Experiment 4

1. Objective:to study RC –oscillators.

2. Equipment: 1 board RC – oscillators
1 dual trace oscilloscope

2 DC power supply

1 function generator

1 frequency counter

3.Theory
3.1. Conditions for oscillations
Oscillations are circuits which built up their own oscillations owing to regenerative feedback. From the
fig below it can be seen that an oscillator network of feedback factor B. consists essentially of an
amplifier of gain A and a positive feedback β.

Vi V0
Amplifiers
A

Feedback network
β

Fig 1.An amplifier and positive feedback

Both stages are connected by the relations

V0
A=
Vi

 V0 =AVi 1)

Vi
β=
V0

 Vi = βV0 2)

Combining equation 1 and 2, the result will become

V0=A βV0

 Aβ=1 3)

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The condition of unity loop gain A β =1 is called Barkhausen Criterion. This condition implies, of
course,

Both that | A β| =1 & That the phase of –A β is zero.

A
The above principles are consistent with the feedback formula Af =
1  A

For if –Aβ = 1,then Af   ,which may be interpreted to mean that there exists an output

voltageevenin the absence of externally applied signal voltage.

Because A and B are complex quantities we may write

|A|ej  |β|ej  =1
From this follow the important balance conditions
Phase balance conditions A + β =0 4)
Amplitude balance conditions |A| |β| =1 5)
As a practical rule |A| |B|> 1 is chosen. Theoretically this would oscillations with infinite
magnitude. I n practice , however , the gain of the amplifier is limited by nonlinear effects so
that |A||B| becomes equal to unity at a particular magnitude. The maximum output voltage of the
amplifier is always limited by the supply voltage ( refer to the results in the exercise operational
amplifier  ).

when the input voltages goes beyond a critical point the output will become distorted.

V0 V0
A A

t f0 f

fig 6a)

V0

AA

t f0 2f0 3f0 4f0 f

fig 6b)

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 V0 V0

A A

t f

f0 2f0 3f0 4f0

fig 6c)
Fig 2.Imput signal and its magnitude spectrum
Fig .2a shows an undistorted output signal if a sinusoidal input signal is applied. The spectrum is
characterized by only one single line .This magnitude is given by the peak value of the input signal. In
fig. 2b the amplifier is slightly overdriven .Beside the original spectrum line which keeps approximately
the same magnitude as in fig 2a additional spectrum lines appear with a small magnitude. When the
amplifier is heavily overdriven (fig 2c) the magnitude of the additional spectrum lines became bigger
and bigger but the magnitude of the original line remains still almost constant at the peak value of the
signal.

In summary, when an amplifier is operated with increasing input and goes into saturation the gain for the
original frequency will decrease. This fact is demonstrated at the curve Voutf(Vin) in fig.3.

Vo

Linear range

Saturation Range

Vi
Fig 3.V0= f (Vi)

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Now it is possible to determine the output voltage by constructing the curves Vout= f(Vin) of both
amplifier and feedback network (as shown in fig.1) in one graph (fig.4)

V0,pp

24

20 Cross point Amplifier with Gain A =5.

16
12
8 Feedback Network with factor β = 0.5
4

4 8 12 16 20 24 26
Fig 4. V0 =f (Vi) of both amplifier and feedback network
In this figure an operational amplifier with a saturation of some  12V(power supply  12V) and a gain
V 1
of A=5 is assumed. The feedback network has a gain of β = i
=0.5 hence the reciprocal is 2.5. The
V0 

cross point gives the Q - point of the oscillator. In this point the product |A||B| becomes exact one.

3.2. Oscillators
The oscillator is an amplifier with positive feedback that generates a number of waveforms usually used
in instrumentation and test equipments. An oscillator that generates a sinusoidal output is called a
harmonic oscillator; the transistor is usually acts in the active region. The output of the relaxation
oscillator is not sinusoidal depending on the transient rise and decay of voltage in RC or RL circuits.
There are two types of RC oscillators:
i. Phase shift oscillators in which the output of an amplifier must be 180o out of
phase with input. A general circuit diagram of a phase shift oscillator is shown in
Fig. (5), where the amplifier is an ideal one. A phase shift network (usually a
resistor-capacitor network) is used to produce an additional phase shift of 180 at
one particular frequency to develop the required positive feedback.

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 C C C

Vi R R R V0

Fig 5. Phase shift network

For lower frequencies high inductances with high costs would be required. That is why the RC -
oscillator is preferred in this frequency range. They are distinguished by their feedback networks..

In fig .5 the phase shift network is shown. For the frequency response the equations (6a) and (6b) can be
derived From the mesh network equations of the feedback network, we find the feedback factor β as,

V0 (w) 1
Β(w) = = 6a)
Vi (w) 1 1 1
  1
(w  ) w   w  
6 4 2

 1  6 w  
2
1
-1  
 w 
.
= –tan  w   w   2 
 5 
6b)


Where   RC

Fig .6 shows the frequency response including gain β and phase of the phase shift network. In the phase
shift oscillator an inverting amplifier (A=1800) is used. Thus only at fo the phase balance condition is
fulfilled (B= –180 o). The resonance frequency fo can be obtained from eqn. (6b) for () =180 0.
 1  6 w  

2
1
0 -1  . 
 180 = –tan  w   w   2 
 5 


 1 1  6 w   
2

 . 
  w   w  2  5  = tan (1800)
 

 1 1  6 w   
2

 . 
  w   w  2  5  =0
 

 1 – 6( w  ) 2 = 0 where   RC

1–6  wRC  where W=W0 =2πf0

2
  0

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 1–6(2πf0RC)2 =0
1
 f0 = 7)
2  RC 6

At this frequency β= 1/29 and it is required that (A) must be at least 29 to satisfy oscillation condition.
The phase shift oscillator is used to the range of frequencies for several hertz to several kilohertz and so
includes the range of audio frequencies. The frequency depends on the impedance elements in the phase
shift network. The phase shift oscillator circuit is not very suitable for generating variable frequency
because the resistors and capacitors must be simultaneously changed to obtain the required frequency
control over a wide range therefore it is used mostly in fixed frequency applications.

f
-90 f0
–180
–270

Fig 6.Frequency Response of the Phase Shift network

3.2.2. The Wien bridge oscillator is used to obtain variable frequency signal. The frequency of
oscillation can be changed by using two gang variable capacitors or two gang variable resistors. The
circuit diagram is shown in Fig.(3). In this circuit, there are two types of feedback:-
a. positive feedback through Z1 and Z2 whose components determine the frequency of oscillation.
b. negative feedback through Rl and R2 whose elements affect the amplitude of oscillation.
The wien bridge is a lead - lag network because the phase angle leads for some frequencies and lags for
other frequencies.

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 R1

R2 – V0
+ C
Z1
R

Z2
R C

Fig 7a) Wien Bridge Oscillator

R C
Vi R C V0

Fig 7b. Wien Bridge Network

The frequency response is given by the equations (8a) and (4b).

V0 1
 w    8a)
2
Vi
 1 
9    w 
 w 

 
 
3
  w   90  tan  
1
8b)
 1 
  w 
 w 

Where   RC

Its corresponding resonance frequency is

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 1
  w   90
0
 f0 = when
2  RC

The corresponding curves are sketched in Fig.8.

f
+90

-90

Fig 8.frequency response of Wien Bridge Network

Similar to 3.2.1. The resonance frequency fo is determined by the phase but here the phase angle is zero
at the resonance frequency. To match the phase balance condition the amplifier has to be a non inverting
one.

4. Preparations

4.1. Prepare graph papers and plot the frequency response (gain and phase) of both RC phase shift
network

and wien bridge in the range =0.1... 10 into the attached semi log graph papers.

4.2. What are the expected oscillator frequencies of both types of oscillators assuming C=10F and

R=3.9k Ω

4.3. Explain briefly the name “lead - lag - network “for the circuit in fig .5.

4.4. Describe briefly the advantages and disadvantages of RC oscillators in general and phase shift

Oscillator and Wien bridge oscillator in particular.

4.5. In fig .5 all resistors of the phase shift network are grounded. In fig .9. the third resistor instead is

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Connected to the negative input terminal of the operational amplifier. Will this affect the performance.

of the phase shift network? Is there a difference between V0 in fig .5 and Vi of the inverting

Operational amplifier?

[Answering this question refer to the basic theory of the operational amplifier.]

5. Procedure

5.1. RC– oscillator .

On the board an operational amplifier(as used in the exercise “operational amplifier “ ) is mounted.

It needs a DC power supply of  12 V.

The other components have the values

R= 3.9Ω

C= 10F

R1=56 KΩ

Rf= 0.. . . 200 KΩ

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5.2. RC - Phase shift oscillator

5.2.1 Connect the RC phase shift oscillator in fig. 9 including the power supply of the operational
amplifier.

R1 is not used in this case.

Rf

+12V
– 7
LM741
+4
3.9kΩ –12V

V0

10nF 10nF 10nF

3.9kΩ 3.9kΩ

Fig 9.
Rf

+12V
R –
LM741
+
Vi V0
+
–12V
~

Fig 10.

5.2.2. Beginning from the maximum gain featured by oscillation with distorted output signal reduce the
gain

( Tuning Rf) until the point where the oscillator generates a stable output wave form with
minimum distortion. Now do not touch the potentiometer until point Rf.

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5.2.3. Read the output frequency f 0 by replacing V0 by frequency counter and also determine the output

voltage.

5.2.4. Disconnect amplifier and feedback network and use the set up in fig. 10 to determine the gain of
the amplifier. Adjust the frequency of the generator to f o obtained in 5.2.2.

5.3. Wien Bridge Oscillator

5.3.1 Connect the Wien bridge oscillator in fig.11 including the power supply for the operational
amplifier

5.3.2 Repeat 5.2.2 for the Wien bridge oscillator.

5.3.3 Repeat 5.2.2.

5.3.4 Repeat 5.2.4 with the set up in fig .12.

5.3.5 Using the same set up plot the curve Vout pp= f ( Vin pp) in the range Vout pp =0 . . . 12V. Do not care
about thewave form.

5.3.6 Investigate the feedbacknetwork in fig.13. Determine the phase shift between Vin and V0 at

f=0.1f0,0.5f0,f0 , 2f0 and 10f0and add the values to the curve prepared in 4.1 Determine the
frequency atwhich the phase shift is zero.

5.3.7.Beside theWien bridge and phase shift oscillators a great variety of circuits for generating
sinusoidal or different wave forms are known. Just as an example infig.14 a square wave
generator is shown.Connect the circuit and observe the output waveform. By varying the
potentiometer determine the minimum and the maximum frequency at which you can get a square
wave form.

Rf

Has no connection

– 7 +Vcc
+ V0
4 –Vcc 3.9kΩ
5.6kΩ 10nF frequency counter.
f

10nF 3.9kΩ
Fig 11

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 Rf R C

R1
+Vcc
–Vin R C Vout

+
–Vcc
Vin Fig 13.

Fig 12.

Rf

+Vcc
–

+
3.9kΩ
10nF –Vcc Vout

3.9kΩ

Fig 14

6. Evaluation

6.1. Discuss differences in the calculated and measured values of f0 and the gain A for both types of
RC oscillators.

6.2. State all differences you have realized in the behavior of both types of oscillators

6.3. Add to the curve obtained in 5.3.5 the curve for the feedback network and determine the Q -point of
the oscillator. Compare with the measured results.

6.4. Very often in practical circuits the operational amplifier is replaced by transistor emitter stages.
How many emitter stages you need at least to built up an RC phase shift oscillator and a Wien bridge
oscillator? Explain.

6.5. Comment any observations you made.

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