Chap10 Analog Systems

7/27/2019 Chap10 Analog Systems

1/39

Jaeger/Blalock7/1/03

Microelectronic Circuit DesignMcGraw-Hill

Chapter 10

Analog Systems

Microelectronic Circuit Design

Richard C. Jaeger

Travis N. Blalock

Chap10 - 1


2/39



Chapter Goals

Develop understanding of linear amplification concepts such as:

Voltage gain, current gain, and power gain,

Gain conversion to decibel representation,

Input and output resistances,

Transfer functions and Bode plots,

Cutoff frequencies and bandwidth,

Low-pass, high-pass, band-pass, and band-reject amplifiers,

Biasing for linear amplification,

Distortion in amplifiers, Two-port representations of amplifiers,

g-, h-, y-, and z-parameters,

Use of transfer function analysis in SPICE.

Chap10 - 2


3/39



Example of Analog Electronic System:

FM Stereo Receiver

Linear functions: Radio and audio frequency amplification, frequency

selection (tuning), impedance matching(75-W input, tailoring audio

frequency response, local oscillator

Nonlinear functions: DC power supply(rectification), frequency

conversion (mixing), detection/demodulation

Chap10 - 3


4/39



Amplification: Introduction

A complex periodic signal can be represented as the sum of many

individual sine waves. We consider only one component with

amplitude VS=1 mV and frequency wSwith 0 phase (signal is used

as reference):

Amplifier output is sinusoidal with same frequency but different

amplitude VO and phase :

tssVsv wsin

)(sin w tsoVov

Chap10 - 4


5/39



Amplification: Introduction (contd.)

Amplifier output power is:

Here,PO = 100 W andRL=8 W

Output power also requires output current which is:

Input current is given by

phase is zero because circuit is purely resistive.

LR

oVoP

12

2

V40810022 L

RoPoV

)(sin w tsoIoi

A58V40

W

LR

oVoI

A81082.1k50k5

V3-10 WW

in

RsRsV

sI

Chap10 - 5


6/39



Amplification: Gain

Voltage Gain:

Magnitude and phase of voltage gain are given by

and

For our example,

Current Gain:

Magnitude of current gain is given by

sVoV

sVoV

svov

vA 0

sVoV

vA vA

4104V310

V40

sVoV

vA

sIoI

sIoI

sioi

iA

0

81075.2A8-101.82

5A

sIoI

iA

Chap10 - 6


7/39



Amplification: Gain (contd.)

Power Gain:

For our example,

On decibel scale, i.e. in dB

iAvA

sIoI

sVoV

sIsV

oIoV

sPoP

PA

22

22

131010.181082.1310

540

P

A

PA

PdBA log10

vAvdBA log20

iA

idBA log20

Chap10 - 7


8/39



Amplifier Biasing for Linear Operation

iv

IV

Iv

VI= dc value ofvI,

vi = time-varying component

For linear amplification- vImust be biased in desired region of outputcharacteristic by VI.

ovOV

Ov

If slope of output characteristic is positive, input and output are in

phase (amplifier is non-inverting).

If slope of output characteristic is negative, input and output signals

are 1800 out of phase (amplifier is inverting).

Chap10 - 8


9/39




(contd.)

IV

IvI

vov

vA

Voltage gain depends on bias

point.Eg: if amplifier is biased at VI=

0.5 V, voltage gain will be +40

for input signals satisfying

If input exceeds this

value, output is distorted due tochange in amplifier slope.

V1.0i

v

Chap10 - 9


10/39




(contd.)

Output signals for 1 kHZ sinusoidal

input signal of amplitude 50 mV

biased at VI= 0.3 V and 0.5V:V)2000sin05.03.0( t

IAv

ForVI=0.3V:

V)2000sin14( tOA

v

ForVI=0.5V:

Gain is 20, output variesabout dc level of 4 V.

V)2000sin05.05.0( t

IB

v

V)2000sin210( tOB

v

Gain is 40, output varies

about dc level of 10 V.

Chap10 - 10


11/39



Distortion in Amplifiers

Different gains for positive and negative values of input cause

distortion in output.

Total Harmonic Distortion (THD) is a measure of signal distortion

that compares undesired harmonic content of a signal to the desired

component.

Chap10 - 11


12/39



Total Harmonic Distortion

...)3

sin3(3

)2

sin2(2

)1

(sin1

)( www toVtoVtoVoVtv

dc desired

output

2nd harmonic

distortion

3rd harmonic

distortion

1

2

%100 2

V

Vn

THD

Numerator= sum of rms amplitudes of distortion terms,

Denominator= desired component

Chap10 - 12


13/39



Two-port Models for Amplifiers

Simplifies amplifier-behavior modeling in complex systems.

Two-port models are linear network models, valid only under

small-signal conditions.

Represented byg-, h-, y- andz-parameters.

(v1, i1) and (v2, i2) represent signal components of voltages and

currents at the network ports.

Chap10 - 13


14/39



g-parameters

Using open-circuit (i=0) and short-

circuit (v=0) termination

conditions,

2

i

221

v

212

v2i

121v

111i

gg

gg

01v2

i2

v

22

02i1v

2v

21

01v2

i1i

12

02i1

v1i

11

g

g

g

g Open-circuit input

conductance

Reverse short-circuit

current gain

Forward open-circuit

voltage gain

Short-circuit output

resistance

Chap10 - 14


15/39



g-parameters:Example

Problem:Findg-parameters.

Approach: Apply specified boundary

conditions for eachg-parameter, use

circuit analysis.

Forg11 andg21: apply voltage v1 toinput port and open circuit output port.

Forg12 andg22: apply current i2 to

output port and short circuit input port.

998.0)k200)(51(

1102i1v

2v

21

S81079.9

)k200(514102

1

02i1

v1i

11

W

WW

gg

g

0196.0k20

391

01

v2i

2v

22

391

k20

51

k200

11

01

v2i1i

12

W

W

W

W

W

g

g

2i21091.3

1v998.0

2v

2i21096.1

1v81079.9

1i

Chap10 - 15


16/39



Hybrid orh-parameters



conditions,

2

v

221

i

212

i2

v121

i111

v

hh

hh

01i2

v2i

22

02

v1i

2i

21

01i2

v1v

12

02

v1i1v

11

h

h

h

h Short-circuit input

resistance

Reverse open-circuit

voltage gain

Forward short-circuit

current gain

Open-circuit output

conductance

Chap10 - 16


17/39



h-parameters:Example

Problem:Find h-parameters for the

same network (used ing-parameters

example).


conditions for each h-parameter, usecircuit analysis.

Forh11 and h21: apply current i1 to input

port and short circuit output port.

Forh12 and h22: apply voltage v2 to

output port and open circuit input port.

51

02

v1i2i

21

4102

02

v1i1

v

11

W

h

h

S6105k200

1

01i2

v2i

22

10

1i2

v1

v

12

W

h

h

2v6105

1i51

2i

2v

1i4102

1v

Chap10 - 17


18/39



Admittance ory-parameters



conditions,

2

v

221

v

212

i2

v121

v111

i

yy

yy

01v2

v2i

22

02

v1v

2i

21

01v2

v1i

12

0

2

v1v1i

11

y

y

y

y Short-circuit input

conductance

Reverse short-circuit

transconductance

Forward short-circuit

transconductance

Short-circuit output

conductance

Chap10 - 18


19/39



y-parameters:Example

Problem:Findy-parameters for the


example).


conditions for eachy-parameter, use

circuit analysis.

Fory11 andy21: apply voltage v1 to input

port and short circuit output port.

Fory12 andy22: apply voltage v2 to

output port and short circuit input port.

S31055.2k20

51

02

v1v2i

21

S5105k20

1

02

v1v1i

11

W

W

y

y

S31056.2391

1

01i2

v1i

22

S5105k20

1

01

v2v1i

12

W

W

y

y

2v31056.2

1v31055.2

2i

2v5105

1v5105

1i

Chap10 - 19


20/39



Impedance orz-parameters



conditions,

2

i

221

i

212

v2i

121i

111v

zz

zz

01i2

i2

v

22

02i1i

2v

21

01i2

i1v

12

0

2

i1i1v

11

z

z

z

z Open-circuit input

resistance

Reverse open-circuit

transresistance

Forward open-circuit

transresistance

Open-circuit output

resistance

Chap10 - 20


21/39



z-parameters:Example

Problem:Findz-parameters for the


example).

Approach: Apply specified

boundary conditions for eachz-

parameter, use circuit analysis.

Forz11 andz21: apply current i1 to

input port and open circuit output

port. Forz12 andz22: apply current i2

to output port and open circuit inputport.

W

WWW

M2.10

0

2

i1i2

v

21

M2.10)k200(51k20

02i1

i1v

11

z

z

W

W

k200

01i2

i2

v

22

k200

01i2

i1

v

12

z

z

2i51000.2

1i71002.1

2v

2i51000.2

1i71002.1

1v

Chap10 - 21


22/39



Mismatched Source and Load

Resistances: Voltage Amplifier

LR

outR

LR

inRsR

inR

A

sVoV

vA

inRsR

inR

LR

outR

LR

A

sv1v

1vov

IfRin >>Rs andRout


23/39



Mismatched Source and Load

Resistances: Current Amplifier

h-parameter representation (h12=0) with Norton equivalent of input source:

LR

outR

outR

inRsRsR

sIoI

iA

inRsRsR

LR

outR

outR

si1i

1ioi

IfRs >>Rin andRout>> RL,

i

A

In an ideal current amplifier,andRin=0

outR

Chap10 - 23


24/39



Amplifier Transfer Functions

ssV

soVsvA

Av(s)=Frequency-dependent voltage gain

Vo(s) and Vs(s) = Laplace Transforms of input and output voltages of

amplifier, w js

3...

21

...21

pspsps

mzszszsKsvA

(-z1, -z2,-zm)=zeros (frequencies for which transfer function is zero)

(-p1, -p2,-pm)=poles (frequencies for which transfer function is infinite)

www jvAjvAjvA

Bode plots display magnitude of the transfer function in dB and the

phase in degrees (or radians) on a logarithmic frequency scale..

(In factorized form)

(In polar form)

Chap10 - 24


25/39



Low-pass Amplifier: Description

Amplifies signals over a range of frequencies including dc.

Most operational amplifiers are designed as low pass amplifiers.

Simplest (single-pole) low-pass amplifier is described by

Ao = low-frequency gain or mid-band gain

wH= upper cutoff frequency or upper half-power point of

amplifier.

H

soA

Hs

HoA

svA

ww

w

1

Chap10 - 25


26/39



Low-pass Amplifier: Magnitude Response

ForwwH:

ForwwH:

Gain is unity (0 dB) atwAowH , called gain-bandwidth product

Bandwidth (frequency range with constant amplification )= wH (rad/s)

22log20log20dB

22

HHoAjvA

H

HoA

Hj

HoA

jvA

wwww

ww

w

ww

w

w

dB)log20( oAoAjvA w

dBlog20log20

HoA

HoA

jvAw

w

w

w

w

dB3)log20(2

oA

oAjvA w

Chap10 - 26


27/39



Low-pass Amplifier: Phase Response

IfAo

positive: phase angle = 00

IfAo negative: phase angle = 1800

At wC: phase =450

One decade below wC: phase =5.70

One decade above wC: phase =84.30

Two decades below wC: phase =00

Two decades above wC: phase =900

H

oA

H

j

oAjvA ww

w

ww1tan

1

Chap10 - 27


28/39



RC Low-pass Filter

Problem: Find voltage transfer

function

Approach: Impedance of the where

capacitor is 1/sC, use voltage

division

H

sRR

R

sVoV

sCR

sCRR

sCR

sCR

sVoV

w1

1

21

2

/12

/2

1

/12

/2

CRRH

21

1w

Chap10 - 28


29/39



High-pass Amplifier: Description

True high-pass characteristic impossible to obtain as it requires

infinite bandwidth.

Combines a single pole with a zero at origin.

Simplest high-pass amplifier is described by

wH= lower cutoff frequency or lower half-power point of amplifier.

Ls

soAsvAw

Chap10 - 29


30/39



High-pass Amplifier: Magnitude and

Phase Response

Forw>>wL :

Forw


31/39

Jaeger/Blalock7/1/03 Microelectronic Circuit DesignMcGraw-Hill

RC High-pass Filter

Problem: Find voltage transfer

function

Approach: Impedance of the where

capacitor is 1/sC, use voltage

division

Ls

s

RR

R

sVoV

RsC

R

R

sVoV

w21

2

2

1

1

2

CRRL

21

1w

Chap10 - 31


32/39


Band-pass Amplifier: Description

Band-pass characteristic obtained by combining highpass and

low-pass characteristics.

Transfer function of a band-pass amplifier is given by

Ac-coupled amplifier has a band-pass characteristic:

Capacitors added to circuit cause low frequency roll-off Inherent frequency limitations of solid-state devices cause

high-frequency roll-off.

1

1)())((

H

sLs

soA

Hs

Ls

HsoAsvA

wwww

w

Chap10 - 32


33/39


Band-pass Amplifier: Magnitude and

Phase Response

The frequency response shows a wide band of operation.

Mid-band range of frequencies given by , whereHLwww oAjvA )( w

)22)(22())((

HL

HoA

Hj

Lj

HjoAjvA

wwww

ww

wwww

ww

w

Chap10 - 33


34/39


Band-pass Amplifier: Magnitude and

Phase Response (contd.)

At both wHand wL, assumingwL


35/39


Narrow-band or High-Q Band-pass

Amplifiers

Gain maximum at center frequency woand

decreases rapidly by 3 dB at wHand wL.

Bandwidth defined aswH - wL, is a small

fraction ofwowith width determined by:

For high Q, poles will be complex and

Phase response is given by:

BWo

f

Lf

Hf

of

LHoQ ww

w

22

oQo

ss

Qos

oAsvA

w

w

w

2211tan090

ww

www

o

oQo

AjvA

Chap10 - 35


36/39


Band-Rejection Amplifier or Notch

Filter

Gain maximum at frequencies far from wo

and exhibits a sharp null at wo.

To achieve sharp null, transfer function has

a pair of zeros onjwaxis at notch frequency

wo , and poles are complex.

Phase response is given by:

22

22

oQ

oss

osoAsvA

www

2211tan22

ww

wwwww

o

oQoo

AjvA

Chap10 - 36


37/39


All-pass Function

Uniform magnitude response at all frequencies.

Can be used to tailor phase characteristics of a signal

Transfer function is given by:

For positive Ao,

osos

oAsvA ww

o

jvA

oAjvA

www

w

1tan2

Chap10 - 37


38/39


Complex Transfer Functions

1

5

1

431

2mid

5431

2

wwww

w

wwww

w

ssss

ssA

ssss

sKssvA

Amplifier has 2 frequency ranges withconstant gain. Midband region is

always defined as region of highest

gain and cutoff frequencies are

defined in terms of midband gain.

2

midA

HjvAL

jvA ww

ww

2 3434BW

ff

Since wH= w4and wL = w3,

Chap10 - 38


39/39

J /Bl l k Mi l t i Ci it D i

Bandwidth Shrinkage

If critical frequencies arent widely spaced, the poles and zeros

interact and cutoff frequency determination becomes

complicated.

Example : for which ,Av(0) =Ao

Upper cutoff frequency is defined by or

2)1

(

2

1w

w

s

oA

svA

2oA

HjvA w 222

1

1

Ao

H

Ao

w

w

Solving forwHyields wH=0.644w1.The cutoff frequency of

two-pole function is only 64% that of a single-pole function.

This is known as bandwidth shrinkage.

Ch 10 39

Date post:	02-Apr-2018
Category:	Documents
Upload:	manh-cuong-tran
View:	226 times
Download:	0 times

Chap10 Analog Systems

Documents