oscullator hand outs

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Oscillator Principles1

An oscillator is a circuit that produces a periodic waveform on its output with only the dc supply voltage as an input.

Oscillator Principles2

Types of Oscillators:

Feedback oscillator – returns a fraction of the output signal to the input with no net phase shift, resulting in a reinforcement of the output signal.

Relaxation oscillator – uses an RC timing circuit to generate a waveform that is generally a square or other non-sinusoidal wave.

Feedback Oscillator3

A feedback oscillator consists of an amplifier for gain and a positive feedback circuit that produces phase shift and provides attenuation.

Positive feedback – condition wherein a portion of the output voltage of an amplifier is fed back to the input with no net phase shift, resulting in a reinforcement of the output signal.


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Conditions for sustained oscillation:


Start-up Conditions: When oscillation starts at t0, the condition

Acl > 1 causes the sinusoidal output voltage amplitude to build up

to a desired level. Then Acl decreases to 1 and maintains the

desired amplitude.

Oscillators with RC Feedback Circuit7

A lead-lag circuit and its response curve:

Vout R(-jX)/(R-jX)

Vin (R-jX) + R(-jX)/(R-jX)=

Simplify:

Vout RX

Vin 3RX + j(R2-X2)=

No j term for a 00 phase angle,

Vout/Vin = 1/3


A lead-lag circuit and its response curve:

fr = 1

2πRC

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The Wien-Bridge Oscillator Basic Circuit:

Two forms of the same circuit.

Acl=R1+R2R2


Notes on the Wien-Bridge Oscillator


Positive Feedback Conditions for Oscillation:


The unity-gain condition in the feedback loop is met

when Acl = 3,

To achieve a closed-loop gain of 3,

R1=2R2

Positive Feedback Conditions for Oscillation:

Acl= = = 3R1+R2 2R2 + R2

R2 R2

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Start-up Conditions:


Start-up Conditions:Self starting Wien-bridge oscillator using back-to-back zener diodes.

Acl= R1+R2+R3

R22R2+R2 + R3

R2=

= 3 +R3R2


Self starting Wien-bridge oscillator using FET in the negative feedback loop.


Example. Determine the resonant frequency for the given Wien-bridge oscillator. Compute for the Rf assuming the internal drain-source

resistance r’ds of the JFET is 500Ω when oscillations are stable.

fr = = 1.59kHz 1

2πRC

Since R1=R2=R and C1=C2=C

Acl= + 1 = +1Rf RfRi R3+r’ds

but Acl= 3therefore Rf = 3kΩ

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The Phase-Shift Oscillator


Since β = 1/29 = R3/Rf

then Rf = 290kΩ

fr = = 6.5kHz1

2π√√√√6 RC

Example. Det. the frequency of oscillation and the value of Rf necessary for the circuit to operate as an oscillator.

From the ckt. R1=R2=R3 = R and C1=C2=C3=C, therefore


The Twin-T Oscillator:

Oscillators with LC Feedback Circuit20

The Colpitts Oscillator:

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Loading of the feedback circuit affects the frequency of oscillation.

1 Q2

2π√√√√LCT Q2+1

fr =


The Basic FET Colpitts Oscillator:


Oscillator loading:

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Example. (a) Determine the frequency for the oscillator. Assume that there is negligible loading on the feedback circuit and that its Q is greater than 10. (b) Find the frequency if the oscillator is loaded to a point where the Q drops to 8.


1

2π√√√√LCT

fr = = 7.46kHz

C1C2

C1+C2

CT= = 0.0091µF

(a)

1 Q2

2π√√√√LCT Q2+1

fr = =7.4kHz

(b)


The Clapp Oscillator : 1

2π√√√√LCT

fr =


The Hartley Oscillator :

where LT = L1 + L2

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The Armstrong Oscillator :1

2π√√√√LpriC1

fr =

Crystal-Controlled Oscillators30



Basic Crystal Oscillators :

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Relaxation Oscillators33

A Triangular-Wave Oscillator:


A Practical Triangular-Wave Oscillator:


Example: Determine the frequency of oscillation of the circuit. To what value must R1 be changed to make the frequency 20kHz?

1 R2

4R1C R3

fr = = 8.25kHz

To make f=20kHz:

1 R2

4fC R3

R1 = = 4.13kΩ


A Sawtooth Voltage-Controlled Oscillator(VCO):

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A Sawtooth Voltage-Controlled Oscillator(VCO):

VP − VF

|VIN| / RiCT =

From f = 1/T

|VIN| 1

RiC VP - VF

f =


Example: (a) Find the amplitude and frequency of the sawtoothoutput in the figure. Assume that the forward PUT voltage, VF, is approximately 1V. (b) Sketch the output waveform.

(a) First, det. the gate voltage at which the PUT turns on.

R4

R3 + R4

VG = (+V)

= 7.5V


This voltage sets the approximate max peak value of the sawtoothoutput (neglecting the 0.7V). VP = 7.5VThe minimum peak value (low point) is:

VF = 1VSo the peak-to-peak amplitude is

VPP = VP – VF = 6.5VDetermine the frequency as follows:

R2

R2 + R1

VIN = (-V)

= -1.92V


|VIN| 1

RiC VP - VF

f = = 628Hz

(b) The output waveform is shown where the period is determined as follows:

1

fT = = 1.59ms

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A Square-wave Oscillator:

The 555 Timer42

The 555 Timer43

Internal diagram of a 555 timer

The 555 Timer44

Astable Operation

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The 555 Timer45

Astable Operation

The 555 Timer46

Astable OperationOperation of the 555 timer in the astable mode

The 555 Timer47

Astable Operation

The 555 Timer48

Astable Operation 1.44(R1+2R2)Cext

fr =

tH = 0.694(R1+R2) Cext

tL = 0.694R2Cext

T = tH+tL=0.694(R1+2R2) Cext

R1+R2(R1+2R2)

Duty Cycle = X100%

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The 555 Timer49

Astable Operation1.44

(R1+R2)Cext

fr =

R1(R1+R2)

Duty Cycle = X100%

The 555 Timer50

Example: A 555 timer configured to run in the astable mode (oscillator) is shown. Determine the frequency of the output and the duty cycle.

1.44(R1+2R2)Cext

fr = = 5.64 kHz

R1+R2(R1+2R2)

Duty Cycle = X100%

= 59.5%

The 555 Timer51

Operation as a Voltage Controlled Oscillator(VCO)

The 555 Timer52

Monostable Mode

trigger

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