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Wein Bridge Oscillators Presentation-Darren

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Wein Bridge Oscillators Presentation-Darren Powered By Docstoc
					Wien-Bridge Oscillator Circuits
By Darren De Ronde May 15, 2002

Why Look At the Wien-Bridge?


It generates an oscillatory output signal without having any input source

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

Basics About the Wien-Bridge


Amplifies the signal with the two negative feedback resistors

Modification to Circuit

Analysis


The loop gain can be found by doing a voltage division
V 1( s )  Z 2( s ) Z 1( s )  Z 2( s )

V o( s )

Analysis


The two RC Networks must have equal resistors and capacitors
Z 1( s ) R 1 sC 1 sC 1 sC

R Z 2( s ) R

Analysis
Need to find the Gain over the whole Circuit: Vo/Vs
Operational amplifier gain G V1( s ) Vs( s ) 1 R2 R1

V o( s )

V 1( s ) 

Z 2( s ) Z 1( s )  Z 2( s )

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

V o( s )

G  V s( s ) 

sRC s  R  C  3 s  R  C  1
2 2 2

Analysis
We now have an equation for the overall circuit gain
T( s ) V o( s ) V s( s )
2 2

s  R  C G s  R  C  3 s  R  C  1
2

Simplifying and substituting jw for s
T j

1  2  R2  C2  3  j    R  C

j   R  C G

Analysis
In order to have a phase shift of zero, 1 R C
2 2 2

0 When RC, T(j) simplifies to:

This happens at RC T j G 3

If G = 3, oscillations occur

If G < 3, oscillations attenuate
If G > 3, oscillation amplify

4.0V

G=3

0V

-4.0V 0s V(R5:2) Time 4.0V 0.2ms 0.4ms 0.6ms 0.8ms 1.0ms

G = 2.9

0V

-4.0V 0s V(R5:2) Time 20V 0.2ms 0.4ms 0.6ms 0.8ms 1.0ms

G = 3.05

0V

-20V 0s 100us V(R5:2) 200us 300us Time 400us 500us 600us

Ideal vs. Non-Ideal Op-Amp



4.0V

Red is the ideal op-amp. Green is the 741 op-amp.

0V

-4.0V 0s V(R1:2) 0.2ms V(R5:2) 0.4ms Time 0.6ms 0.8ms 1.0ms

Making the Oscillations Steady




Add a diode network to keep circuit around G = 3 If G = 3, diodes are off

Making the Oscillations Steady


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

Making the Oscillations Steady


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

Results of Diode Network


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

4.0V

0V

-4.0V 0s V(D2:2) Time 0.2ms 0.4ms 0.6ms 0.8ms 1.0ms

Frequency Analysis


By changing the resistor and capacitor values in the positive feedback network, the output frequency can be changed.
R  10k   f  1 RC  2  C  1nF   1  10
5 rad

sec

f  15.915kHz

Frequency Analysis
Fast Fourier Transform of Simulation
4.0V (15.000K,2.0539) 2.0V

0V 0Hz V(D2:2)

10KHz

20KHz Frequency

30KHz

40KHz

Frequency 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 indefinitely


				
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