Positive Feedback (Oscillator) LECTURE CONTENTS • • • • • • 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. THANK YOU
