6/9/2015. amplifiers and oscillators 6/9/2015

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AMPLIFIERS AND OSCILLATORS

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Differential amplifier

An amplifier which amplifies the difference

between the two input signals is called

differential amplifier Fig 9.1.

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CLASSIFICATION

Differential amplifier is classified as

1 Voltage amplifier

2. Current amplifier

3. Tran conductance amplifier

4. Tran resistance amplifier

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CHARACTERISTICS

The characteristics of differential amplifier

1.Infinite input resistance

2.Zero output resistance

3.Infinite gain

4.Infinite Bandwidth

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BASIC DIFFERENTIAL AMPLIFIER

Basic differential amplifier has two input voltage signals.

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Cont

VO=Ad(V1-V2)

Where Ad is called differential gain, (V1-V2)

Is called differential voltage Vd= (V1-V2) .

Ad =VO / Vd if V1=V2 then Vd, and VO would

become zero. In practical differential

Amplifier depends on both difference

and average common level of two inputs.

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Cont….

Common mode signal is VC = (V1+V2) /2

VO=AC(V1+V2) /2 or VO=AC VC.

Where VC is called as common mode gain.

The Total output voltage is

VO= Ad Vd +AC VC.

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COMMON MODE REJECTION RATIO

The ability of a differential amplifier to reject a common mode signal is expressed by

CMMR =ρ=20 log Ad/ AC

Differential amplifier has two modes

Common mode and differential mode.

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DIFFERENTIAL MODE

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Cont…

The input signals are same in magnitude

but 180 degree phase shift Q1 is positive going and Q2 is negative

going. RE has same current but 180 degree

phase shift ,so AC current in RE is zero.

The output of collector one and two are equal but opposite in phase.

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Cont…..

Therefore the total output voltage of

differential mode is Vd= (V1-V2)

V1 = Voltage across collector one

V2 = Voltage across collector two

Therefore total output voltage is twice as

large as signal voltage from either collector

to ground.

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COMMON MODE

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Cont..

The input signals of Q1 and Q2 are equal

in magnitude and phase.

RE carries current (2IE) and provides negative feedback.

The output voltage at collector one and two are same in phase and magnitude

Therefore Vo= 0 in common mode .

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Feedback amplifiers and oscillators

Feedback is a process of injecting some energy from the output and then return it to the input .The amplifier which uses this is called feedback amplifier.

Classification of feedback amplifiers

1. Positive Feedback amplifiers

2. Negative Feedback amplifiers

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

Types of feedback connectionsTypes of feedback connections

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Feedback amplifiers and oscillators

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Oscillators

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Oscillator

• Introduction of Oscillator

• Linear Oscillator– Wien Bridge Oscillator– RC Phase-Shift Oscillator– LC Oscillator

• Stability

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Oscillators

Oscillation: an effect that repeatedly and regularly fluctuates about the mean value

Oscillator: circuit that produces oscillation

Characteristics: wave-shape, frequency, amplitude, distortion, stability

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Application of Oscillators

• Oscillators are used to generate signals, e.g.– Used as a local oscillator to transform the RF

signals to IF signals in a receiver;– Used to generate RF carrier in a transmitter– Used to generate clocks in digital systems;– Used as sweep circuits in TV sets and CRO.

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Linear Oscillators

1. Wien Bridge Oscillators

2. RC Phase-Shift Oscillators

3. LC Oscillators

4. Stability

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Integrant of Linear Oscillators

For sinusoidal input is connected “Linear” because the output is approximately sinusoidal

A linear oscillator contains:- a frequency selection feedback network- an amplifier to maintain the loop gain at unity

+

+Amplifier (A)

Frequency-SelectiveFeedback Network ()

Vf

Vs VoV

PositiveFeedback

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Basic Linear Oscillator

+

+

SelectiveNetwork(f)

Vf

Vs VoV

A(f)

)( fso VVAAVV and of VV

A

A

V

V

s

o

1

If Vs = 0, the only way that Vo can be nonzero is that loop gain A=1 which implies that

0

1||

A

A (Barkhausen Criterion)

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Wien Bridge OscillatorFrequency Selection NetworkLet

11

1

CX C

and

111 CjXRZ 2

2

1

CX C

22

22

1

222

11

C

C

C jXR

XjR

jXRZ

Therefore, the feedback factor,

)/()(

)/(

222211

2222

21

2

CCC

CC

i

o

jXRXjRjXR

jXRXjR

ZZ

Z

V

V

222211

22

))(( CCC

C

XjRjXRjXR

XjR

Vi Vo

R1 C1

R2C2

Z1

Z2

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can be rewritten as:

)( 2121221221

22

CCCCC

C

XXRRjXRXRXR

XR

For Barkhausen Criterion, imaginary part = 0, i.e.,

02121 CC XXRR

Supposing,

R1=R2=R and XC1= XC2=XC,

2121

2121

/1

11or

CCRR

CCRR

)(3 22CC

C

XRjRX

RX

0.2

0.22

0.24

0.26

0.28

0.3

0.32

0.34

Fee

dbac

k fa

ctor

-1

-0.5

0

0.5

1

Phas

e

Frequency

=1/3

Phase=0

f(R=Xc)

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Example

Rf

+

R

R

C

CZ1

Z2

R1

Vo

By setting , we get

Imaginary part = 0 and

RC

1

3

1

Due to Barkhausen Criterion,

Loop gain Av=1

where

Av : Gain of the amplifier

1

131R

RAA f

vv

21

R

R fTherefore, Wien Bridge Oscillator

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RC Phase-Shift Oscillator

+

Rf

R1

R R R

C C C

Using an inverting amplifier The additional 180o phase shift is provided by an RC

phase-shift network

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Applying KVL to the phase-shift network, we have

R R R

C C CV1 Vo

I1 I2 I3Solve for I3, we get

)2(0

)2( 0

)(

32

321

211

C

C

C

jXRIRI

RIjXRIRI

RIjXRIV

C

C

C

C

C

jXRR

RjXRR

RjXRR

jXRR

VRjXR

I

20

2

000

02

3

1

)2(])2)[(( 222

21

3CCC jXRRRjXRjXR

RVI

Or

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The output voltage,

)2(])2)[(( 222

31

3CCC

o jXRRRjXRjXR

RVRIV

Hence the transfer function of the phase-shift network is given by,

)6()5( 2323

3

1 CCC

o

XRXjRXR

R

V

V

For 180o phase shift, the imaginary part = 0, i.e.,

RC

R

XXRX CCC

6

1

6X

(Rejected) 0or 0622

C

23

and,

29

1

Note: The –ve sign mean the phase inversion from the voltage

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LC Oscillators

+

~

Av Ro

Z1 Z2

Z3

12

Zp

The frequency selection network (Z1, Z2 and Z3) provides a phase shift of 180o

The amplifier provides an addition shift of 180o

Two well-known Oscillators:• Colpitts Oscillator• Harley Oscillator

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+

~Av Ro

Z1 Z2

Z3

Zp

Vf Vo

321

312

312

)(

)//(

ZZZ

ZZZ

ZZZZ p

For the equivalent circuit from the output

po

pv

i

o

p

o

po

iv

ZR

ZA

V

V

Z

V

ZR

VA

or

Therefore, the amplifier gain is obtained,

)()(

)(

312321

312

ZZZZZZR

ZZZA

V

VA

o

v

i

o

oof VZZ

ZVV

31

1

Zp AvVi

Ro

+

+

Vo

Io

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The loop gain,

)()( 312321

21

ZZZZZZR

ZZAA

o

v

If the impedance are all pure reactances, i.e.,

332211 and , jXZjXZjXZ

The loop gain becomes,)()( 312321

21

XXXXXXjR

XXAA

o

v

The imaginary part = 0 only when X1+ X2+ X3=0 It indicates that at least one reactance must be –ve (capacitor) X1 and X2 must be of same type and X3 must be of opposite type

2

1

31

1

X

XA

XX

XAA vv

With imaginary part = 0,

For Unit Gain & 180o Phase-shift,1

2 1X

XAA v

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R L1

L2

C

RC1

C2

L

Hartley Oscillator Colpitts Oscillator

T

oLC

1

1

2

RC

Cgm

21

21

CC

CCCT

CLLo

)(

1

21

2

1

RL

Lgm

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Colpitts Oscillator

RC1

C2

L

Equivalent circuit

In the equivalent circuit, it is assumed that: Linear small signal model of transistor is used The transistor capacitances are neglected Input resistance of the transistor is large enough

+

V

gmVR C1C2

L

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)(11 LjiVV At node 1,

where,

VCji 21

)1( 22

1 LCVV

Apply KCL at node 1, we have

0111

2 VCjR

VVgVCj m

01

)1( 122

2

Cj

RLCVVgVCj m

0)(1

213

212

2

CLCCCj

R

LC

Rgm

For Oscillator V must not be zero, therefore it enforces,

+

V

gmVR C1C2

Lnode 1

I1

I2I3

I4

V

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Imaginary part = 0, we have

0)(1

213

212

2

CLCCCj

R

LC

Rgm

T

oLC

1

1

2

RC

Cgm

21

21

CC

CCCT

Real part = 0, yields

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THANK U

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