0
\$\begingroup\$

I am trying to make this particular Wien Bridge Oscillator (WBO) circuit using LTspice.enter image description here Ideally I should get an output sine signal. Although I can't seem to make it work. Anything wrong with what I did here? It only shows this straight line with no change in Amplitude. enter image description hereenter image description here

\$\endgroup\$
6
  • 3
    \$\begingroup\$ Please have a look at both schematic it seems that you swapped inverting and the non-inverting input of the op-amp. Additionally not sure what "arrow" in LTSpice is. Is that model of rheostat? Also, the top schematic has a single GND bottom seems to have 3 (try to keep GND pointing down it is a good practice). \$\endgroup\$
    – Wintermute
    Feb 12, 2021 at 12:52
  • 1
    \$\begingroup\$ Also note that in the real world, an oscillator starts on noise and/or small imbalances in the circuit. These are (usually) not present in a simulator, everything is more ideal. That often means that the simulator won't start the oscillation. Solve that by injecting a current pulse at some point in the circuit. That often "kick starts" the oscillation. \$\endgroup\$
    – Bimpelrekkie
    Feb 12, 2021 at 13:11
  • 1
    \$\begingroup\$ That R1 is a simple resistor across C2, so delete that arrow you've drawn and instead of ground at the end of R1 make the connection across C2. Then the values of R1 and/or R3 can be changed, since they are potentiometers in the original schematic. Also, there's no need to .lib opamp.sub since you've used real models (LT1001). \$\endgroup\$
    – a concerned citizen
    Feb 12, 2021 at 13:49
  • \$\begingroup\$ In simulation you need to kick it to start. Instead of a DC supply for V1 you can use a step from 0 to +5V at T=0. \$\endgroup\$
    – user16324
    Feb 12, 2021 at 14:10
  • \$\begingroup\$ In nearly each relevant textbook you will find a circuit diagram for a WIEN oscillator - with fixed resistor values (and perhaps some non-linear components for amplitude limiting). More than that, you do not need the shown buffer stage at the output. Did you realize that both diagrams you have shown are quite different? \$\endgroup\$
    – LvW
    Feb 12, 2021 at 14:57

2 Answers 2

Reset to default
2
\$\begingroup\$

A main issue you are having is understanding the potentiometers.

The arrow does not simply denote an arrow of some sort as a hint, it is an electrical contact which you have to model in your simulation as well.

You can think of a potentiometer as two resistors. The sum of the two resistor values is the value of the potentiometer, and the slider (arrow) will choose the value of the two single resistors. Like in this example:

schematic

simulate this circuit – Schematic created using CircuitLab

So I don't know of a component in LT Spice which handles this internally, so I model this manually. If you want to get fancy you could use formulas to recreate this behaviour.

If the potentiometer only has a two connections used, you can replace it with a single resistor as the other one would be dangling in the air (one side unconnected, which doesn't help the simulation)

Another mistake is the polarity of the opamp inputs, you have reversed them from the original schematic. You can mirror components horizontally if you don't want to mess with the wires, but then you have to change your supply rails accordingly (because those will be flipped as well).

If your circuit does not start to oscillate or behaves strange, it can sometimes help to select "startup voltages" (or something like that, my LT Spice doesn't open for some reason, sorry) in the transient simulation tab.

\$\endgroup\$
1
\$\begingroup\$

enter image description here

You need to fix this and take much more care when copying a circuit.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.