Wien-Bridge Oscillator CircuitsWhy Look At the Wien-Bridge?Basics About the Wien-BridgeSlide 4Modification to CircuitAnalysisSlide 7Slide 8Slide 9Slide 10PowerPoint PresentationIdeal vs. Non-Ideal Op-AmpMaking the Oscillations SteadySlide 14Slide 15Results of Diode NetworkFrequency AnalysisSlide 18Slide 19ConclusionsWien-Bridge Oscillator CircuitsWhy Look At the Wien-Bridge?It generates an oscillatory output signal without having any input sourceBasics About the Wien-BridgeUses two RC networks connected to the positive terminal to form a frequency selective feedback networkCauses Oscillations to OccurBasics About the Wien-BridgeAmplifies the signal with the two negative feedback resistorsModification to CircuitAnalysisThe loop gain can be found by doing a voltage divisionVos( ) V1s( )Z2s( )Z1s( ) Z2s( )AnalysisThe two RC Networks must have equal resistors and capacitorsZ1s( ) R1s CZ2s( )R1s CR1s CAnalysisOperational amplifier gainGV1s( )Vss( )1R2R1Vos( ) V1s( )Z2s( )Z1s( ) Z2s( )Need to find the Gain over the whole Circuit: Vo/VsVos( ) G Vss( )s R Cs2R2 C2 3 s R C 1Solve G equation for V1 and substitute in for above equ.AnalysisT s( )Vos( )Vss( )s R C Gs2R2 C2 3 s R C 1We now have an equation for the overall circuit gainT j j  R C G1 2R2 C2 3 j  R CSimplifying and substituting jw for sAnalysisIn order to have a phase shift of zero,1 2R2 C2 0This happens at  1RC When RC, T(j) simplifies to:T j G3If G = 3, oscillations occurIf G < 3, oscillations attenuateIf G > 3, oscillation amplifyTime0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(R5:2)-4.0V0V4.0VG = 3 Time0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(R5:2)-4.0V0V4.0VG = 2.9 Time0s 100us 200us 300us 400us 500us 600usV(R5:2)-20V0V20VG = 3.05Ideal vs. Non-Ideal Op-AmpRed is the ideal op-amp. Green is the 741 op-amp. Time0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(R1:2) V(R5:2)-4.0V0V4.0VMaking the Oscillations SteadyAdd a diode network to keep circuit around G = 3If G = 3, diodes are ofMaking the Oscillations SteadyWhen output voltage is positive, D1 turns on and R9 is switched in parallel causing G to dropMaking the Oscillations SteadyWhen output voltage is negative, D2 turns on and R9 is switched in parallel causing G to dropResults of Diode Network Time0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(D2:2)-4.0V0V4.0VWith the use of diodes, the non-ideal op-amp can produce steady oscillations.Frequency AnalysisBy changing the resistor and capacitor values in the positive feedback network, the output frequency can be changed.R 10k C 1nF1R C  1 105radsecf2  f 15.915 kHzFrequency Analysis Frequency0Hz 10KHz 20KHz 30KHz 40KHzV(D2:2)0V2.0V4.0V(15.000K,2.0539)Fast Fourier Transform of SimulationFrequency AnalysisDue to limitations of the op-amp, frequencies above 1MHz are unachievable.ConclusionsNo Input Signal yet Produces Output OscillationsCan Output a Large Range of FrequenciesWith Proper Configuration, Oscillations can go on