSSRRSS,eq=CCSS{R{RSS||[1/(g||[1/(gm11+g+gmb1b1)])]

• τD=CDRD,eq=CDRD

A ( + )R /(1+( + )R )

• A=(gm1+gmb1)RD/(1+(gm1+gmb1)RS)

• Vout(s)/Vin(s)=A/[(1+sτS)(1+sτD)]

(40)

High-Frequency Response- High-Frequency Response-

L24

Common-Source Amplifier

Common Source Amplifier

(41)

High Frequency Model of CS Amplifier

• CGD creates a Miller Effect.

• It is essential to assume a nonzero signal g

source resistance RS, otherwise signal voltage changes appear directly across Cg pp y GSGS

(42)

C Miller Effect on input C

GD

Miller Effect on input

L A b h l f i l i f

• Let AV be the low frequencies voltage gain of the CS amplifier, such as AV≈-gmRD.

C i dd d C i C (1 A )

• Capacitance added to CGS is CGD(1-AV).

• τin=RS[CGS+CGD(1-AV)]

(43)

C Miller Effect on output C

GD

Miller Effect on output

C i dd d C i C (1 A 1) C

• Capacitance added to CDB is CGD(1-AV-1)≈CGD.

• τout = RD(CGD+CDB)

• If τin and τout are much different, then the largest one is valid and the smaller one is invalid.

(44)

Common Source (must be in Saturation Mode)

Neglecting input/output interaction,

fp,in = 1

2πRSS

[ [

CGSGS + (1+ g( gmmRDD )C) GDGD

] ]

fp,out = 11

2π

[ (

CGD + CDB

)

RD

]

(45)

CS Amplifier’s Bandwidth if R

S S

large

( ) ( )

• τout = RD(CGD+CDB) ; τin=RS[CGS+CGD(1-AV)] ; AV=-gmRD

• If all transistor capacitances are of the same order of

magnitude and if R is of the same order of magnitude of R magnitude, and if RS is of the same order of magnitude of RD (or larger) then AV(s)≈ -gmRD/(1+sτin) : Input capacitance dominates. BW=fh=1/2πτin

(46)

CS Amplifier’s Bandwidth if V

in

is almost a perfect voltage source

• τout = RD(CGD+CDB) ; τin=RS[CGS+CGD(1-AV)] ; AV=-gmRD

• If all transistor capacitances are of the same order of

i d d if R i ll A ( ) R /(1 )

magnitude, and if RS is very small : AV(s)≈-gmRD/(1+sτout) : Output capacitance dominates. BW=fh=1/2πτout

(47)

High Frequency Response of CS Amplifier – “Exact” Analysis

• Use small-signal diagram, and replace every capacitor by its impedance 1/sC ; Neglect ro

• Result: Only two poles, and one zero in Voutout(s)/V( ) inin(s)!( )

(48)

CS HF Response: Exact Result CS HF Response: Exact Result

) (

)

( GD m D

out s sC g R

V

s

⎟ s

s2 s

[ (1 ) ( )] 1

) (

) (

) (

) (

2 + + + + + + + +

=

DB GD

D GS

S GD

D m S

DB GD SB

GS GD

GS D

S

D m GD

in out

C C

R C

R C

R g R

s C

C C

C C

C R R s

g s

V

Assume D = s

ωp1 +1

s

ωp2 +1

⎟ = s

ω p1ωp 2 + s

ωp1 +1 , ωp2 >>ωp1

fp,in = 1

2π

( (

RS

[ [

CGS + (1+ gmRD)CGD

] ]

+ RD(CGD + CDB )

) )

(49)

CS Exact Analysis (cont ) CS Exact Analysis (cont.)

RS (1+ g RD )CGD + RSCGS + RD (CGD + CDB ) f p,out = RS (1+ gmRD )CGD + RSCGS + RD (CGD + CDB )

2πRS RD(CGSCGD + CGSCDB + CGDCDB)

fp,out 1

2πR (C + C ) , for large CGS 2πRD (CGD + CDB )

C R R g

DB GD

DB GS

GD GS

D S

GD D

S m out

p

g

C C

C C C

C R

R

C R R f g

) (

, 2 + +

π

GD DB

GS

m C

C C

g , for large )

( 2

+ π

(50)

CS Amplifier RHP Zero CS Amplifier RHP Zero

Right half plane zero, from the numerator of Vout(s)/Vin(s)

[ (1 ) ( ] 1

) (

) (

) (

) (

2 + + + + + + + +

= GD m D

out

C C

R C

R C

R g R

s C

C C

C C

C R R s

R g sC

s V

s V

C

[ (1 ) ( ] 1

) (

)

( S D GS GD + GS SB + GD DB + S + m D GD + S GS + D GD + DB) +

in s s R R C C C C C C s R g R C R C R C C

V

sCGD − gm

. . . → fz = +gm 2πCGD

(51)

Zero Creation Zero Creation

• A Zero indicates a frequency at which outputA Zero indicates a frequency at which output becomes zero.

• The two paths from input to output may create e t o pat s o put to output ay c eate signals that perfectly cancel one another at one specific frequency.

(52)

Zero Calculation Zero Calculation

• At frequency at which output is zero (s=sAt frequency at which output is zero (s szz) , the) , the currents through CGD and M1 are equal and

opposite. Can assume that at this frequency (only) output node is shorted to ground.

• V1/(1/szCGD)=gmV1 Æ Zero expression results.

(53)

RHP Zero Effect on the Frequency Response

• RHP Zero (like LHP zero) contributes positivelyRHP Zero (like LHP zero) contributes positively to the slope of magnitude frequency response curve Æ reduces the roll-off slope.

• RHP Zero (contrary to LHP zero) adds NEGATIVE phase shift.

(54)

RHP Zero Impact RHP Zero Impact

• It is often negligible, as it happens in very high g g , pp y g frequencies.

• It becomes important if CS amplifier is part of multi-stage p p p g amplifiers creating a very large open-loop gain. It may affect stabilization compensation.

(55)

Input Impedance of CS stage

Input Impedance of CS stage

(56)

High-Frequency Response- High-Frequency Response-

L25

Source Follower

Source Follower

(57)

High-Frequency Response of Source Follower – No Miller

• No Miller Effect – CNo Miller Effect CGDGD is not a bridge capacitoris not a bridge capacitor (as Drain side is grounded).

• C is a combination of several capacitances:

• CL is a combination of several capacitances:

CSB1, CDB,SS, CGD,SS and Cin of next stage.

(58)

High-Frequency Response of Source Follower – No solution “by inspection”

• Because of CBecause of CGSGS that ties input to output it isthat ties input to output, it is impossible to find various time constants, one capacitor at a time – combined effects of CGS capacitor at a time combined effects of CGS and CL.

(59)

Source Follower HF Response Derivation Assumptions

• Use small-signal diagram and 1/sC termsUse small signal diagram and 1/sC terms.

• Simplifying assumptions: We neglect effects of ro and γ (body effect)

and γ (body effect).

(60)

Source Follower HF Response Derivation

• Sum currents at output node (all functions of s):Sum currents at output node (all functions of s):

• V1CGSs+gmV1=VoutCLs , yielding:

V [C /( C )]V

• V1=[CLs/(gm+CGSs)]Vout

• KVL: Vin=RS[V1CGSs+(V1+Vout)CGDs]+V1+Vout

(61)

Source Follower Transfer Function

(

S GD GD GS

)

L GD GD

GS L

GS S

GS m

i out

g C

C C

R g s C

C C

C C

C R

s

sC g

s V

s V

+ +

+ +

+ +

= +

) (

) (

) (

2 S GS L GS GD GD L

(

m S GD GD GS

)

m

in s s R C C C C C C s g R C C C g

V ( ) ( + + )+ + + +

(62)

Idea for detecting “dominant pole”:

Idea for detecting dominant pole :

1 1

1

1 1

) (

) 1

)(

1

( 1 2 1 2 2 1 21 2 2 + 1 +

+ +

= + +

+ s

τ

s

τ τ τ

s

τ τ

s

τ τ

s

τ

s

If Æ C fi d b i i f

If τ

1

>> τ

2

Æ Can find τ

1

by inspection of

the s coefficient

(63)

Source Follower Dominant Pole (if it exists…)

( )

GS m

out s g sC

V ( ) = +

2 S GS L GS GD GD L ( m S GD GD GS) m

in s s R C C C C C C s g R C C C g

V ( ) 2 ( + + )+ + + +

fp1 gm

2π(g R C + C + C ) , assuming fp2 >> fp1

p 2π(gmRSCGD + CL + CGS) p p

= 1

C + C

2π RSCGD + CL + CGS gm

(64)

Source Follower Dominant Pole (if R

S

is very small)

( )

GS m

out s g sC

V ( ) = +

2 S GS L GS GD GD L ( m S GD GD GS) m

in s s R C C C C C C s g R C C C g

V ( ) 2 ( + + )+ + + +

( ) 1

2

1

1 2

⎟⎟

⎜⎜

+

+ =

GS GS L

L m p

g C C C

C f g

π π

gm

(65)

Source Follower Input Impedance Source Follower Input Impedance

• At low frequencies we take R

in

of CS,

Source Follower, Cascode and Differential , amplifiers to be practically infinite.

• At higher frequencies we need to study

• At higher frequencies, we need to study the Input Impedance of the various

f

amplifiers.

• Specifically for a Source Follower: Is Z Specifically for a Source Follower: Is Z

inin

purely capacitive?

(66)

Source Follower Input Impedance Derivation

C

Laying CGD aside, as it appears in parallel to the

i i f Z

remaining part of Zin:

1 ) 1 ||

( g C s sC

I I g

sC

VX IX X m X ⎟⎟

⎜⎜

+

+

sCGS sCGS gmb CLs

(67)

Source Follower Input Impedance

Neglecting C

Neglecting CGD , Zin 1

+ 1+ gm

1

in sCGS sCGS

gmb + sCL

| sC

>>|

g s, frequencie low

At mb L

(

1 ( / )

)

1/

1 g g g

Z sC m mb mb

GS

in + +

) (

)

/(g g C same as Miller g

C

Cin = GS mb m + mb + GD

(68)

“Miller Effect” of C Miller Effect of C

GS

Low frequency gain is gmm/(gmm+gmbmb)

CG,Miller=CGS(1-AV)

| sC

>>|

g s, frequencie low

At mb L

(

1 ( / )

)

1/

1 g g g

Z sC m mb mb

GS

in + +

) (

)

/(g g C same as Miller g

C

Cin = GS mb m + mb + GD

Effect not important: Only a fraction of C Effect not important: Only a fraction of CGS is added to Cin

(69)

Source Follower HF Z Source Follower HF Z

in

At high frequencies g << | sC | At high frequencies, gmb << | sCL | Zin 1

C + 1

C + gm

2C C sCGS sCL s CGSCL

At a particular frequency, input impedance includes CGD in parallel with series

combination of CGSGS and CLL and a negative resistance equal to

-gm/(CGSCLω2).

(70)

Source Follower Output Impedance Derivation Assumptions

Body effect and CSB contribute an

contribute an

impedance which is a parallel portion of a parallel portion of Zout – we’ll keep it in mind We also

mind. We also neglect CGD.

?

/ =

= X X

OUT V I

Z

(71)

Source Follower Output Impedance Derivation

I V

g s

C

V + = −

X S

GS

X m

GS

V V

sR C

V

I V

g s

C V

= +

= +

1 1

1 1

X S

GS 1

1

X X

OUT

V I

Z = /

GS S

sC g

C sR

+

= + 1

GS m

sC

g +

(72)

Source Follower Output Impedance - Discussion

I V

Z = /

C C sR

I V

Z

GS S

X X

OUT

1

/

= +

=

s frequencie low

at g

sC g

m

GS m

, /

1

+

s frequencie high

at RS ,

We already know that at low frequencies We already know that at low frequencies Zout≈ 1/gm

At high frequencies, CGS short-circuits the Gate-Source, and that’s why Zout depends on the Source Follower’s driving signal

source.

(73)

Source Follower Output Impedance - Discussion

I V

Z = /

C C sR

I V

Z

GS S

X X

OUT

1

/

= +

=

s frequencie low

at g

sC g

m

GS m

, /

1

+

s frequencie high

at RS ,

(74)

Source Follower Output Impedance - Discussion

Typically RS > 1/gm Therefore |Zout(ω)| is increasing with ω

increasing with ω.

(75)

Source Follower Output Impedance - Discussion

If |Zout(ω)| is increasing with ω, then it must have an INDUCTIVE

component!

(76)

Source Follower Output Impedance – Passive Network Modeling?

Can we find values for R1, R2 and L such that Z1=Zout of the Source Follower?

that Z1 Zout of the Source Follower?

(77)

Source Follower Output Impedance – Passive Network Modeling Construction

Clues:

At ω=0 Zout=1/gm ; At ω=∞ Zout=RS At ω=0 Z1=R2 ; At ω=∞ Z1=R1+R2

Let R2=1/g and R1=RS-1/g (here is where Let R2 1/gm and R1 RS 1/gm (here is where we assume RS>1/gm !)

(78)

Source Follower Output Impedance – Passive Network Modeling Final Result

R 1/ g R2 =1/ gm

R1 = RS −1/gm L = CGS

gm

(

RS −1/gm

)

Output impedance inductance dependent on source impedance dependent on source impedance, RS (if RS large) !

(79)

Source Follower Ringing Source Follower Ringing

O tp t ringing d e to t ned circ it formed Output ringing due to tuned circuit formed with CL and inductive component of

output impedance It happens especially if output impedance. It happens especially if load capacitance is large.

(80)

High Frequency Response- High Frequency Response-

L26

Common Gate Amplifier

Common Gate Amplifier

(81)

CG Amplifier neglecting r CG Amplifier neglecting r

o

• CCSS=CCGS1GS1+C+CSB1SB1 Capacitance “seen” from SourceCapacitance seen from Source to ground.

• C =C +C Capacitance “seen” from Drain to

• CD=CDG+CDB Capacitance seen from Drain to ground.

(82)

CG Amplifier if r negligible CG Amplifier if r

o

negligible

• Can determine (by inspection) the two polesCan determine (by inspection) the two poles contributed, one by CS and another by CD.

• This is easy to do only if we neglect rs s easy to do o y e eg ect 0101.

• Strategy: Find equivalent resistance “seen” by each capacitor, when sources are nulled.p ,

(83)

CG Amplifier Transfer Function if r

oo

is negligible

• ττSS=CCSSRRSS,eq=CCSS{R{RSS||[1/(g||[1/(gm11+g+gmb1b1)])]

• τD=CDRD,eq=CDRD

A ( + )R /(1+( + )R )

• A=(gm1+gmb1)RD/(1+(gm1+gmb1)RS)

• Vout(s)/Vin(s)=A/[(1+sτS)(1+sτD)]

(84)

CG Bandwidth taking r

o o

into account

Can we use Miller’s theorem for the bridging ro resistor?

(85)

CG Bandwidth taking r

o o

into account

Can we use Miller’s theorem for the bridging ro resistor? Well, no.

Effect of ro at input: Parallel resistance ro/(1-AV).

Recall: AVV is large and positive Æ Negative g p g

resistance! Æ How to compute time constants?

(86)

CG Bandwidth taking r

o o

into account

Need to solve exactly, using impedances 1/sC, and Kirchhoff’s laws.

(87)

Example: r

oo

effect if R

D

is replaced with a current source load

• Current from VCurrent from Viin through Rthrough RSS is: Vis: VouttCCLLs+(-Vs+( V11)C)Ciinss

• Adding voltages: (-VoutCLs+V1Cins)RS+Vin=-V1 V [ V C R V ]/(1 C R )

• V1= -[-VoutCLsRS+Vin]/(1+CinRSs)

• Also: ro(-VoutCLs-gmV1)-V1=Vout

(88)

Example: r

oo

effect if R

D

is replaced with a current source load – final result

V t(s)/Vi (s)=(1+g r )/D(s) where:

Vout(s)/Vin(s) (1+gmro)/D(s), where:

D(s)=roCLCinRSs2+[roCL+CinRS+(1+gmro)CLRS]s+1

Th ff t f b dd d i l b

The effect of gmb can now be added simply by replacing gm by gm+gmb.

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