Download - wien bridge oscillator lab
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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