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6/2/2011
M. Latina 1
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.
Feedback Oscillator4
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Feedback Oscillator5
Conditions for sustained oscillation:
Feedback Oscillator6
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
Oscillators with RC Feedback Circuit8
A lead-lag circuit and its response curve:
fr = 1
2πRC
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Oscillators with RC Feedback Circuit9
The Wien-Bridge Oscillator Basic Circuit:
Two forms of the same circuit.
Acl=R1+R2R2
Oscillators with RC Feedback Circuit10
Notes on the Wien-Bridge Oscillator
Oscillators with RC Feedback Circuit11
Positive Feedback Conditions for Oscillation:
Oscillators with RC Feedback Circuit12
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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Oscillators with RC Feedback Circuit13
Start-up Conditions:
Oscillators with RC Feedback Circuit14
Start-up Conditions:Self starting Wien-bridge oscillator using back-to-back zener diodes.
Acl= R1+R2+R3
R22R2+R2 + R3
R2=
= 3 +R3R2
Oscillators with RC Feedback Circuit15
Self starting Wien-bridge oscillator using FET in the negative feedback loop.
Oscillators with RC Feedback Circuit16
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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Oscillators with RC Feedback Circuit17
The Phase-Shift Oscillator
Oscillators with RC Feedback Circuit18
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
Oscillators with RC Feedback Circuit19
The Twin-T Oscillator:
Oscillators with LC Feedback Circuit20
The Colpitts Oscillator:
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Oscillators with LC Feedback Circuit21
The Colpitts Oscillator:
Oscillators with LC Feedback Circuit22
The Colpitts Oscillator:
Loading of the feedback circuit affects the frequency of oscillation.
1 Q2
2π√√√√LCT Q2+1
fr =
Oscillators with LC Feedback Circuit23
The Basic FET Colpitts Oscillator:
Oscillators with LC Feedback Circuit24
Oscillator loading:
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Oscillators with LC Feedback Circuit25
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.
Oscillators with LC Feedback Circuit26
1
2π√√√√LCT
fr = = 7.46kHz
C1C2
C1+C2
CT= = 0.0091µF
(a)
1 Q2
2π√√√√LCT Q2+1
fr = =7.4kHz
(b)
Oscillators with LC Feedback Circuit27
The Clapp Oscillator : 1
2π√√√√LCT
fr =
Oscillators with LC Feedback Circuit28
The Hartley Oscillator :
where LT = L1 + L2
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Oscillators with LC Feedback Circuit29
The Armstrong Oscillator :1
2π√√√√LpriC1
fr =
Crystal-Controlled Oscillators30
Crystal-Controlled Oscillators31
Crystal-Controlled Oscillators32
Basic Crystal Oscillators :
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Relaxation Oscillators33
A Triangular-Wave Oscillator:
Relaxation Oscillators34
A Practical Triangular-Wave Oscillator:
Relaxation Oscillators35
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
Relaxation Oscillators36
A Sawtooth Voltage-Controlled Oscillator(VCO):
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Relaxation Oscillators37
A Sawtooth Voltage-Controlled Oscillator(VCO):
VP − VF
|VIN| / RiCT =
From f = 1/T
|VIN| 1
RiC VP - VF
f =
Relaxation Oscillators38
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
Relaxation Oscillators39
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
Relaxation Oscillators40
|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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Relaxation Oscillators41
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