Unit 3 Oscillators

7/31/2019 Unit 3 Oscillators

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Unit 3:

Oscillators


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OSCILLATOR

The positive feedback results in oscillations in electronic circuits.

-is an electronic circuit that produces a repetitive electronic signal, often a sine wave or asquare wave.


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1. The frequency at which a sinusoidal oscillator will operate is the frequency for whichthe total phase shift introduced, as a signal proceeds from the input terminals, throughthe amplifier and feedback network and back again to the input, is precisely zero (or anintegral multiple of 2 . In other words, the frequency of a sinusoidal oscillator isdetermined by the condition that the loop phase shift is zero.

2. Oscillations will not be sustained if, at the oscillator frequency, the magnitude of theproduct of the transfer gain of the amplifier and the magnitude of the feedback factor ofthe feedback network is less than unity.

The condition of unity loop gain - A=1 is called the Barkhausen Criterion. This conditionimplies that A=1 and that the phase of A is zero. The above principles areconsistent with the feedback formula:

If A = 1, A f which means that there exists an output voltage even in the absenceof an externally applied signal voltage.

A A

A f 1

Barkhausen Criterion

The condition for oscillations can be explained by the Barkhausen Criterion.


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SINUSOIDAL OSCILLATOR

- produces a cosine function cos (2 f c t) where f c is pre-specified.

Noninverting Amplifier

PositiveFeedback

Circuit

V out

In phase

Selective inphase

A

B

V out

V in

The principle of a sinusoidal oscillator

t

The steady state of the oscillation


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LC OSCILLATOR

L

Ci

An LC circuit

LC f c

2

1t

Oscillation in the idealized LC circuit

Decayed oscillation due to the existence of resistance in the LC-circuit

This uses a resonant LC-circuit to generate a relatively pure frequency (little harmonics).

Disadvantages are the nonlinear relationship between LC and frequency and the lack ofvariable L or C with sufficient range (either L or C needs to be tuned in a 1:100 range to getfrequencies between 10 kHz and 100 kHz).

Using an LC-oscillator at high frequencies with the double heterodyne method solves bothproblems to some extent.


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RC OSCILLATOR

The input is shifted 180 o through the amplifier stage and 180 o again through a secondinverting stage giving us "180 o + 180 o = 360 o" of phase shift which is the same as 0 o therebygiving us the required positive feedback.

RC Phase-Shift Network

-the frequency is inversely proportional to the RC product. The nonlinear (inverselyproportional) relationship between the RC product and the frequency is not veryconvenient. This can be alleviated somewhat by using a logarithmic potentiometer, but in

that case the clockwise position of the potmeter corrresponds to the lowest frequency.


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RC OSCILLATOR

It is stable and provide a well-shaped sine wave output with the frequencybeing proportional to 1/RC and therefore, a wider frequency range is possible

when using a variable capacitor.

However, it is restricted to frequency applications because of their bandwidthlimitations to produce the desired phase shift at high frequencies.


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Example

Determine the frequency of oscillations of a RC Oscillator circuit having 3-stageseach with a resistor and capacitor of equal values. R = 10k and C = 500pF.


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PHASE-SHIFT OSCILLATOR-is an oscillator circuit that follows the basic development of a feedback circuit.

The output of the op-amp is fed to a three-stage RC network, which provides the needed 180 of phase shift (at an attenuation factor of 1/29).

If the op-amp provides gain (set by resistors R i and R f ) of greater than 29, a loop gain greater than unity results and the circuit acts as an oscillator with anoscillator frequency is given by:

29

1

62

1

RC f

The amplifier must supply enough gain to compensate for losses. The overall gain mustbe unity.


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Solution:

Let C=0.1F

R=3.25K

To prevent the loading of amplifier R 1 10R.

Therefore R 1=33K

Therefore R F=29(33K )=957K

Select R F=1MPot

Examples:Design a phase shift oscillator for the oscillation frequency 200Hz.

)1.0)(200(62

1

F R


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WIEN BRIDGE OSCILLATOR

-is an oscillator circuit that follows the basic development of a feedback circuit.

12

21

43

C C

R R

R R

2;

2

1

2

1

4

3

2211

R

R

RC

f

C RC R f

o

o

The amplifier must supply enough gain to compensate for losses. The overall gain mustbe unity.

The feedback resistors are R 3 and R 4.The phase-shift components are R 1, C 1 and R 2, C 2.


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1. Calculate the resonant frequency of the Wien bridge oscillator.

Examples page 843

2. Design the RC elements of a Wien bridge oscillator shown above for operation at f o =10 kHz.


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Tuned Oscillator Circuits

Tuned oscillators use a parallel LC resonant circuit (LC tank) to providethe oscillations.

Two Common Types

1. Colpitts The resonant circuit is an inductor and two capacitors.

2. Hartley The resonant circuit is a tapped inductor or two inductorsand one capacitor.


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COLPITTS OSCILLATOR

eqo

LC f

2

1

It achieves positive feedback by using an inverting amplifier plus the 180 phase shift across a parallel resonant circuit.

2121C C

C C C eq

2121

1

C C C LC

o

Frequency of Oscillation

Resonant Frequency


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Example

For the Colpitts oscillator shown below has the following circuit values:C 1 =750 pF, C 2 =2500 pF, and L=40 H. Determine the circuit oscillation frequency.


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HARTLEY OSCILLATOR

eqC o

L f

2

1

Leq = L1 + L2 + 2M

where: M=mutual coupling

Frequency of oscillation

)(

1

21 L LC o

It can also be suitably used for generating RF signals.

The frequency can be easily varied by varying the inductances which can be done by

making the core movable.

Another method of varying frequency is of varying capacitance.

It is not suitable for low frequency work because at low frequency, the value of inductancerequired becomes large.

Resonant Frequency


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Example

For the Hartley oscillator shown below has the following circuit values:C=250pF, L 1=1.5 mH, L 2 =1.5 mH and M=0.5 mH. Determine the oscillationfrequency of the circuit.


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CRYSTAL OSCILLATOR

Electrical Equivalent Circuit of a Crystal

The crystal appears as a resonant circuit.

The crystal has two resonant frequencies:

1. Series resonant condition RLC determine the resonantfrequency The crystal has a low impedance

2. Parallel resonant condition RL and C M determine the resonantfrequency

The crystal has a high impedance

The series and parallel resonant frequencies are very close, within 1% of each other.

smsm

p p

p C C C C

C LC

f

:Frequency Resonant Parallel

2

1

ss LC

f :Frequency Resonant Series

2

1 R

L

C s

C m


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Example

A crystal has these values: L=3H, C s=0.05pF, R=2k and C m=10pF. What arethe series and parallel resonant frequencies of the crystal?


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CRYSTAL OSCILLATOR

It is basically a tuned-circuit oscillator using piezoelectric crystal as a resonanttank circuit.

It has a high gain so that an output of the square-wave signal results.

A pair of zener diodes is shown at the output to provide output amplitude atexactly the Zener voltage (V Z).

121 2

2

QQ

LC f

:Frequency Resonant

r

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Unit 3 Oscillators