How to remove this problem (Bubba oscillator - Proteus error)

[Iiest2021]

Hi,

I want to generate a sine wave by using Bubba oscillator. But when I do the simulation then showing this error in Proteus.

Please help me !!

pls help me .

 

[Easy peasy]

for 2v5 sitting at the + pin - the (-) pin must be 2v5 as well, assuming the DC average input is 2v5 as well, there should be no net average DC in Rg or Rf, so yes, 2v5 offset seems the better choice for 5V supply - with additional 360k resistor for input current balance ...

One has to assume a typo in the picture, it would be much better to have split supplies and a sine wave about zero

Buffers on the cap output points would be a good idea ....

 

[BradtheRad]

Bubba oscillator simulation. Bipolar AC waveforms.

Each op amp causes delay 1/8 of a cycle. Then the first op amp inverts the waveform coming to it from the last op amp.

At first sight this looks like a phase shift oscillator, but notice this has capacitors to ground (each causing a phase delay), while the phase shift type has caps in series (each causing phase advance).

 

[KlausST]

Hi,

A sine oscillator needs:
* gain = 1
* phase shift = 0° (360°)
(While an inverter is considered to cause a phase shift of 180°)

In your case there is one inverter and 4 equal "shift" stages.
Thus means the 4 equal stages need to provide 180°, or in other words each stage needs to provide 180° /4 = 45°
In an RC with 45° you have
* xc = r
* gain = 0.707

Focussed on gain: 0.707 ^4 = 0.25
To get a loop gain of 1 the amplifier needs to be set to 4.
"4" is idealistic. Any value lower than 4 will stop oscillation any value bigger than 4 (increasing) will cause distortion.
(If the gain is very high, you rather get a square wave instead of a sine.)

Phase shift 45°:
R = Xc: xc = 1 / (2 x Pi x f x c)..
R= 31k, C= 100n
--> solved for f: f = 51.34 Hz

....
And if you want two signals "sine" and "cosine" then just use the outputs of two stages with a distance of "2". (Not crossing the gain stage)
Mind that each filter provides a signal attenuation of 0.707, thus the amplitude ratio will be 1:0.5 or 2:1

Klaus

But first he needs to get rid of all hus schematic errors... (post#3l