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University Mhamed Bougara ,Boumerdes

Institute of Electrical and Electronic Engineering

Department of Electronics

Done by

-Oussama Gassab -hammouya houssam - Achraf Djerida

Groupe 01

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Objectives :

The purpose of this set of experiments is to enable us to design a

RC oscillators . the important design parameters for this design

are :

The output signal frequency The frequency stability The signal purity The signal amplitude

We have the following circuit of Wien Bridge oscillators :

A B

i2i1

i2

i1

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Answers to questions :

Question I :

- Finding the frequency of oscillation and the relationbetween R1 R3 when the bridge is at the equilibrium

We have VA = 2( || 1) VA = 2 +1And VB = 11So we have VA

VB = 0

1 =

2

1+1

(1)

by taking the outer upper side loop we got the following

VB + VA +

2

1

+

2

3

1 = 0 By using equation (1) and

some simplification we got the following result

VBVA = ()22 + 2 31 + 1( + 1) 2

At the required frequency

0 we have VB

VA = 0

So at S=j0 the difference must be zero so we have()2(0)2+2310+10(0+1) = 0 that is mean()202 + 2 31 0 + 1 = 0So we have

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1 ()202 + 2 32 0 = 0 this implies that

1

(

)2

0

2 = 0

2 310 = 0

0 = 1

RC31 = 2

So the frequency of oscillator is 0 =1

RC and the relation

between 3 1 31 = 2Question II :

We have the open loop gain circuit, we have obtained by cutting

off the feedback loop

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By applying the divider rule we have

VA = 1 + 1 + 1

So we have

VA =

1

2 + 3 + 1()2VA = 0 2 + 30 + 02

And we have

VB = 1

1+3

VAVB = R1R1 + R3

S2 + 0 2 R3R1S + 022 + 30 + 02 Since =

R3

R1 2 we got the following resultVAVB = 1

3 + S2 0S + 022 + 30 + 02 So

0 =

VA

VB

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The open loop gain is = 0 so

= A3 + S2 0S + 022 + 30 + 02Finding the poles and zeros of this open loop gain

The zeros : S2

0

S +

0

2 = 0

1 = 0(24)2 2 = 0(+24)2 The zeros are real when

R3

R1> 4 the zeros are complex when

0 021 3 > 0 so 1

< 5931 > 32 so

32

< 1 < 59 3 since

3 = 20 we have 0 < 1 5

since 2 6 so we can write

-5

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4 =

(0 + 5) ; 0 < 5

0 ;

0 >

5

When we take R0 = 111 = 1.1 11 = 12.1 R6 = 1011 = 110 So we have R3 = 2R0 = 22 K by using this value

We will find that = 3(5

3

9

1

)

21 3 = 11And = 5( +4) = 5 4 = 3.54b) implementation

We have built the following circuit

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We have got the following results

So the result is perfect

sinusoidal wave with

very small distortion

with frequency

994.278 1kHZSo the saturation of the

op-amp is theamplitude of the

resulting wave

When we choose vcc15

for the am-amp will got

sinusoidal wave with

amplitude 15 v

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e have learnt how to design and build a perfect

sinusoidal wave with specified amplitude and

frequency by using the feedback theory and the

stabilization concept .

o do that we have dealt with mathematical calculation

by using some physical theories , and we have verified

the results by using simulation program .

n order to make this world work with us we have to

understand its phenomenas and its realities to do that we

have to use its language which is the mathematic .

W

T

I