CNU EE 8.2-1

Microelectronic Circuits II

Ch 8 : Frequency Response

8.6 High-Frequency Response of CG & Cascode Amplifier8.7 High-Frequency Response of Source Follower8.8 High-Frequency Response of Differential Amplifier8.9 Multistage Amplifier Examples

CNU EE 8.2-2

High-Frequency Response of CG Amplifier§ High-Frequency Response

- CG amplifier w/ Cgs & Cgd- CL || Cgd- grounded 3 capacitors à No Miller effectà much wider bandwidth than CS, especially

when Rsig is large§ when ro is neglected (b)

- input is isolated from the output à 2 poles- one pole at the input side w/ fP1

- the other pole at the output w/ fP2

- usually fP2 < fP1 à fP2 is dominant- fP1 & fP2 in CG >> frequency of the

dominant input pole in the CS stage

( )21

2Pgd L L

fC C Rp

=+

÷÷ø

öççè

æ=

msiggs

P

gRC

f1||2

11

p

CNU EE 8.2-3

§ when ro is taken into account - 3-dB frequency fH by the method of open-circuit time constants - Rgs & Rgd seen by Cgs & (Cgd + CL), respectively :

- fH

- input resistance Rin & output resistance Ro of the CG amplifier

- CG circuit : wide bandwidth, low Rin , very low overall midband gain à Cascode (CG + CS) :: high Rin, high gain of CS + wide bandwidth of CG amplifier

( )1

2H

gs gs gd L gd

fC R C C Rp

=é ù+ +ë û

insiggs RRR ||= oLgd RRR ||=

High-Frequency Response of CG Amplifier

( ) sigomsigooom

Loin RrgRrR

rgRrR ++=

++

=1

CNU EE 8.2-4

Impedance transformation properties of CG amplifier

1

2

3

4

LR 2inR 1dR 1vA 2vA vACase¥ ¥ 0r 0rgm- 0rgm

20 )( rgm-

20 )(

21 rgm-0rgm)(

21

0rgm-00 )( rrgm 0r 2/0r

0rmg

2mg

2 )(21

0rgm )( 0rgm-

mg10 1- 0 0

2-

mg1

RL = infinity à ideal current-source loadRL = (gmro)ro à cascode current-source loadRL = ro àsimple current source loadRL = 0 à signal short circuit at the ground

CNU EE 8.2-5

High-Frequency Response of the MOS Cascode

§ 3dB frequency fH by open-circuit time constant

§Effective time constant tH

siggs RC >1

( ) 11111 1 dsigdmgdgd RRRgRC ++=>

22

2121121 ||||

om

Looinodgsdb rg

RrrRrRCC +==+ >

( ) 122122 ,|| oomooooLLgd rrgrrRRRCC ++=+ >

( )[ ] ( ) ( ) )||(1 212111111 oLLgddgsdbdsigdmgdsiggsH RRCCRCCRRRgCRC +++++++=t

( )[ ] ( ) ( )( )LgdoLgsdbgdddmgdgssigH CCRRCCCRRgCCR +++++++= 221111111 ||1t

Cascode : very high Rin & voltage gain, Ao2 (Ao=gmro)

in the midband à Trade off high midband gain forwider bandwidth

- Cdb1 & Cgs2 are in parallel- CL & Cgd2 are in parallel

12H

H

fpt

@

§ Large Rsig à dominant 1st term à large Miller multiplier (1+gm1Rd1) : à reduced BWwhen RL ~ Aoro à larger Rin2 (=ro)& large proportion of gain at the 1st stage, Q1

for wider BW, RL ~ ro à lower Rin2 (=2/gm) & Rd1 à insignificant Miller effect, but dc gain of cascode ~ Ao

:: Therefore, dc gain is the same as CS amplifier but bandwidth is greater

CNU EE 8.2-6

Comparison of CS and Cascode§ Small Rsig à minor Miller effect in Q1

Therefore, large RL ~ Aoro à large dcgain Ao

2 à dominant 3rd term in tH

§ when Rsig = 0

- same form as CS amplifier w/ Rsig = 0but (RL||Ro) >> RL

/ by a factor of Aoà fH of cascode < fH of CS amplifier

by a factor of Ao- Cascoding increases dc gain by Ao

while keeping unity-gain frequencyunchanged

- To achieve the high gain, the loadresistance must be increased by thefactor of AOEffect of cascoding on gain & bandwidth in the case of Rsig = 0

( )( )oLgdLH RRCC ||2+»t

( )( )oLgdLH RRCC

f||2

1

2+=

p

( )22 gdL

mt CC

gf+

»p

CNU EE 8.2-7

High-Frequency Response of Source Follower§ Source follower (Common-Drain amplifier : CD)

- voltage gain ~ 1, high Rin, low Ro (à output stage of a multistage amplifier), wide bandwidth- Excellent high-frequency response because none of internal capacitances suffers from the Miller effect

§ High-frequency model of source follower- low-frequency voltage gain AM & output resistance Ro (all C à 0)

( )( ) ( ) o

mo

moL

oLM r

gR

grRrRA ||11||

||=

+=

CNU EE 8.2-8

Frequency Response of Source Follower§ Simplified equivalent circuit- too complicate transfer function w/ 3 Capacitances- At first, locate the transmission zeros, then

open –circuit time constant for 3-dB frequency f3dB- 2nd order transfer function : capacitors form a

continuous loop (three capacitances : Cgd, Cgs, CL)à two transmission zeros

- Vo = 0 ß CL has a zero impedance & thus acts as ashort across the output, which is w or s = infinity

- Vo = 0 ß current into impedance RL/||CL = 0 à

(gm+sCgs)Vgs = 0 at s=sz

- wT= gm/(Cgs+Cgd) & Cgd << Cgs à fZ ~ fT

mZ

gs

gs

C= - m

Zgs

gC

w =

Factor (1+gmRL/) in Rgs reduces the effective resistance with which Cgs interacts

§ Estimate of fH by open-circuit time constant method- Rgd seen by Cgd

- Rgs seen by Cgs (C)

- RCL seen by CL

gd sigR R='

'1sig L

gsm L

R RR

g R+

=+

( )1 1 22 LH gd sig gs gs L C

H

f C R C R C Rppt

= = + +

moLC grRRL

1||||=Low RCL due to 1/gm

oLL rRR ||/ =

(c)

CNU EE 8.2-9

High-Frequency Response of the Differential Amp.§ Analysis of the Resistively Loaded MOS Amplifier

Differential Half Circuit

Common ModeHalf Circuit

- Variation w/ frequency of both differential gain & common-mode gain, and hence the CMRR- Total impedance between node S and ground : ZSS=RSS||CSS à affects common-mode gain & CMRR- RSS : output resistance of current source QS - CSS : Cdb & Cgd of QS, Csb1, CSb2- (b) Differential half-circuit : differential gain Ad (s) is identical to the CS amplifier - (c) Common-mode half-circuit : CSS/2 & 2RSS à real-axis zero in the common-mode gain function is

lower than the other poles & zeros à dominates the frequency dependence of Acm & CMRR- If output of differential amplifier is taken single-endedly à common-mode gain Vocm/Vicm

Resistively loaded MOS differential pair

CNU EE 8.2-10

- Common mode gain when there is a mismatch ΔRDbetween two drain resistances

- Frequency dependence of Acm by replacing RSS by ZSS

High-Frequency Response of the Differential Amp.

÷÷ø

öççè

æ D-=

D-=

D

D

SS

D

SS

Dcm R

RRR

RRA

22

( )SSSSD

D

SS

DSS

SSD

DD

SSD

DD

D

D

SS

Dcm

RsCRR

RRsC

RRRR

YRRR

RR

ZRA

+÷÷ø

öççè

æ D-=÷÷

ø

öççè

æ+÷÷

ø

öççè

æ D-=

÷÷ø

öççè

æ D-=÷÷

ø

öççè

æ D-=

12

121

21

2 Zero

SSSSZ RC

1=w

SSSS

ZZ RC

fpp

w2

12

==

Product of Vocm/Vicm & per-unit mismatch DRD/RD

- Zero : fZ << otherpoles & zeros

Common-mode gain

Differential gain

CMRR w/ frequency

- Acm increases at the rate of +6dB/octave starting at fZ (a)

à CMRR begins to decrease (c)

- CMRR in dB = difference between Ad & Acm

- trade-off between the need to reduce dc voltage across QS and

the need to keep CMRR reasonably high at higher freq.

- Small VDS at QS à low VOV à for a given I, larger W/L ratio

(i.e., wider transistor) à higher CSS à lower fZ à CMRR

gain deteriorates at a relatively low frequency

(c)

(b)

(a)

CNU EE 8.2-11

High-Frequency Response of the Differential Amp.

§ Need for high CMRR at higher frequency- power-supply voltage VDD is corrupted with high-frequency noise

- Drains of Q1 & Q2 = VDD -(I/2)RD à same high frequency noise as VDD at vD1 & vD2

- Vcm w/ high frequency noise à 2nd differential stage by Q3 & Q4

- Perfectly matched 2nd differential stage à differential output Vo is free of high-frequency noise

- In practice, finite Acm2 by unmatched 2nd stage & increased Acm2 with frequency by zero formed by RSS & CSS

à some noise in VO

CNU EE 8.2-12

§ Frequency response of the current-mirror-loaded MOS differential-pair circuit- Cm : Total capacitance at the input node of the current mirror- CL : Total capacitance at the output node

Cx : actual load capacitance and/or input capacitance of a subsequent stage- Two capacitances à dependence of Ad (s) on frequency§Overall transconductance Gm- Differential signal Vid in a balanced fashion & output node is shorted- CL has no effect on Gm

43311 gsgsdbdbgdm CCCCCC ++++=

xgddbdbgdL CCCCCC ++++= 4422

Analysis of Active-loaded MOS Amplifier

CNU EE 8.2-13

mm

idmg sCg

VgV+

-=3

32

33

34344 1

22 43

mm

idmgg

mm

idmmgmd gsC

VgsCgVggVgI

mm

+=

+=-=

=

( )21

2

324 idm

mm

idmddo Vg

gsCVgIII +

+=+=

Analysis of Active-loaded MOS Amplifier- Q1 conducts a drain current signal of gmVid /2 à

diode-connected transistor Q3 à (1/gm3)||Cm, wherero1 & ro3 >> 1/gm3, thus ro1 & ro3 are neglected

- Total output current IO at the output node

- Gm at high-frequency has pole & zero :

Since Cm ~ Cgs3+Cgs4=2Cgs

- Mirror pole and zero occur at very high frequency but their effect is significant

- Gm : à Gm =gm of Q1 & Q2 at the low-frequency3

3

121

mm

mmm

id

om gsC

gsCgVIG

++

=º

m

mp C

gfp2

32 =

m

mZ C

gfp2

2 3= Zero frequency is twice that of pole, fp2

( ) 222233

2T

gs

m

m

mp

fC

gC

gf »»=pp TZ ff @ fT : unity-gain

frequency of Q3

CNU EE 8.2-14

§Output voltage Vo by short-circuit output current IO & total impedance between output node & ground

where, Ro = ro2||ro4

( ) ( ) ÷÷ø

öççè

æ++

+=

+=

+=

3

3

121

1111

mm

mm

oL

oom

id

o

oL

oidm

Lo

oo gsCgsC

RsCRRg

VV

RsCRVG

sCR

IsV >

Pole, fP1

Pole, fP2

Zero, fZ

DC gain

Analysis of Active-loaded MOS Amplifier§Overall transconductance Gm versus frequency- Transfer function of signal current path by Q1 is

different from that of signal current by Q2- Q1 signal current encounters Cm & mirror pole- At low frequencies, Id1 is replicated as Id4à Id4 + Id2 provides a factor-of-2 increase in Gm

(Gm ~ gm : double the value w/o current mirror)At high frequencies, Cm acts as a short circuità Vg3 = 0 à Id4 = 0à reducing Gm to gm/2

- Gain of differential amplifier has pole & zero of Gm + a pole w/ fp1 :

- Output pole w/ fp1 is often dominant, especially when a large load capacitance is presentoL

p RCf

p21

1 =

CNU EE 8.2-15

§ CS amplifier w/ a source degeneration resistance Rs (a) - Voltage gain reduction, but extension of bandwidth- At low-frequency, controlled-current source GmVi + output resistance Ro (b)

- Source degeneration reduces Gm & increases Ro by the same factor (1+gmRs)- low-frequency voltage gain :

Wideband Amplification by Source Degradation

( )smoosm

mm RgrR

RggG +»

+» 1

1

( ) /|| LmLomsig

oM RGRRG

VVA -=-=º

CNU EE 8.2-16

§ High-frequency response of the source degenerated amplifier (c)- open-circuit time constants for estimate of 3-dB frequency fH

Wideband Amplification by Source Degradation

- Determine Rgd : resistance seen by Cgd (d)

where

(c)

(d)

oLL RRR ||/ =/|| LoLC RRRR

L==

÷÷ø

öççè

æ+

+

+»

Lo

osm

ssiggs

RrrRg

RRR

1

( ) //1 LLmsiggd RRGRR ++=

CNU EE 8.2-17

§ When Rsig is relatively large, frequency response is dominated by the Miller multiplication of Cgdà CgdRgd is the largest of the three open-circuit time constants that make up tH :

gdgdCLgdgdgsgsH RCRCRCRCL»++=t

gdgdH RC

fp2

1@

( ) sigMsigLmLLmsiggd RARRGRRGRR =»++= '''1

MsiggdH ARC

fp2

1=then,

§ Gain-Bandwidth product remains constant at

siggdHM RC

fAproductbandwidthGainp2

1==-

§ Trade-off between gain and bandwidth- As Rs increase, |AM|=GmRL

/ decrease à Rgd decrease à fH increase- Expression for Rgd by assuming GmRL

/ >> 1 & GmRsig >> 1,

Wideband Amplification by Source Degradation

shows the gain-bandwidth trade-off

CNU EE 8.2-18

§ Equivalent circuit of two-stage CMOS op amp - Gm1 : transconductance of input stage (Gm1 = gm1 = gm2)- R1 : output resistance of the 1st stage (R1 = ro2||ro4)- C1 : total capacitance at the interface between

the 1st & 2nd stage

- Gm2 : transconductance of 2nd stage (Gm2 = gm6)- R2 : output resistance of the 2nd stage (R2 = ro6||ro7)- C2 : total capacitance at the output node

- CL : load capacitance (CL >> the other C’s, C2 >> C1 )- CC : equip op amp w/ a uniform -6-dB/octave slope

à how to select a value of CC ?- Cgd6 || CC & CC >> Cgd6 à Cgd6 can be neglected

Frequency Response of Two-Stage CMOS Op Amp

622441 gsdbgddbgd CCCCCC ++++=

Lgddbdb CCCCC +++= 7762

CNU EE 8.2-19

§ Equivalent circuit of two-stage CMOS op amp - To determine Vo, a node equation at node D2 :

- a node equation at node D6 :

- Vi2 in terms of Vo & eliminate Vi2 and then determine Vo in terms of Vid

- First, for s = 0 (i.e. dc) V0/Vid = (Gm1R1)(Gm2R2)- Second, a transmission zero at s = sZ :

à positive real frequency wZ

Frequency Response of Two-Stage CMOS Op Amp

( ) 02211

21 =-+++ oiCi

iidm VVsCVsC

RVVG

( )( )[ ] ( )[ ] 212121

2212122211

2121

11 RRCCCCCsRRRRGCRCRCsRRsCGG

VV

CmC

Cmm

id

o

+++++++++-

=

( ) 0222

22 =-+++ ioCoo

im VVsCVsCRVVG

C

mZ

C

mZCZm C

GCGsCsG 22

2 0 ===- w>

CNU EE 8.2-20

- Two poles = roots of denominator polynomial à frequencies of two poles wP1 & wP1 :

- if dominant pole is at wP1 then, wP2 >> wP1 & approximated D(s)- Frequency of the dominant pole wP1 :

- 1st term : R1 (Ro of 1st stage) & total Capacitance (C1 + Miller capacitance CC(1+Gm2R2)CC(1+Gm2R2) : CC is connected in the negative feedback path of 2nd stage whose gain is Gm2R2

- Since R1 & R2 are comparable, 1st term >> 2nd term à approximate wP1 :because C1 << Miller capacitance

Gm2R2 >> 1

- 2nd, nondominant pole : Since C1 << C2 & C1 << CC

- Op amp w/ a uniform gain rolloff of -20dB/decade down to 0 dB à compensation capacitor CC selectionà unity-gain frequency wt must be lower than wZ & wP2 .

Frequency Response of Two-Stage CMOS Op Amp( )

( )[ ] ( )[ ] 2121212

212122211

2121

11 RRCCCCCsRRRRGCRCRCsRRsCGG

VV

CmC

Cmm

id

o

+++++++++-

=

( )1 2

1 1P P

s sD sw w

æ öæ ö= + +ç ÷ç ÷è ø è ø

2

1 2 1 2

1 11P P P P

ssw w w wæ ö

= + + +ç ÷è ø

( )2

1 1 2

1P P P

s sD sw w w

@ + +

( ) ( )[ ] ( )CmCmCP CCRRGCCRRRRRGCRCRC ++++

=+++++

=222211212122211

1 11

11w

( )[ ] 2211

22111

11

1RGCRRGCCR mC

PmC

P »++

» ww >

( ) 2

2

2121

22 C

GCCCCC

CG m

C

CmP »

++=w

( )2

22

2112211 C

GCG

CGRGRG m

PC

mZ

C

mPmmt »=== wwww