Positive Feedback
(Oscillator)
LECTURE CONTENTS
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Nyquist criteria , Barkhausen criteria
CR phase shift oscillator
Hartley oscillator
Colpit oscillator
Weign bridge oscillator
Piezo electric crystal
•.
Amplifier , Application & Classification
• Oscillator is an electronic circuit that generates a periodic waveform on its
output without an external signal source. It is used to convert dc to ac.
• Oscillators are circuits that produce a continuous signal of some type
without the need of an input. These signals serve a variety of purposes such
as communications systems, digital systems (including computers), and test
equipment make use of oscillators.
• Oscillation: an effect that repeatedly and regularly fluctuates about the
mean value.Oscillator is a circuit that produces oscillation.
• Characteristics: wave-shape, frequency, amplitude, distortion, stability
Coutesy : Slide Player
Oscillator
•
An oscillator is a circuit that produces a repetitive signal from a dc voltage.
•
The feedback oscillator relies on a positive feedback of the output to maintain the
oscillations.
Sine wave
Square wave
Sawtooth wave
Nyquist Criterion
• The Nyquist plot combines two bode plots of gain vs frequency and phase shift vs frequency
on a single plot. A nyquist plot is used to quickly show whether an amplifier is stable for all
frequencies and how stable the amplifier is relative to some gain and phase shift criteria.
• The amplifier is unstable if the nyquist
curve (encloses (encircles) the -1
point and it is stable otherwise.
Barkhausen criterion.
Vs
+
V
A(f)
Vo
Vo AV A(Vs V f ) and
+
Vf
V f Vo
Vo
A
Vs 1 A
SelectiveNetwork
(f)
If Vs = 0, the only way that Vo can be nonzero is that loop gain A=1 which implies that
| A | 1
A 0
(Barkhausen Criterion)
Barkhausen criterion.
• Oscillation will not be sustained if, at the oscillator frequency the magnitude of the
product of the transfer gain of the amplifier and the magnitude of the feedback network
are less than unity . 𝛽𝐴 = 1 ;
• The frequency at which a sinusoidal oscillator will operate is the frequency for which the
total phase shift introduced as a signal proceeds from input terminals through the
amplifier and feedback network and back again to input is precisely zero or an integer
multiple of 2𝜋 . 𝛽𝐴=0, 2𝜋𝑛
Classification
1. RC oscillators
• Wien Bridge
• Phase-Shift
2. LC oscillators
• Hartley
• Colpitts
• Crystal
Phase-Shift Oscillator
• Oscillation occurs at the frequency
where the total phase shift through the
three RC feedback circuits is 180°.
• The inversion of the op-amp itself
provides the another 180° phase shift to
meet the requirement for oscillation of a
360° (or 0°) phase shift around the
feedback loop.
.
𝐴𝑛𝑔𝑙𝑒(𝛽𝐴) =0, 2𝜋𝑛
Phase-Shift Oscillator
𝛽=
Mathematical Problem-1
𝑅𝑓
= 29
𝑅3
𝑅𝑓 = 29 𝑋 10 𝐾Ω = 290𝐾Ω
𝑓=
1
2𝜋𝑅𝐶 6
= 6.5𝐾Hz
Wein Bridge Oscillator
Wein Bridge Oscillator
Oscillation occurs at the particular frequency when ac balance is
obtained in the Wein Bride. At the balanced condition of the bridge
we can write,
𝛽=
Wein Bridge Oscillator
Wein Bridge Oscillator
Mathematical Problem-2
Design the weign bridge oscillator to produce 100 KHz output frequency with an amplitude
± 9𝑉. Design the amplifier to have a closed loop gain of 3 .Determine the slew rate.
Here,
𝑅1 = 𝑅2 = 𝑅 ;
𝐶1 = 𝐶2 = 𝐶;
Assume , 𝐶 = 1000 𝑃𝐹
𝑅3 = 2𝑅4
𝐶1 = 𝐶2 = 1000 𝑃𝐹
𝑅1 = 𝑅2 = 1.59 𝐾Ω
Assuming, 𝑅4 = 1.59 𝐾Ω
𝑅3 = 2𝑅4 = 3.18 𝐾Ω
𝑆𝑙𝑒𝑤 − 𝑟𝑎𝑡𝑒 ∶ 𝑆𝑅 = 2𝜋𝑓𝑉 = 5.65 𝜇𝑉/𝑠𝑒𝑐
Colpitts Oscillator
The following figure shows the circuit diagram of the
Colpitts oscillator. Oscillation occurs at the frequency
where the L-C feedback circuits is at resonance.
Colpitts Oscillator
Colpitts Oscillator
Colpitts Oscillator
Hartley Oscillator
Oscillation occurs at the
frequency where the C-L
feedback circuits is at
resonance.
Hartley Oscillator
Hartley Oscillator
Mathematical Problem - 3
Design a Hartley oscillator to produce 5 KHz output frequency. Use 0.1 𝜇𝐹 Capacitor and an op-amp with ±10 𝑉
Supply.Make necessary assumptions and consider the coils are wound on seperate core .
𝑓 = 5 𝐾𝐻𝑧
𝐶 = 0.1 𝜇 𝐹
𝐿 𝑇 = 10.132 𝑚𝐻
𝑀=0
Let , 𝐿2 = 10𝐿1
𝐿1 = 0.921 𝑚𝐻
𝐿2 = 9.21 𝑚𝐻
Let , 𝑅1 = 10𝑋𝐿1 = 10 ∗ 5 ∗ 103 ∗ 0.921 *10−3 Ω = 46.05 Ω
𝐴𝐶𝐿 =
𝐿2 𝑅𝑓
=
= 10
𝐿1 𝑅1
𝑅𝑓 = 460.5 Ω
Crystal Oscillator
Crystal Oscillator
Neglecting Resistance 𝑅, the equivalent impedance :
𝑗 𝜔2 − 𝜔𝑠2
𝑍 = 𝑍1 ||𝑍2 = −
. 2
′
𝜔𝐶 𝜔 − 𝜔𝑝2
𝜔𝑠 = series resonant frequency =
𝜔𝑝 = Parallel resonant frequency =
𝜔𝑠 < 𝜔< 𝜔𝑝
1
𝐿𝐶
1 1
(
𝐿 𝐶
1
+ ′
𝐶
)
:The reactance is inductive and outside the range the value will be capacitive.
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