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Wein Bridge Oscillators Presentation-Darren

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Wien-Bridge Oscillator Circuits By Darren De Ronde May 15, 2002
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Page 1: Wein Bridge Oscillators Presentation-Darren

Wien-Bridge Oscillator Circuits

By Darren De RondeMay 15, 2002

Page 2: Wein Bridge Oscillators Presentation-Darren

Why Look At the Wien-Bridge? It generates an

oscillatory output signal without having any input source

Page 3: Wein Bridge Oscillators Presentation-Darren

Basics About the Wien-Bridge Uses two RC

networks connected to the positive terminal to form a frequency selective feedback network

Causes Oscillations to Occur

Page 4: Wein Bridge Oscillators Presentation-Darren

Basics About the Wien-Bridge Amplifies the

signal with the two negative feedback resistors

Page 5: Wein Bridge Oscillators Presentation-Darren

Modification to Circuit

Page 6: Wein Bridge Oscillators Presentation-Darren

Analysis The loop gain

can be found by doing a voltage division

Vo s( ) V1 s( )Z 2 s( )

Z 1 s( ) Z 2 s( )

Page 7: Wein Bridge Oscillators Presentation-Darren

Analysis The two RC

Networks must have equal resistors and capacitors

Z1 s( ) R1

s C

Z2 s( )

R1

s C

R1

s C

Page 8: Wein Bridge Oscillators Presentation-Darren

Analysis

Operational amplifier gain

GV1 s( )

Vs s( )1

R2

R1

Vo s( ) V1 s( )Z 2 s( )

Z 1 s( ) Z 2 s( )

Need to find the Gain over the whole Circuit: Vo/Vs

Vo s( ) G V s s( )s R C

s2R2 C

2 3 s R C 1

Solve G equation for V1 and substitute in for above equ.

Page 9: Wein Bridge Oscillators Presentation-Darren

Analysis

T s( )Vo s( )

V s s( )

s R C G

s2R2 C

2 3 s R C 1

We now have an equation for the overall circuit gain

T j j R C G

1 2R2 C

2 3 j R C

Simplifying and substituting jw for s

Page 10: Wein Bridge Oscillators Presentation-Darren

Analysis

In order to have a phase shift of zero,

1 2R2 C

2 0

This happens at RC When RC, T(j) simplifies to:

T j G

3

If G = 3, oscillations occur

If G < 3, oscillations attenuate

If G > 3, oscillation amplify

Page 11: Wein Bridge Oscillators Presentation-Darren

Time

0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(R5:2)

-4.0V

0V

4.0V

G = 3

Time

0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(R5:2)

-4.0V

0V

4.0V

G = 2.9

Time

0s 100us 200us 300us 400us 500us 600usV(R5:2)

-20V

0V

20V

G = 3.05

Page 12: Wein Bridge Oscillators Presentation-Darren

Ideal vs. Non-Ideal Op-Amp Red is the ideal op-amp. Green is the 741 op-amp.

Time

0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(R1:2) V(R5:2)

-4.0V

0V

4.0V

Page 13: Wein Bridge Oscillators Presentation-Darren

Making the Oscillations Steady

Add a diode network to keep circuit around G = 3

If G = 3, diodes are off

Page 14: Wein Bridge Oscillators Presentation-Darren

Making the Oscillations Steady

When output voltage is positive, D1 turns on and R9 is switched in parallel causing G to drop

Page 15: Wein Bridge Oscillators Presentation-Darren

Making the Oscillations Steady

When output voltage is negative, D2 turns on and R9 is switched in parallel causing G to drop

Page 16: Wein Bridge Oscillators Presentation-Darren

Results of Diode Network

Time

0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0msV(D2:2)

-4.0V

0V

4.0V

With the use of diodes, the non-ideal op-amp can produce steady oscillations.

Page 17: Wein Bridge Oscillators Presentation-Darren

Frequency Analysis By changing the resistor and

capacitor values in the positive feedback network, the output frequency can be changed.

R 10k C 1nF

1

R C 1 10

5rad

sec

f

2 f 15.915kHz

Page 18: Wein Bridge Oscillators Presentation-Darren

Frequency Analysis

Frequency

0Hz 10KHz 20KHz 30KHz 40KHzV(D2:2)

0V

2.0V

4.0V(15.000K,2.0539)

Fast Fourier Transform of Simulation

Page 19: Wein Bridge Oscillators Presentation-Darren

Frequency Analysis Due to limitations of the op-

amp, frequencies above 1MHz are unachievable.

Page 20: Wein Bridge Oscillators Presentation-Darren

Conclusions No Input Signal yet Produces

Output Oscillations Can Output a Large Range of

Frequencies With Proper Configuration,

Oscillations can go on indefinitely


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