analog (low freqency)

8/9/2019 Analog (low freqency)

1/143

Unit-1The transistor at low

frequency


2/143

Graphical Analysis of the

CE Configuration


3/143

CE configuration


4/143

Notation

S. No. Base(Collector )voltage withrespect toemitter

Base(Collector )current towardelectrode fromexternal circuit

1 Instantaneous total value vB ( vC) iB (iC)

2 Quiescent value VB (VC) IB (IC)

3 Instantaneous value of

varying component

vb (vc) ib (ic)

4 Effective value of varyingcomponent

Vb (Vc) Ib (Ic)

5 Supply voltage VBB (VCC ) --------


5/143

Total signal = DC component(Quiescent

value) + timevarying signal

ic = Ic+ ic


6/143

Out put nonlinear distortion

Input nonlinear distortion

Distortions


7/143

How to reduce thedistortion?

If the amplifier is biased so that Q is near thecenter of the iC-vCE plane there will be less

distortion if the excitation is a sinusoidal basevoltage than if it is a sinusoidal base current.

The dynamic load curve can be approximated by astraight line over a sufficiently small line segment,and hence, if the input signal is small, there will benegligible input distortion under any condition of

operation(current-source or voltage-source driver).


8/143

TWO PORT DEVICE AND THE

HYBRID MODEL


9/143

Two port network

Two portactive

device

i1 i2

v1 v2

+ +

- -

Out put portInput port

v1 = h11 i1 + h12 v2

i2 = h21 i1 + h22

v2

h11 , h12 , h21 and h22 are called the h or hybrid parameter.


10/143

1. hi = h11 = v1/i1 for v2 = 0; input resistance

with out put short-circuit(in ohms)

2. hr = h12 = v1/v2 for i1 = 0; reverse open

circuit voltage amplification

3. hf = h21 = i2/i1 for v2 = 0; forward current

transfer ratio or short-circuit current gain

4. ho = h22 = i2/v2 for i1 = 0; outputconductance with input short circuit


11/143

h-Parameter Models (small signalmodel)

v1 = hi i1 + hr v2i2 = hf i1 + ho v2

+

v2

-

+

v1

-

+

-

hi

hr v2hf i1 ho

i1 i2

Linear Active Network


12/143

The hybrid small signal models forcommon-emitter configuration

vb = hie ib + hre vc

ic = hfe ib + hoe vc

C +

vc

E -

+ B

vb

- E

+

-

hie

hre vchfe ib hoe

ib ic


13/143

e y r sma s gnamodels for common-

collector configuration

E +

vc

C -

+ B

vb

- C

+

-

hic

hrc vchfc ib hoc

ib ie

vb = hic ib + hrc ve

ie = hfc ib + hoc ve


14/143

Typical h-parameter values for atransistor(at IE = 1.3 mA)

S.No Parameter CE CC CB

01 hi = h11 1100 1100 21.6

02 hr = h12 2.5*10-4 ~1 2.9*10-4

03 hf = h21 50 -51 -0.98

04 ho = h22 24 A/V 25 A/V 0.49 A/V

05 1/ho 40K 40K 2.04 M


15/143

A basic amplifier circuit

Two-port active

Network (transistor)

Rs

VsV1

I2

IL

ZL

+

-

V2

Yo

I1

Zi

+

-

+

-

1

1

2

2

t t


16/143

na ys s o a trans storamplifier circuit using h-

parameter

Vo

hi

I1

hrV2hfI1 ho

I2

Vs

+

-

+

-

Rs

+

Vin

-

ZL

2

2

1

1

IL


17/143

Small-signal analysis of atransistor amplifier

S No Quantities

1 Current gain AI =-hf/(1+hoZL)

2 Voltage gain Av = AIZL/Zi

3 Input impedance Zi=-hi+ hr AI ZL

4 Output admittance Yo =ho- hfhr/( hi +Rs)

= 1/Zo

5 Over all voltage gain Avs = Av Zi/( Zi +Rs)

= AIZL/( Zi +Rs)

6 Over all current gain gain Avs = AI Rs/( Zi +Rs)

= AVs Rs/ ZL


18/143

Emitter Follower Circuit


19/143

Characteristics of emitter followercircuit

Voltage gain is close to unity. (AV = 0.997)

Input resistance is very high (hundreds of kilo ohms).(Ri = 409K)

Output resistance is very low (tens of ohms). (Ro = 41.2

ohms) It is use as a buffer stage which performs the resistance

transformation over a wide range of frequencies, withvoltage gain close to unity

Emitter follower increases the power level of the signal.

There is no phase shift between output and input ineither voltage or current.

ompar son o trans stor


20/143

ompar son o trans storconfigurations (RL=3K,

Rs=3K)SNo

Quantity CE CC CB

01 AI High(-46.5) High(47.5)Low(0.98)

02 AV High(-131) Low(0.99) High(131)

03 Ri Medium(1065)

High(144K)

Low(22.5)

04 Ro Medium(45.5K Low(80.5 High(1.72M


21/143

Miller's theorem

The Millers theorem establishes that in a linear circuit, ifthere exists a branch with impedanceZ, connecting twonodes with nodal voltages V1and V2, we can replace thisbranch by two branches connecting the correspondingnodes to ground by impedances respectivelyZ/ (1-K) and

KZ/ (K-1), where K= V2 / V1.

N

N1

N2

N1 N2

N


22/143

Miller's dual theorem

If there is a branch in a circuit with impedanceZconnectinga node, where two currents I

1and I

2converge, to ground, we

can replace this branch by two conducting the referredcurrents, with impedances respectively equal to (1- A

I)Z

and ( AI-1 )Z/ AI , where AI = -I2 / I1 .

Z1=

Z/(1-AI) Z2= (AI-1)Z/AI

N

N


23/143

Q.1 For the amplifier shown in fig calculate Ri, Ri,

AV, AVs and

AI=-I2/I1.

Vcc

RL=10K

R=10K

Rs=10K

I1

Ri

I2

Ri

+

-

Vo


24/143

Steps

Millers theorem is applied to the 200K ohmresistance R.

K=AV is the voltage gain from base to collector.

Assuming that gain is much larger than unity (-AV>>1

), then

200/(1-(1/AV)) 200

Independent DC voltage source is replaced by ashort circuit.


25/143

On applying MillersTheorem

Rs=10K

Z2=200/(1-(1/AV))

RL=10K

RiRi

Z1=200/(1-AV)

I1

I2


26/143

Simplified common-emitterhybrid model

Ib

Ie

Ic

hfeIb

B

E

C


27/143

Approximate CE model

It is valid if hoeRL < 0.1

Rs

VsVb

hie

Ib

hfe Ib

Ic

RLVc=Vo

B C

E E


28/143

Simplified hybrid model for thecommon-collector circuit

Vs

Rs

Ib

Vb

Ri

Ie

RoRo

RL

hfe Ib

(1+hfe)Ib

B

C

C

E


29/143

Bootstrapped Darlingtonequivalent circuit

hfe1 Vce1

+ -

hfe1 Vce1

hie1

hoe1

hie2

Re=Rc1 Re2

hfe2 Ib2

E1, B2

Ib1 Ib2

B1

N

C1

E1

C2

I

Vi


30/143

The low-frequency small-signalFET model

gm vgsrd vds

id

S -

G +

D

+

vgs

S -


31/143

The CS and the CDconfiguration

-VDD -VDD

Rd

vivo

vo

Rs


32/143

UNIT-2

TRANSISTOR AT

HIGHFREQUENCIES

The hybrid model for a


33/143

The hybrid- model for atransistor in the CE

configuration

rbb

rbe = 1/gbegm Vbe

rbc = 1/gbc

rce = 1/gce

Cc

Ce

B

E

C

E

B

Vbe

+

-

The hybrid model for a


34/143

The hybrid- model for atransistor in the CE

configuration at low frequencyrbcrbb

rbe = 1/gberce = 1/gce

gm Vbe

Ib Ic

Vbe

B

E

B

E

C


35/143

The hybrid small signal models forcommon-emitter configuration

C

E

B

E

+

-

hie

hre Vcehfe Ib hoe

Ib Ic


36/143

HYBRID- CONDUCTANCES

S.NO HYBRID-CONDUCTANCES

EQUATION

1 TransistorTransconductance gm =|Ic|/VT

2 The Input Conductance rbe = hfe/gm = hfeVT/|Ic|

Or gbe = gm/hfe

3 The FeedbackConductance rbc = rbe /hre Orgbc = hre/rbe

4 The Base SpreadingResistance

rbb = hie rbe

5 The OutputConductance gce = hoe (1+hfe)gbc = 1/rce


37/143

HYBRID- CAPACITANCE


38/143

Minority-carrier chargedistribution in the base region

Emitter Collector

p(0)

QB

x = Wx = 0

x

h i i


39/143

CE Short-Circuit Currentgain

The load RLon this stage is the

collector-circuit so that Rc= R

L.

For short circuit current gain RL = 0.


40/143

The hybrid- circuit for a singletransistor with a resistive load RL

rbe = 1/gbe

rbc =

1/gbc

rce = 1/gcegm Vbe

rbb

Ce

Cc

RL

B+

B

E-

C+

E-

Vce

VbeVbe

IL

Approximate equivalent circuit for the


41/143

Approximate equivalent circuit for thecalculation of the short-circuit CE

current gain

B C

EE

IL

Ii

gbe Ce + CC

gm Vbe

C i i h i i


42/143

Current gain with resistiveload

For current gain, RL 0.


43/143

The hybrid- circuit for a singletransistor with a resistive load RL

rbe = 1/gbe

rbc =

1/gbc

rce = 1/gcegm Vbe

rbb

Ce

Cc

RL

B+

B

E-

C+

E-

Vce

VbeVbe

IL

Apply millers theorem at node B


44/143

Apply miller s theorem at node Band C.

Ce

gbegbc (1-

k)

Cc (1-k)

gmVbeRL

B

E-

C+

E-

Cc (k-1)/kgbc (1-

k)/k

gce

Ii

F th i lifi ti f th


45/143

Further simplification of theequivalent circuit

Cc (1-k)CegbegmVbe

RL

E

C

IL

Ii

E

B

Si l t CE t i t


46/143

Single-stage CE transistoramplifier response

In the preceding section the Transistor is driven from an idealcurrent source which had infiniteresistance. Now we use a voltage

source Vs has a finite resistance Rs.

Equivalent circuit for frequency analysis of


47/143

Equivalent circuit for frequency analysis ofCE amplifier stage driven from a voltage

source

B B C

E

rbe

rbb

Vs

Rs

Ce

Cc

gmVbe RL Vo

Equivalent circuit for frequency


48/143

Equivalent circuit for frequencyanalysis of CE amplifier stage using

the miller effect

Vs

E

CB

Rs =Rs+ rbb

rbeCe

Cc (1-k)

gmVbe

Cc (k-1)/kRLVo


49/143

Neglecting the out put time constant

Rs =Rs+ rbb

Cc (1-k)gmVbe

Cerbe RLVs

B C

E

E itt f ll t hi h


50/143

Emitter follower at highfrequency

Vs

Rs RL CL

Vo

VCC

Hi h f i l t


51/143

High frequency equivalentcircuit of emitter follower

rbe

Ce

Vs

Rs

CLRLCc gmVbe

rbb EB

C

Vo

B

Vi Vi

Ri

Ib

e equ va ent c rcu t o


52/143

e equ va ent c rcu t oemitter follower using

millers theorem Vo

RLgbe (1-k)

Ce (1-k)

Cc

RsE

B

C

rbb

CL

B

gmVbe

gbe (k-

1)/k

Ce (k-1)/k

Vi

Yi

Ib


53/143

UNIT-3

MULTISTAGEAMPLIFIER

CLASSIFICATION OF


54/143

CLASSIFICATION OFAMPLIFIER

Amplifier are described in many ways,according to their

Frequency range

The method of operation

The ultimate use

The type of loadThe method of interstage coupling

etc..

CLASSIFICATION OF


55/143

CLASSIFICATION OFAMPLIFIERS BASED ON THE

FREQUENCY RANGE DC Amplifier(from zero frequency) Audio Amplifier(20 Hz to 20kHz)

Video or Pulse Amplifier(up to few mega

Hz)

Radio frequency Amplifier(few kilo Hz toHundreds of mega Hz)

Ultra high-frequency Amplifier( hundredsor thousands of mega Hz)


56/143

AMPLIFIER BASED ON

METHOD OF OPERATIONMETHODS OF OPERATION ARE DEPENDUPON Q-POINTAND THE EXTENT

OF THE CHARACHTERISTICS

CLASS A Amplifier

CLASS B Amplifier

CLASS AB Amplifier

CLASS C Amplifier


57/143

Class A amplifier


58/143

Output Stages And Power Amplifiers

Low Output Resistance no loss of gain

Small-Signal Not applicable

Total-Harmonic Distortion (fraction of %)

Efficiency

Temperature Requirements


59/143

Collector current waveforms for transistors operating in (a) class A, (b) class B, (c) class

AB, and (d) class C amplifier stages.

An emitter follower (Q1) biased with aClass A


60/143

(Q1)

constant current I supplied by transistorQ2.

Class A

Transfer Characteristics

Transfer characteristic of the emitter follower. This linearcharacteristic is obtained by neglecting the change invBE1

with iL. The maximum positive output is determined by

the saturation ofQ1.

In the negative direction, the limit of

the linear region is determined either byQ1turning off or

by Q2saturating, depending on the values ofIandR

L.

Class A


61/143

Crossover distortion can be eliminated by biasing the transistors at a small,

non-zero current.

A bias Voltage VBB is applied between Qn and Qp.

For vi = 0, vo = 0, and a voltage VBB/2 appears across the base-emitter junctionof each transistor.

iN iP IQ IS e

VBB

2 VT

VBB is selected to result the required quiscent current IQ

vo viVBB

2+ vBEN

iN iP iL+

vBEN vEBP+ VBB VT lniN

IS VT ln

iP

IS +2 VT ln

iQ

IS

iN2

IQ2

iN2

iL iN IQ2

0

Class A

Transfer

Characteristics

Class A


62/143

Class A

Transfer CharacteristicsFrom figure 9.3 we can see that

vomax VCC VCE1sat

In the negative direction, the limite of the linear region is

determined either by Q1 turning off

vOmin I RL

or by Q2 saturating

vOmin VCC VCE2sat+

Depending on the values of I and RL. The absolutely lowest

output voltage is that given by the previous equation and is

achieved provided that the bias current I is greater than the

magnitude of the corresponding load current

IVCC VCE2sat+

RL

Class A


63/143

Class A

Transfer Characteristics

Exercises D9.1 and D9.2

Class A


64/143

Class A

Signal Waveforms

0 5 1 01

0

1

v o t( )

t

0 5 1 00

1

2

v c E 1t( )

t

0 5 1 00

1

2

i c 1t( )

t0 5 1 0

0

0 . 5

1

p D 1t( )

t


65/143

Class A


66/143

Class A

Power Conversion Efficiency

load_power PL( )

supply_power PS( )

Vo average voltagePL1

2

Vo2

RL

PS 2 VCC I

1

4

Vo2

I RL VCC

1

4

Vo

I RL

Vo

VCC

Vo VCC Vo I RL

maximum efficiency is obtained when

Vo VCC I RL

Class A


67/143

Class A

Exercise 9.4

Vopeak 8:= I 100 10 3:= RL 100:= VCC 10:=

PL

Vopeak

2

2

100:= P

L0.32=

Pplus VCC I:= Pplus 1=

Pminus VCC I:= Pminus 1= PS Pplus Pminus+:=

PL

PS:= 0.16=


68/143

Biasing the Class B Output

No DC current is used to bias this configuration.

Activated when the input voltage is greater than the Vbefor the transistors.

npn Transistor operates when positive, pnp when negative.

At a zero input voltage, we get no output voltage.

Class A


69/143

CLASS A

Many class A amplifiers use the same transistor(s) for both halves of the audio waveform.

In this configuration, the output transistor(s) always has current flowing through it, even if it

has no audio signal (the output transistors never 'turn off'). The current flowing through it is

D.C.

A pure class 'A' amplifier is very inefficient and generally runs very hot even when there isno audio output. The current flowing through the output transistor(s) (with no audio signal)

may be as much as the current which will be driven through the speaker load at FULL

audio output power. Many people believe class 'A' amps to sound better than other

configurations (and this may have been true at some point in time) but a well designed

amplifier won't have any 'sound' and even the most critical 'ear' would be hard-pressed to

tell one design from another.

NOTE: Some class A amplifiers use complimentary (separate transistors for positive and

negative halves of the waveform) transistors for their output stage.

Class A

Power Conversion Efficiency

Class B CLASS 'B'


70/143

Class B output stage.

Class B

Circuit Operation

CLASS 'B'

A class 'B' amplifier uses complimentary transistors for

each half of the waveform.

A true class 'B' amplifier is NOT generally used for audio. In

a class 'B' amplifier, there is a small part of the waveform

which will be distorted. You should remember that it takes

approximately .6 volts (measured from base to emitter) to

get a bipolar transistor to start conducting. In a pure class

'B' amplifier, the output transistors are not "biased" to an

'on' state of operation. This means that the the part of the

waveform which falls within this .6 volt window will not bereproduced accurately.

The output transistors for each half of the waveform

(positive and negative) will each have a .6 volt area in

which they will not be conducting. The distorted part of the

waveform is called 'crossover' or 'notch' distortion.

Remember that distortion is any unwanted variation in asignal (compared to the original signal). The diagram below

shows what crossover distortion looks like.

Class B


71/143

Transfer characteristic for the class B output stage in Fig. 9.5.

Class B

Circuit Operation


72/143


73/143

vb

vs

1+( ) re par RL ro,( )+( )RS 1+( ) re par RL ro,( )+( )+

vo

vb

par ro RL,( )

re par ro RL,( )+

Rs will be small for most

configurations, so the vb/vs will be a

little less than unity. The same is true

for re, so vo/vb will be a little less thanunity making our vo/vs a little less

than unity.

Emitter Follower Configuration (Chapter 4)

Characteristics of the Emitter Follower:

High Input Resistance

Low Output Resistance

Near Unity Gain


74/143

Transfer Characteristic


75/143

Push-Pull Nature of Class B

Push: The npn transistor will push the current to ground

when the input is positive.

Pull: The pnp transistor will pull the current from the

ground when the input is negative.


76/143

Crossover Distortion

The Crossover Distortion is due to the dead band of input

voltages from -.5V to .5V. This causes the Class B outputstage to be a bad audio amplifier. For large input signals,

the crossover distortion is limited, but at small input

signals, it is most pronounced.


77/143

Illustrating how the dead band in the class B transfercharacteristic results in crossover distortion.

Graph of Crossover Distortion

Power Efficiency


78/143

Power Efficiency

PL

1

2

Vop2

RL

Load Power: Since each transistor is only conducting for

one-half of the time, the power drawn from

each source will be the same.

Ps1

Vop

RL

VCC

PL

2 Ps

1

2

Vop2

RL

21

Vop

RL

VCC

This efficiency will be at a max when

Vop is at a max. Since Vop cannot

exceed Vcc, the maximum efficiency

will occur at pi/4.

4

Vop

VCC

max4

This will be approximately 78.5%,

much greater than the 25% for

Class A.

Class AB


79/143

C ass

Circuit Operation

Crossover distortion can be eliminated by biasing the transistors at a small,

non-zero current.A bias Voltage VBB is applied between Qn and Qp.

For vi = 0, vo = 0, and a voltage VBB/2 appears across the base-emitter junction

of each transistor.

iN iP IQ IS e

VBB

2 VT

VBB is selected to result the required quiscent current IQ

vo vi

VBB

2+ vBEN

iN iP iL+

vBEN vEBP+ VBB VT lniN

IS

VT ln

iP

IS

+ 2 VT ln

iQ

IS

iN2

IQ2

iN

2iL

iN

IQ

2

0

Class AB


80/143

Output Resistance

Class AB


81/143

Exercise 9.6

Calvin College - ENGR 332

Class AB Output Stage Amplifier

Consider the class AB circuit (illustrated below) with Vcc=15 V, IQ=2 mA, RL=100 ohms.

Determine VBB. Determine the values of iL, iN, iP, vBEN, vEBP, vI , vO/vI, Rout, and vo/vi versusvO for vO varying from -10 to 10V.

Note that vO/vI is the large signal voltage gain and vo/vi is the incremental gain obtained as

RL/(RL+Rout). The incremental gain is equal to the slope of the transfer curve.

Assume QN and QP to be matched, with IS=10E-13.

Class AB


82/143

Exercise 9.6

under quiescent conditions iN=iP=IQ vO=vI=0

Solving for VBB

VBB 1:= IS 1013:= VT 0.025:= IQ 2 10

3:= RL 100:=

Given

IQ IS e

VBB

2

VT

VBB Find VBB( ):= i 0 100..:= VBB 1.186=

Class AB


83/143

Exercise 9.6

vOi

10i

5+:= iL

i

vOi

RL:=

10 0 10

0iLi

vOi

Class AB

Solving for iN


84/143

Exercise 9.6

vBENi

VTln

iNi

IS

:=

10 5 00.01

0.1

1

10

1001 .10

3

iPi 1000

vOi

iPi

iNi

iLDi

:=

iN10

4.997 105

=

10 5 0 5 100.01

0.1

1

10

100

1 .103

iNi 1000

vOi

iNi

iNN IQi

iLi

,( ):=

iLDi

iLi

:=IQi

0.002:=i 0 100..:=

iNN IQ iLD,( ) Find iN( ):=

iN

2

iLDiN IQ2

0

Given

IQ 0.002:=iLD 0.02:=iN 0.02:=initial guesses

Class AB


85/143

Exercise 9.6

10 5 00.01

0.1

1

10

100

1 .103

iPi 1000

vOivBEN

iVT ln

iNi

IS

:=

10 5 00.5

0.6vBENi

vOi

vEBPi

VTln

iPi

IS

:=

10 5 0

0.6vEBPi

Class AB


86/143

Exercise 9.6 vIi

vOi

vBENi

+VBB

2:=

10 5 010

0

vIi

vOi

vOvIi

vOi

vIi

:=

10 5 00

0.5vOvIi

vOi

Class AB


87/143

Exercise 9.6

10 5 00

0.5vOvIi

vOiRout

i

VT

iPi iNi+:=

10 5 0 5 10

0

5

Rout i

vOivovii

RL

RL Routi

+:=

0.95

1

vovii

UNIT 4


88/143

UNIT 4

FEEDBACK AMPLIFIER

Fee ac o Amp erCi it I


89/143

89

Circuits I Feedback is to return part of the output to the input for

a circuit/system (amplifiers in our context)

Feedback is very useful in Control Theory and Systemsand is well researched

Amplifier circuit can have negative feedback andpositive feedback. Negative feedback returns part ofthe output to oppose the input, whereas in positive

feedback the feedback signal aids the input signal. Both negative feedback and positive feedback are used

in amplifier circuits

Negative feedback can reduce the gain of the amplifier,but it has many advantages, such as stabilization of

gain, reduction of nonlinear distortion and noise,control of input and output impedances, and extensionof bandwidth

Concept of amplifier feedback


90/143

90

gainloop:

tcoefficienfeedback:

amplifiertheofgainloopopenthe:

amplifiertheofgainloopclosedthe:

feedbacknegativethen,if,1

A

A

A

AAA

A

x

xA

f

f

s

of

> fsissf xxxxA

AxxA

Thus, the closed-loop gain would be muchmore stable and is nearly independent of changes of open-loop gain

Thus, in a negativefeedback amplifier, the output takes the value to drive the amplifier

input to almost 0 (this is summing point constraints).

./1then,1If >> fAA

Amplifier negative feedback: reduce nonlineardistortion


91/143

91

010-for98.9

100for99.9


92/143

92

2

22

2

21

1

21

21

11

221

2

)()(

)(

1)(

1)()(

)()(

)()()(

)()()(

Ax

xSNR

AA

Atx

AA

AAtxtx

txAtx

txtxAtx

txtxtx

noise

s

noiseso

o

noise

os

=

++

+=

=

+==

2

2

11

)(

)(

)()()(

noise

s

noiseso

x

xSNR

AtxAtxtx

=

+=

If an amplifier (assumed to be noisefree or very low noise) is placed beforethe noisy amplifier, then the Signal-to-Noise (SNR) ratio is greatly enhanced(by a factor equal to the precedingamplifier gain)

As a summary, negative feedback is very useful in amplifier circuits. Itcan help stabilize the gain, reduce nonlinear distortion and reduce noise.

Also, as will be shown later, negative feedback in amplifiers can alsocontrol input and output impedance.

Amplifier negative feedbacktypes


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93

types

Amplifier negative feedbacktypes


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94

types If the feedback network samples the output voltage, it is voltage

feedback. If it samples the output current, it is current feedback.

The feedback signal can be connected in series or in parallel with thesignal source and the amplifier input terminals, so called seriesfeedbackandparallel feedback.

So, there are four types of negative feedback in amplifier circuits: Series voltage feedback (corresponding to (a) in previous slide)

Series current feedback (corresponding to (b) in previous slide)

Parallel voltage feedback (corresponding to (c) in previous slide) Parallel current feedback (corresponding to (d) in previous slide)

In voltage feedback, the input terminals of the feedback network are in parallelwith the load, and the output voltage appears at the input terminals of thefeedback block.

Whereas in current feedback, the input terminals of the feedback network are

in series with the load, and the load current flows through the input of thefeedback block.

As a result, a simple test on the feedback type is to open-circuit or short-circuitthe load. If the feedback signal vanishes for an open-circuit load, then it iscurrent feedback. If the feedback signal vanishes for a short-circuit load, it isvoltage feedback.

Effect of negative feedback ongain


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95

gain

vvvffA

AA

A

AA +=+= 1so,1,generalIn

m

mmff

G

GG

A

AA

+

=

+

=1

soamplifier,uctancetranscondabymodeledisthis,1

generalIn

m

mmff

R

RR

A

AA

+=

+=

1soamplifier,stancetransresiabymodeledisthis,

1generalIn

In series voltage feedback, input signal is voltage and output voltageis sampled, so it is natural to model the amplifier as a voltageamplifier.

Amplifier employing series current feedback is modeled as atransconductance amplifier.

Amplifier employing parallel voltage feedback is modeled as atransresistance amplifier.

Amplifier employing parallel current feedback is modeled as a currentamplifier.

i

iiff

A

AA

A

AA

+=

+=

1soamplifier,currentabymodeledisthis,

1generalIn

Negative feedback on input impedance


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96

)1( ARR iif += iR

)1/( ARR iif +=

For series feedback, the following model can be used for analysis ofinput impedance (the output x could be either voltage or current)

If the input impedance of the open-loop amplifier is Ri, then theclosed-loop impedance is

so, series feedback (either current or voltage) increase the inputimpedance

Similarly, the effect of parallel feedback on input impedance can beanalyzed using a similar model, the closed-loop input impedancewould then be

so, parallel feedback decrease the input impedance

feedbacknegativefor1notice),1( >>+= AARR iif

Negative feedback on output impedance


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97

)1/( ARR oof +=

For voltage feedback, (it could be either series or parallel feedback),the closed-loop impedance is

so, voltage feedback decrease the output impedance Similarly, for current feedback (either series or parallel feedback), the

closed-loop impedance is

so, current feedback increase the output impedance

As a summary, negative feedback tends to stabilize and linearizegain, which are desired effects.

For a certain type of amplifier, negative feedback tends to produce an

ideal amplifier of that type. For example, series voltage feedback increases input impedance,

reduces output impedance, which gets closer to an ideal voltageamplifier.

So, negative feedback should be used in amplifiers circuits.

)1( ARR oof +=

Some practical feedback network inamplifiers

In practice negative feedback network consists of resistor or capacitors


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98

p In practice, negative feedback network consists of resistor or capacitors ,

whose value is much more precise and stable than active devices (suchas transistors). Then amplifier characteristics mainly depends on feedbacknetwork, thereby achieving precision and stability.

Design of negative feedback amplifiers


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99

A few steps to design negative feedback amplifiers:

Select the feedback type and determine feedback ratio

Select an appropriate circuit configuration for thefeedback network (adjustable resistor can be used sothat feedback ratio can be set precisely)

Select appropriate values for resistance in the feedback

network (this could be a difficult step due to varioustradeoffs)

E.g., in series voltage feedback (like the non-inverting amplifier),we do not want the feedback resistance too small because it loadsthe output of the amplifier, on the other hand, we do not wantfeedback resistance too large because it would cause part of thesource signal to be lost).

Verify the design using Computer Simulations (realcircuits could be very different from the ideal case)


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Unit 5

Oscillators

An oscillator is any measurable


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yquantity capable of repetition.

Examples:

Volume of a loudspeaker

Brightness of a bulbAmount of money in a bank account

The air pressure near your eardrum

The oscillator parameteris the


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pquantity that repeats. In ourexamples, the oscillator parameterhas units of sound intensity, lightintensity, dollars, and pressure,respectively.

Question: Is the x-component of theelectric field, at a particular location in

space, an oscillator? If not, why not? Ifso, what are the units of theoscillation parameter?

The period (denoted T measured in


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The period (denoted T, measured inseconds) is the time required for

one cycle.

The amplitude (A) of the oscillation


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p ( )is the maximum value of the

oscillation parameter, measuredrelative to the average value ofthat parameter.

Many naturally occurring


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Many naturally occurringoscillators are of the class called

resonators. They oscillate freelyat one particular frequency. Thiskind of oscillation is called simple

harmonic motion.

identify the oscillator parameter


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identify the oscillator parameterfor the below resonators


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F o r a f r e er e s o n a t o r, t h e o s c i ll a t o r p a r am e t e r

v a r i e s a n d c a n be e x p r e s s ed a ss i n u s o i d a l l yx ( t ) =x t0 c o s ( ) +

x

Some definitions for simple


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Some definitions for simpleharmonic motion.

x t x t

x

t

o r

( ) c o s ( )= +

+

0

0

a m p l i t u d ef u n c t

a m p l i t u d e

p h a s e

p h a s e a n g le p h a s e c o ns t a n t


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( )

T h e o s c i l la t o r p a r a me t e r r e p e at s i t s e l f

e v e r y t i m et h e p h a s e+ c h a n g e s b y2T h i s h a p p en s e v e r y ti m ec h a n g e s by o n e p e r io d ( T )

T h e f r e q u en c y o f t h em o t i o n i s

e x p r e s s e di n u n i t s of H e r t z= o s c i l l a t io n s p e r s

T h e p h a s ec h a n g e s a tt h e r a t er a d i a n s p er s .

T h i s i s c al l e d t h e ci r c u l a r f re q u e n c y o ra n g u l a r

f r e q u e n c y .

tt

t = T

fT

.

2

1

2

=

= =


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The phase constant specifies the oscillator parameter at t= 0.

x x x( ) c o s ( ) c o s ()0 00 0= + =

x 0 c o s ()


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demo


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v

d x

d t x t

v x x= = +

0

0 0

c o s ( )

v a r i e s f r om - t o +


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Heres a way of thinking about simple harmonic motion that will be very useful later. Suppose a

particle is in uniform circular motion on a circle of radiusA and with angular speed T. The

projection of that motion onto any axis is a point moving in simple harmonic motion with

amplitudeA and circular frequency T.

[ ]2

22 21 1 1 1

The total energy of a mass-spring system is

l ( )dx

k k


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[ ]

[ ] [ ]

22 21 1 1 102 2 2 2

2 21 1

0 02 22 2 2

2 20 0

2

total energy cos( )

sin( ) cos( )

sin ( ) cos ( )2 2

Now, for a spring, so that

t

dxmv kx m k x t

dt

m x t k x t

m x k xt t

k

m

= + = + +

= + + +

= + + +

=

2 2 2 22 20 0otal energy = sin ( ) cos ( )

2 2

m x m xt t + + + =

The total energy of a simple harmonic oscillator is constant in time

and proportional to the square of the amplitude

kinetic and potential energies (as fraction of total energy) for x = x0 cos(Tt)

U / (K+U) K/(K+U)


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U / (K+U), K/(K+U)

The frequency of a resonator isdetermined by the way energy


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determined by the way energyshifts back and forth betweenkinetic and potential terms.

2 21 1

2 2

2

2

2 2

2 2

0

since . Using gives

0 0

E K U mv k x

dE dv dx dv

mv k x v m k xdt dt dt dt

dx dv d xv

dt dt dt

d x d x k m k x x

dt dt m

= + = +

= = + = +

= =

+ = + =

T h e s o l u t io n t o t h ee q u a t i o n i s

2d x kx 0+ =


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These are called sinusoidaloscillations, or simple harmonicmotion.

T h e s o l u t io n t o t h ee q u a t i o n i s

w h e r ea n da r e

c o n s t a n t st h a t d e p e nd o n t h e po s i t i o n a nd v e l o

o f t h e m a ss a t t i m et = 0 .

0

d t mx

x t x km

t x

2

0

0+ =

= +

( ) c o s

All oscillators can exhibit simple


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harmonic motion if they are drivenharmonically.

When a speaker oscillates sinusoidally, itcauses the pressure at any point in a

nearby column of air to vary sinusoidally


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What happens if more than one


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pposcillatory signal is sent to the

light?

V t V t V t( ) ( ) ( )= +1 2


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V t V t V t ( ) ( ) ( ) +1 2

Suppose the two signals have the


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pp gsame amplitude and frequency but

differ in phase.


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For a mass-and-spring system, the motioni i l h i t l


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is simple harmonic at an angularfrequency of

The motion is always simple harmonicwhenever the restoring force is a linear

function of the oscillation parameter.Another way to say this is that the motionis simple harmonic whenever thepotential energy is a quadratic function of

the oscillation parameter.

k m

For example, a pendulum has a


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p pkinetic energy of K = mv2 and a

potential energy of U = mgh.

( )x L vd x

d tL

d

d t

d d

x= = =s i n c o s


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( )

( )

( )

y L vd x

d tL

d

d t

K m v m v v

m L Ld

d tm L

d

d t

x

x y

= = =

= = +

= +

=

c o s s i n

c o s s i n

12

2 12

2 2

12

2 2 2 2

2

12

2

2

( )U m g hm gL= = 1 c o s


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( )

U m gL

1

1

2

2

12

2, c o s

f o r s m a l la n g l e ss o t h a t

d

2

2

2


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K m Ld

d t

U m gL=

12

2 12

2

,

K = m L2(d/dt)2 , U = mgL 2C t


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Compare to

K = m (dx/dt)2

, U = kx2

The circular frequency of themass-spring system is (k /

m)1/2

= ( k /m)1/2

By analogy, the circular frequencyof the pendulum is [ mgL / m

L2

]1/2

= [g / L]1/2

SIGNAL GENERATORS


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130

SIGNAL GENERATORS

/OSCILLATORSA positive-feedback

loop is formed by an

amplifier and a

frequency-selective

network

In an actual oscillator circuit, no input signal

will be present

Oscillator-frequency


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131

Oscillator frequency

stability

Limiter Ckt Comparator


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132

Limiter Ckt Comparator

Wien-bridge oscillator


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November 2, 2007133

Wien bridge oscillator

without amplitude stabilization.

Wien bridge w/ Amp


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November 2, 2007134

Wien bridge w/ Amp.

Stabil.limiter used for amplitude control.

Alternate Wien bridge


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135

Alternate Wien bridge

stabil.

Phase Shift Oscillator


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136

Phase Shift Oscillator

Phase Shift. Osc. W/ Stabil.


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137

Phase Shift. Osc. W/ Stabil.

Quad Osc. Circuit


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138

Quad Osc. Circuit

Active Tuned Osc.


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139

Active Tuned Osc.

Colpitts and Hartley


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140

Colpitts and Hartley

Oscillators

Equiv. Ckt


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141

Equiv. CktTo simplify the analysis, negtlect Cm and rp

ConsiderCp to be part ofC2, and include ro inR.

Collpits Oscillator


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142

Collpits Oscillator

Piezzoelectric Crystal


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e oe ec c C ys a

analog (low freqency)

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