Problem with Wien's Bridge basic working and simulation in multisim

I am very new to the concept of Wien's bridge and am trying to make a basic circuit of it but didnt get the expected output (noise only).Tried to simulate on Multisim but got the same output.I have gone through many documents on the web but they are either too technical or incomplete.

Here is my circuit: I believe since both my attempts gave me the same output I must be doing some mistake. Also I have problems understanding the concept of gain, why it should be slightly greater than 3 and the requirement of initial noise for the circuit to work.

EDIT:

SOLUTION

I made a couple of changes in the circuit and it was working well enough.

First of all, I changed the OPAMP from "AJ741JN" to "741". Second I added an extra R5 (22k) resistance to the circuit and changed the connection from pin 2 (see new figure).

After making these changes the circuit starts to produce a signal of 50 Hz frequency but the wave form is a distorted sine wave. Adding a couple of "1N4007G" diodes as shown eliminates this problem as well (dont know how). If we let R1 = R2 = R and C1 = C2 = C which is usually the case for this type of oscillator (same values). We can see that C1, C2, R1 and R2 form a potential diver.

The impedance of the top part R2 in series with C2 is:

$$Z_{top} = R + \frac{1}{j \omega C} = \frac{1+j\omega C R}{j\omega C}$$

The impedance of the bottom part R1 in parallel with C1 is:

$$Z_{bot} = \frac{R \frac{1}{j \omega C}}{R + \frac{1}{j \omega C}} = \frac{R}{1 + j \omega C R}$$

So the gain of this potential divider is:

\begin{align} G & = \frac{Z_{bot}}{Z_{top}+Z_{bot}} \\ & = \frac{\frac{R}{1 + j \omega C R}}{\frac{1+j\omega C R}{j\omega C}+\frac{R}{1 + j \omega C R}} \\ & = \frac{\frac{R}{1 + j \omega C R}}{\frac{(1 + j \omega C R)^2 + j \omega C R}{j \omega C(1+j\omega C R)}} \\ & = \frac{j \omega C R}{1 + 3j \omega C R -\omega^2 C^2 R^2} \\ & = \frac{\omega C R}{-j(1 + 3j \omega C R -\omega^2 C^2 R^2)}\\ & = \frac{\omega C R}{(\omega^2 C^2 R^2 -1)j + 3 \omega C R} \end{align}

If we chose a frequency such that the imaginary (j) part is zero then the Gain G is 1/3. Now for an oscillator to work it needs a gain of exactly one so the gain of the amplifier needs to be 3 since 3 times 1/3 = 1. Typically there is a non linear element such as a PTC or bulb in the feedback path R6, R3 to provide a simple auto gain control to ensure this.

The frequency when this will oscillate is:

$$\omega^2 C^2 R^2 - 1 = 0 \Rightarrow \omega = \frac{1}{C R} \Rightarrow f = \frac{1}{2 \pi C R}$$

Finally consider what would happen if the output were exactly zero volt. All inputs to the opamp would be zero and the circuit would never start. However there is always some noise in the system to start it going.

Edit:

If you are simulating this Remove R4 and change R3 to 20k. This should make the gain exactly 3 as required. You will also need to add an initial charge to either C1 or C2 to get it started otherwise the inputs will be at 0V and it wont start.

If you are testing with real hardware you will need to play with the setting of the pot to much or too little gain and it wont work.

• answered really nice..cleared all my doubts.Really appreciate it.There's your +1 :) – akki Jan 17 '14 at 12:30
• one more thing...can you tell me what is the problem with the circuit I made.I still cant find why its not working – akki Jan 18 '14 at 12:28
• @akki I've edited my post – Warren Hill Jan 18 '14 at 15:17
• sorry for taking so long to come back..Thanks for all the efforts, they really helped me getting the right direction. – akki Jan 29 '14 at 12:20

From memory (meaning it's a long time since I used them) you need a gain of 3 for the Wien bridge to oscillate, and you have lower gain than that : 1 + R6/R3 gives a gain of around 1.2. If you reduce R4 far enough (to about 5k) it ought to start oscillating.

This page shows a better way to set the gain (scroll down to "non-inverting amplifier").

As for why you need a gain of 3 - work out the impedance of the series arm (RC in series) at the resonant frequency : you'll find it's sqrt(2) * the resistor. Now work out the resistance of the shunt arm (RC in parallel) - it's R/sqrt(2). Thus the input voltage is 1/3 of the output voltage meaning you need a gain of 3 to sustain the oscillation.

• can you please clarify what memory is. – akki Jan 15 '14 at 18:26
• thank you very much..the source you provided helped me a lot with the theoretical part. – akki Jan 30 '14 at 9:40