chap17 frequency response

7/27/2019 Chap17 Frequency Response

1/61

Jaeger/Blalock2/9/04

Microelectronic Circuit DesignMcGraw-Hill

Chapter 17

Frequency Response

Microelectronic Circuit Design

Richard C. Jaeger

Travis N. Blalock

Chap 17- 1


2/61



Chapter Goals

Review transfer function analysis and dominant-pole approximations

of amplifier transfer functions.

Learn partition of ac circuits into low and high-frequency equivalents.

Learn short-circuit and open-circuit time constant methods to estimate

upper and lower cutoff frequencies. Develop bipolar and MOS small-signal models with device

capacitances.

Study unity-gain bandwidth product limitations of BJTs and

MOSFETs.

Develop expressions for upper cutoff frequency of inverting, non-inverting and follower configurations.

Explore gain-bandwidth product limitations of single and multiple

transistor circuits.

Chap 17- 2


3/61



Chapter Goals (cont.)

Understand Miller effect and design of op amp frequency

compensation.

Develop relationship between op amp unity-gain frequency and slew

rate.

Understand use of tuned circuits to design high-Q band-pass

amplifiers.

Understand concept of mixing and explore basic mixer circuits.

Study application of Gilbert multiplier as balanced modulator and

mixer.

Chap 17- 3


4/61



Transfer Function Analysis

Av(s) N(s)

D(s)

a0

a1sa

2s2 ...a

msm

b0

b1sb

2s2 ...b

nsn

Av(s) AmidFL(s)FH(s)

Amid is midband gain between upper

and lower cutoff frequencies.

FL

(s)s Z1L

s Z2L

...s ZkL

s P1L

s P2L

...s PkL

FH(s)

1 s

Z1

H

1 s

Z2

H

...1 s

Zl

H

1 sP1H

1 s

P2H

...1 s

PlH

FH

(j) 1 for ZjL ,

PjL , j =1,k

AH

(s)Amid

FH

(s)

Chap 17- 4


5/61



Low-Frequency Response

FL

(s) ss

P2

L

P2

Pole P2 is called the dominant low-

frequency pole (> all other poles) and

zeros are at frequencies low enough tonot affect L.

If there is no dominant pole at low

frequencies, poles and zeros interact to

determine L.

AL

(s)Amid

FL

(s)Amid

sZ1 sZ2 s

P1 sP2

For L

s jL

, A(j

L)

Amid

2

1

2

L

2 Z1

2 L2 Z22 L

2 P1

2 L2 P22

12

1Z1

2 Z2

2 L

2Z1

2 Z2

2 L

4

1P1

2 P2

2 L

2P1

2 P2

2 L

4

Pole L > all other pole and zero frequencies

In general, forn poles and n zeros,

L P12 P22 2Z12 2Z22

L

Pn

2

n

2 Zn

2

n

Chap 17- 5


6/61



Transfer Function Analysis and

Dominant Pole Approximation Example Problem: Find midband gain,FL(s) andfL for

Analysis: Rearranging the given transfer function to get it in standard form,

Now,

Zeros are at s = 0 and s = -100. Poles are at s = -10, s =-1000

All pole and zero frequencies are low and separated by at least a decade. Dominant pole is

at = 1000 andfL = 1000/2p= 159 Hz. For frequencies > a few rad/s:

AL

(s) 2000s

s

1001

0.1s1 s1000

AL

(s) 200s s100

s10 s1000 F

L(s) s(s100)

(s10)(s1000)

AL

(s)Amid

FL

(s) Amid

200

fL

1

2p

102 10002 2 02 1002

158 Hz

AL

(s) 200 ss1000

Chap 17- 6


7/61



High-Frequency Response

FL

(s) s

1 sP3

H

P3

Pole P3 is called the dominant high-

frequency pole (< all other poles).

If there is no dominant pole at low

frequencies, poles and zeros interact to

determine H.

AH

(s)Amid

FH

(s)

AH

(s)Amid 1(s/Z1) 1(s/Z2) 1(s/P1

) 1(s/P2)

For=H

s = jH

, A(j

H)

Amid

2

1

2

1(H

2 /Z1

2 ) 1(H2 /Z22 ) 1(

H

2 /P1

2 ) 1(H2 /P22 )

1

2

1

H

2

Z1

2

H

2

Z2

2

H

4

Z1

2 Z2

2

1

H

2

P1

2

H

2

P2

2

H

4

P1

2 P2

2

Pole H < all other pole and zero frequencies

In general,

H

11

P12 1

P22 2

Z12 2

Z22

H

11

Pn

2n

21

Zn

2n

Chap 17- 7


8/61



Direct Determination of Low-Frequency

Poles and Zeros: C-S Amplifier

Vo(s)I

o(s)R

3 g

mVgs(s)

RD

R3

(1/sC2)R

3

R3

gm

(R3RD

) s

s 1C

2(R

DR3)

Vgs(s)

Vg(s) sC1RGsC1(R

IR

G)1

Vi(s)

Vgs(s)V

g-V

s

s(1/C3R

S)

s 1C

3(1/g

m) R

S

Vg(s)

Av(s)Vo(s)

Vi(s)

AmidFL(s)

Amid

gm

(R3RD

)RG

RG

RI

Chap 17- 8


9/61



Direct Determination of Low-Frequency

Poles and Zeros: C-S Amplifier (cont.)

The three zero locations are: s = 0, 0, -1/(RSC3).

The three pole locations are:

Each independent capacitor in the circuit contributes one pole and one

zero. Series capacitors C1

and C2

contribute the two zeros at s = 0 (dc),

blocking propagation of dc signals through the amplifier. Third zero due

to parallel combination ofC3 andRSoccurs at frequency where signal

current propagation through MOSFET is blocked (output voltage is zero).

FL(s)

s2 s(1/C3R

S)

s 1C

1(R

IR

G)

s 1

C3

(1/gm

) RS

s 1C

2(R

DR

3)

s 1C

1(R

IR

G), 1

C3

(1/gm

)RS

, 1C

2(R

DR

3)

Chap 17- 9


10/61



Short-Circuit Time Constant Method to

Determine L

Lower cutoff frequency for a

network with n coupling and

bypass capacitors is given by:

whereRiSis resistance at

terminals ofith capacitorCi with

all other capacitors replaced by

short circuits. ProductRiSCi isthe short-circuit time constant

associated with Ci.

L

1

RiS

Ci

i 1

n

Midband gain and upper and lower

cutoff frequencies that definebandwidth of amplifier are of more

interest than complete transfer function.

Chap 17- 10


11/61



Estimate ofL for C-E Amplifier

Using SCTC method for C1:

R1S

RI

(RB in

CER )R2 (RB rp)

ForC2

,

R2S

R3(R

C outCER )R3 (RC ro)R3 RC

ForC3,

R3S

RE out

CCR RErp

Rth

o 1

RE

rp

(RIRB

)

o 1

L

1R

iSC

ii 1

3

Chap 17- 11


12/61



Estimate ofL for C-S Amplifier

Chap 17- 12


13/61



Estimate ofL for C-S Amplifier

Using the SCTC method for C1,

R1S

RI

(RG in

CSR )RSRG

For C2,

R2S

R3(R

D outCSR )R3 (RD ro)

R2S

R3R

D

For C3,

R3S

RS out

CGR RS1

gm

Chap 17- 13


14/61



Estimate ofL for C-B Amplifier

Apply the SCTC method :

For C1,

R1S

RI

(RE in

CBR )RI(RE1

gm

)

For C2,

R2S

R3(R

C outCBR )R3 RC

Chap 17- 14


15/61



Estimate ofL for C-G Amplifier

Chap 17- 15


For C1,

R1S

RI

(RS in

CGR )RI(RS1

gm

)

For C2,

R2S

R3(R

D outCGR )R3 RD


16/61



Estimate ofL for C-C Amplifier

Chap 17- 16


For C1,

R1S

RI

(RB in

CCR )

R1S

RI

RB

rp

o

1

R

ER

3

For C2,

R2S

R3(R

E outCCR )R3 RE

rp

Rth

o

1


17/61



Estimate ofL for C-D Amplifier

Chap 17- 17


For C1

,

R1S

RI

(RG in

CDR )RIRG

For C2,

R2S

R3R

S outCDR R3 RS

1

gm


18/61



Frequency-dependent Hybrid-Pi Model

for BJT

Capacitance between base andcollector terminals is:

Cmo is total collector-base junctioncapacitance at zero bias, Fjc is thejunctions built-in potential.

Cm

C

mo

1(VCB

/jc

)

Capacitance between base andemitter terminals is:

tFis the forward transit-time of the

BJT. Cpappears in parallel with rp.

As frequency increases, for a given

input signal current, impedance of

Cp reduces vbe and thus reduces thecurrent in the controlled source at

transistor output.

Cp

gm

tF

Chap 17- 18


19/61



Unity-gain Frequency of BJT

Ic(s) (g

msC

m)V

be(s)

Ic(s) (g

msC

m)I

b(s)

rp

s(Cp

Cm)rp

1

(s)I

c(s)

Ib(s)

o

1 sCmg

m

s(Cp

Cm)rp

1

The right-half plane transmission zero Z= +

gm/Cmoccurring at high frequency can be

neglected.

= 1/ rp(Cm+ Cp ) is the beta-cutoff frequency

where

andfT = T /2p is the unity gain-bandwidth

product. Above fT, the BJT has no current gain.

(s)

o

s(Cp

Cm

)rp

1

o

(s/

)1

(s)

o

s

T

s

T o

o(C

pC

m)rp

g

mCp

Cm

Chap 17- 19


20/61



Unity-gain Frequency of BJT (cont.)

Current gain is o =gmrpat low

frequencies and has single pole roll-

off at frequencies >f, crossing

through unity gain at T. Magnitude

of current gain is 3 dB below itslow-frequency value atf.

Cp

g

m

T

Cm

40I

C

T

Cm

Chap 17- 20

Cp is calculated using


21/61



High-frequency Model of MOSFET

Id(s) (g

msC

GD)V

gs(s)

Id(s)I

b(s)

(gm

sCGD

)

s(CGS

CGD

)

(s)I

d(s)

Ig(s)

T

s1 s

T

1(CGS

/CGD

)

T

g

m

CGS

CGD

fT

mnCox

" W

LVGSVTN

(2/3)Cox

"WL

3

2

mn

VGSVTN L

2

Chap 17- 21


22/61



Limitations of High-frequency Models

Above 0.3fT, behavior of simple pi-models begins to deviatesignificantly from the actual device.

Also, Tdepends on operating current as shown and is not constant asassumed.

For given BJT, a collector currentICMexists that yields maximum

fTmax. For the FET in saturation, CGSand CGD are independent of Q-point

current, so

T

gm

ID

Chap 17- 22


23/61



Effect of Base Resistance

on Midband Amplifiers

Base current enters the BJT through

external base contact and traverses a

high resistance region before entering

active area. rx models voltage drop

between base contact and active areaof the BJT.

To account for base resistance rx is absorbed

into equivalent pi model and can be used to

transform expressions for C-E, C-C and C-B

amplifiers.

igm

vgm

rp

rprxvbe

gm

' vbe

gm' g

m

rp

rp

rx

o

rp

rx

rp' r

pr

x

o'

o

Chap 17- 23


24/61



Direct High-Frequency Analysis

C-E Amplifier

The small-signal model can be simplified by

using a Norton source transformation.

RL

R3

RC

100k 4.3k RB

R1R

2 30k10k

vth viR

BR

IR

B

Rth

RIR

BR

IR

B

is

v

th

Rth

rx

rpo

rp

(Rth

rx

)

Chap 17- 24


25/61




C-E Amplifier (Pole Determination)

From nodal equations for the circuit in

frequency domain,

High-frequency response is given by 2

poles, one finite zero and one zero at

infinity. Finite right-half plane zero, Z=+gm/Cm> T can easily be neglected.

For a polynomials2 +sA1 +A0 with roots

a and b, a A1 and b A0/A1.

V2(s)I

s(s)

(sCm-g

m)

s2

Cp CmCL

CpCL

s Cpg

LC

mg

Lg

mg

p

C

Lgpo

g

Lgpo

CT

Cp

Cm

1gm

RL

R

L

rpo

C

L

RL

rpo

P1

A

0

A1

1rpo

CT

P2

g

m

Cp

1(CL

/Cm)

C

L

Smallest root that gives first pole limits

frequency response and determines H.

Second pole is important in frequencycompensation as it can degrade phase

margin of feedback amplifiers.

Chap 17- 25


26/61




C-E Amplifier (Overall Transfer Function)

Vo(s)

Vth(s)

Rth

rx

(sCm-g

m)

gLgpo

1(s/P1

)

1(s/

P2)

Vo(s)

Vth(s)

Rth

rx

(gmRLrpo

)1(s/

Z)

gLgpo

1(s/P1

)

1(s/

P2)

Vo(s) -

Vth(s)

Rth

rx

gmRLrpo

1(s/P1

)

Avth

(s)Vo(s)

Vth(s)

Amid

1(s/P1

)

Amid

oR

L

Rth

rx

rp

P1

1rpo

CT

Chap 17- 26

Dominant pole model at high

frequencies for C-E amplifier is as

shown.


27/61




C-E Amplifier (Example) Problem: Find midband gain, poles, zeros andfL.

Given data: Q-point = ( 1.60 mA, 3.00 V),fT= 500 MHz, o = 100, Cm= 0.5

pF, rx = 250,CL0

Analysis:gm= 40IC= 40(0.0016) = 64 mS, rp= o/gm =1.56 k

Cp g

m

2pfT

Cm19.9 pF

RL

R3

RC

100k 4.3k 4.12k

Rth

RB

RI

7.5k1k 882

rpo

rp

(Rth

rx

) 656

CT

Cp

Cm

1gmR

L

RL

rpo

C

L

RL

rpo

156 pF

fP1

1

2prpoCT

1.56 MHz

P2

1RLCm

gm

Cp

1 1gmrpo

1gmRL

fP2

P2

2p

603MHz fZ

gm

2pCm

20.4 GHz

Avth

oRL

Rth

rx

rp

153

Overall gain is reduced to -135 as vth = 0.882vs.

Chap 17- 27


28/61



Gain-Bandwidth Product Limitations of

C-E Amplifier

IfRth is reduced to zero in order to increase bandwidth, then rpo would

not be zero but would be limited to approximately rx.

IfRth = 0, rx


29/61



High-Frequency Analysis: C-S Amplifier

Rth

RI

RG

RL

RD

R3

vth

vi

RG

RI

RG

CT CGSCGD 1gmRL

RLR

th

Z

g

m

CGD

P1

1R

thC

T

P2

1R

LC

GD

gmC

GS

1 1g

mR

th

1g

mR

L

Chap 17- 29


30/61



Miller Multiplication

Vo(s)AV

1(s) I

s(s)sC V

1(s)V

o(s)

Y(s)I

s(s)

V1(s)

sC(1A)

Total input capacitance = C(1+A) because

total voltage across Cis vc = vi(1+A) due to

inverting voltage gain of amplifier.

For the C-E amplifier,

CT

Cp

Cm(1A) C

pC

m(1g

mR

L)

Chap 17- 30


31/61

Jaeger/Blalock2/9/04 Microelectronic Circuit DesignMcGraw-Hill

Miller Integrator

Assuming zero current in input

terminal of amplifier,

V1 VinR

sC(Vin

Vo)

Vo

AVin

Av(s)

Vo

V1

1

RC

A

1A

s 1RC(1A)

A

o

so

o

1

RC(1A)where

For frequencies >> o, assumingA >> 1,

which is the transfer function of an

integrator.

Av(s)

Ao

s

1

sRC

Chap 17- 31


32/61


Open-Circuit Time Constant Method to

Determine HAt high frequencies, impedances of

coupling and bypass capacitors are small

enough to be considered short circuits.

Open-circuit time constants associated

with impedances of device capacitances

are considered instead.

H

1

Rio

Ci

i 1

m

whereRio is resistance at terminals of

ith capacitorCi

with all other

capacitors open-circuited.

For a C-E amplifier, assuming CL = 0

Rpo

rpo

Rmo

vx

ix

rpo

(1

gmRL

R

L

rpo

)

H

1R

poCp

Rmo

Cm

1rpo

CT

Chap 17- 32


33/61


Gain-Bandwidth Trade-off Using

Emitter Resistor

Amid

oR

L

Rth

rx

rp

(o

1)RE

R

L

RE

rp

Rth

rxfor and

gain decreases as emitter resistanceincreases and bandwidth of stage will

correspondingly increase.

To find bandwidth using OCTC method:

gmR

E1

Req

vx

ix

Rth

rx

RE

1gm

RE

Rpo

rp

Req

R

thr

xR

E

1gmR

E

Chap 17- 33


34/61


Gain-Bandwidth Trade-off Using

Emitter Resistor (cont.)

Test source ix is first split into two

equivalent sources and then

superposition is used to find vx = (vb - vc).

Assuming that o >> 1 and

Rth rx rp(o 1)RE

Rmo

vxix

(Rth

rx

)1 gmRL1gmRE

RLRthrx

H

1

(Rthrx)Cp

1gmRE

1RE

Rthrx

C

m1gmRL

1gmRE

RL

Rthrx

Chap 17- 34


35/61


Dominant Pole for C-B Amplifier

Rth RE RI RL RC R3

Rpo

rp

vx

ix

rp

Rth

rx

1gmR

th

R

thr

x

1gmR

th

Using split-source transformation assuming

that o >>1 and rx


36/61


Dominant Pole for C-G Amplifier

Rth

R4

RI R

LR

DR

3

RGSo

R

th

1gmRth 1

Gth gm

RGDo

RL

H

1

CGS

Gth

gm

CGDRL

1C

GD

RL

Chap 17- 36


37/61


Dominant Pole for C-C Amplifier

Rth RB RI RL RE R3

Rpo

rp

vx

ix

rp

Rth

rx

RL

1gmR

L

R

thr

xR

L

1gm

RL

Rmo

(Rth

rx

) inCCR (Rth rx ) rp(o 1)RL

Rmo (Rth rx )

H

1

(Rth

rxR

L)

Cp

1gmR

L

(Rth

rx

)Cm

A better estimate is obtained if we setRL = 0

in the expression forRpo.

H

1

(Rthrx)Cp

1gmRL

Cm

GBW (1)H

1

Cmrx

Chap 17- 37


38/61


Dominant Pole for C-D Amplifier

Substituting rpas infinite and rx as

zero in the expression for the emitter

follower,

Rth

RG

RI R

LR

SR

3

H

1R

th

1gmRL

CGS

CGDR

th

1

Rth

CGS

1gmRL

CGD

Chap 17- 38


39/61


Differential Amplifier

Frequency Response

CEE is total capacitance at emitter node

of the differential pair.

Differential mode half-circuit is similar

to a C-E stage. Bandwidth is determined

by the product. As emitter is avirtual ground, CEEhas no effect on

differential-mode signals.

For common-mode signals, at very low

frequencies,

Transmission zero due to CEEis

T

C

o

r

p

Acc

(0) R

C

2REE

1

s Z

1C

EER

EE

Chap 17- 39


40/61


Differential Amplifier

Frequency Response (cont.)

Common-mode half-circuit is similar to a

C-E stage with emitter resistor 2REE.OCTC forCpand Cm is similar to the C-E

stage. OCTC forCEE/2 is:

REE O

2REE

rprx

o1

1g

m

P

1

rx

Cp

12gm

REE

12R

EE

rx

C

m1

gm

RC

12gm

REE

R

C

rx

CEE

2gm

AsREEis usually designed to be large,

P

1CpCEE

2gm

Cm(RCrx)

1

Cm(RCrx)

Chap 17- 40


41/61


Frequency Response

Common-Collector/ Common-Base Cascade

REE is assumed to be large and

neglected.

Sum of the OCTC ofQ1 is:

outCC1R

rp1

rx1

o1

1 1

gm1

inCB2

R

rp2

rx2

o2 1

1

gm2

rx1

Cp1

1gm1

1

gm2

Cm1

rx1C

p1

2 Cm1

Chap 17- 41


42/61


Frequency Response

Common-Collector/ Common-Base Cascade (cont.)

Sum of the OCTC ofQ2 is:

Combining the OCTC forQ1 and Q2, and assuming that transistors

are matched,

rx2

Cp2

1gm2

1

gm1

Cm2

1gmR

C

1gm2

1

gm1

Cm2R

Cr

x2

Cp2

2C

m21

gmR

C

2R

C

rx2

H

1

rx Cp Cm 2gm

RC

2 RCrx

Chap 17- 42


43/61


Frequency Response

Cascode Amplifier

OCTC ofQ1 with load resistor 1/gm2 :

AsIC2 =IC1,gm2 =gm1, gain of first stage is

unity. Assuminggm2

rpo1

>>1,

OCTC ofQ1, a C-B stage forro1 >>RL and

mf>>1:

Assuming matched devices,

Rpo1C

p1R

mo1C

m1r

po1CT1

rpo1

Cp1

Cm1

1gm1

gm2

1

gm2rpo1

rpo1

CT1

rpo1

Cp1

2Cm1

Rpo2

Cp2

Rmo2

Cm2

Cp2

gm2

(rx2 RL)Cm2

H

1

rpo1

Cp

2Cm

r

xR

L

Cm

Chap 17- 43


44/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Frequency Response

MOS Current Mirror

rpo

1g

m1

RL

ro2

Cp

CGS1

CGS2

Cm

CGD2

P1

1

2CGS

gm1

2CGD2ro2

1

2CGD2

ro2

For matched transistors,

Chap 17- 44


45/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Frequency Response

Multistage Amplifier

Problem:Use open-circuit and short-circuit time constant methods to

estimate upper and lower cutoff frequencies and bandwidth.

Approach: Coupling and bypass capacitors determine low-frequency

response, device capacitances affect high-frequency response.

At high frequencies, ac model for multi-stage

amplifier is as shown.

Chap 17- 45


46/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Frequency Response

Multistage Amplifier (Estimate ofL)

SCTC for each of the six independent coupling and bypass capacitors has

to be determined.

R1S

RI

(RGRin1

)10k1M

R1S

1.01M

R2SRS1 1

gm1

200 10.01S

66.7

R3S

RD1

RO1

R

B2Rin2

R3S

RD1

ro1

R

B2rp2

2.69 k

Rth2 RB2 RD1 ro1 571

R4S

RE2

Rth2

rp2

o2

119.4

R5S

RC2

RO2

R

B3R

in3

R5S

RC2

ro2

R

B3rp3

(o3

1)(RE3

RL

)

R5S18.4kR

th3R

B3R

C2r

o2 3.99 k

R6S

RL

RE3

Rth3

rp3

o3

1 311

L 1

RiS

Cii

1

n 3300 rad/s

fL

L

2p 530 Hz

Chap 17- 46


47/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Frequency Response

Multistage Amplifier (Estimate ofH)

OCTC for each of the two capacitors associated with each transistor has to

be determined.

ForM1,

ForQ2,

RL1

RI12

rp2

478

RthCT1Rth CGS1CGD1 1gm1RL1

RL1

Rth

Rth2

RI12

ro1

570

rpo2

rp2

(Rth2

rx2

) 610

RL2 RI23 Rin3

RL2

RI23

rp3

(o3

1)(RE3

RL

)

RL2

3.54 k

rpo2CT2

rpo2

Cp2

Cm2

1gm2RL2

RL2

rpo2

rpo2CT2

1.74107s

ForQ3,

Rth3 RI23 ro2 3.99 k

Rp3OCp3R

m3OCm3(Rth3

rx3

)

1gm3REE

Cp3(Rth3

rx3

)Cm3

Rp3OCp3R

m3OCm31.51108 s

H

1

RioCi

i 1

m

3.3810

6

rad/s

fHH

2p538 kHz

Chap 17- 47


48/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Single-pole Op Amp Compensation

Frequency compensation forces overall amplifier to have a single-pole

frequency response by connecting compensation capacitor around second gain

stage of the basic op amp.

Av(s)

Ao

B

sB

T

sB

Chap 17- 48


49/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Three-stage MOS Op Amp Analysis

Input stage is modeled by its Norton

equivalent- current source Gmvdm and output

resistanceRo. Second stage has gain of

gm5ro5= mf5 and follower output stage is a

unity-gain buffer. Vo(s) = Vb(s) = -Av2Va(s)

Av(s)

Vo(s)

Vdm

(s)

-Av2

Va(s)

Vdm

(s)

Gm

RoA

v2

1sRoC

C(1A

v2)

Av(s)

T

sB

A

o

B

sB

B 1R

oC

C(1A

v2) T GmAv2

CC

(1Av2

)

For large Av2

, T

G

m

CC

Chap 17- 49


50/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Transmission Zeros in FET Op Amps

Incorporating the zero determined bygm5 in the analysis,

This zero cant be neglected due to lowratio of transconductances ofM2 andM5.Zerocan be canceled by addition ofRZ=1/gm5.

Avth

(s) (gm5

ro5

)1(s/

Z)

1(s/P1

)

Z

gm5

CCCGD5

T

gm5

gm2

P1

1

RoCT

CT

CGS5

(CC

CGD5

)1mf5

r

o5

Ro

Z

1(1/g

m5)R

Z

C

C

Chap 17- 50

Note error

in Eq.

17.182.


51/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Bipolar Amplifier Compensation

Bipolar op amp can be compensated in

the same manner as a MOS amplifier

Transmission zero occurs at too high a

frequency to affect the response due to

higher transconductance of BJT that

FET for given operating current.

Unity gain frequency is given by:

Z

g

m5

CC

T

IC5

IC2

T

g

m2

CC

40I

C2

CC

20I

C1

CC

Chap 17- 51


52/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Slew rate of Op Amp

Slew-rate limiting is caused by limitedcurrent available to charge/discharge

internal capacitors. For very largeAv2,

amplifier behaves like an integrator:

For CMOS amplifier,

For bipolar amplifier,

IC1

CC

dvB

(t)

dt C

C

dvo(t)

dt

SRdv

o(t)

dtmax

I1

CC

T

Gm

/I1

SR TG

m/I

1

T I1K

n2

SR

T

Gm

/I1

T

20

Chap 17- 52


53/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Tuned Amplifiers

Amplifiers with narrow bandwidth are often required in RF

applications to be able to select one signal from a large number of

signals.

Frequencies of interest > unity gain frequency of op amps, so active

RC filters cant be used.

These amplifiers have high Q (fHandfL close together relative to center

frequency)

These applications use resonant RLC circuits to form frequency

selective tuned amplifiers.

Chap 17- 53


54/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Single-Tuned Amplifiers

RLC network selects thefrequency, parallel combination

ofRD,R3 and ro set the Q and

bandwidth.

Neglecting right-half plane

zero,

Av

(s)V

o(s)

Vi(s)

sCGD

gm

GPs(CCGD)(1/sL)G

Pg

oG

DG

3

Av(s) Amid

s

o

Q

s2 soQ

o

2

o

1L(CC

GD)

QoR

P(CC

GD)

RP

oL

Chap 17- 54


55/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Single-Tuned Amplifiers (contd.)

At center frequency,s =jo,

Av =Amid.

AmidgmRPgm(ro RD R3)

BW

o

Q 1

RP

(CCGD

)

o

2L

RP

Chap 17- 55

f d d A


56/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Use of tapped Inductor- Auto

Transformer

CGD and ro can often be small enough todegrade characteristics of the tuned

amplifier. Inductor can be made to work

as an auto transformer to solve this

problem.

These results can be used to transform the

resonant circuit and higherQ can be

obtained and center frequency doesnt

shift significantly due to changes in CGD.

Similar solution can be used if tuned

circuit is placed at amplifier input instead

of output

Vo(s)I2(s) nV1(s)Is(s)/n n

2 V1(s)Is(s) Zs(s) n2Zp(s)

Chap 17- 56


57/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Multiple Tuned Circuits

Tuned circuits can be placed at both inputand output to tailor frequency response.

Radio-frequency choke (an open circuit atthe operating frequency) is used for biasing.

Synchronous tuning uses two circuits tunedto same center frequency for high Q.

Stagger tuning uses two circuits tuned toslightly different center frequencies torealize broader band amplifiers.

Cascode stage is used to provide isolationbetween the two tuned circuits andeliminate feedback path between them dueto Miller multiplication.

BWn

BW1

21/ n 1

Chap 17- 57


58/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Mixers: Conversion Gain

Amplifiers discussed so far have always been assumed to be linear and gainexpressions involve input and output signals at same frequency.

Mixers are nonlinear devices whose output signal frequency is different from

the input signal frequency.

A mixers conversion gain is the ratio of phasor representation of output signalto that of input signals, the fact that the two signals are at different frequenciesis ignored.

Chap 17- 58


59/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Single-Balanced Mixer

Eliminates one of the two inputsignals from the output.

No signal energy appears at1 , but

2 appears in output spectrum, so

circuit is single-balanced.

Up-conversion uses component (2-1) and down-conversion uses

(2+1) component.

iEE

IEE

I1sin

1t

v2(t)

4

npnodd sinn2t

Vo(t) 4

npnodd

IEE

RC

sinn2t

I

1R

C

2cos(n

2

1)t

I1R

C

2cos(n

2

1)t

Chap 17- 59

D bl B l d Mi / M d l


60/61

Jaeger/Blalock

2/9/04


McGraw-Hill

Double-Balanced Mixer/ Modulator

The Gilbert Multiplier

Double-balanced mixers dontcontain spectral components ateither of the two input frequencies.

Modulator applications give doublesideband suppressed carrier outputsignal. Amplitude-modulated signal

can also be obtained if

iC1

IBB

V

m

2R1

sinm

t iC2

IBB

V

m

2R1

sinm

t

vo(t)V

mRCR

1

4npnodd

cos(nc

m

)t cos(nc

m

)t

vo(t)Vm

RC

R1

4

npnodd sinn

ctM

2cos(ncm)tM

2cos(ncm)t

v1 Vm(1Msin

mt)

Chap 17- 60


61/61

End of Chapter 17

chap17 frequency response

Documents