feedbackinverteroscillator
According to Barkhausen's criteria, for oscillation the feedback should be made after phase difference of 2pi and gain has to be >=1.
Now when I consider a ring oscillator if 2 inverters are connected as in:
Here the phase change after 2 inverters is 2pi. But this would actually not oscillate since output will not change. This is a contradiction right?
Inside the data sheet you will see the basic circuit of a CD4049. Here: -
If Vin is somewhere about halfway between Vcc and Vss, both transistors are conducting and the voltage at Vout will be neither 1 nor 0. In effect both transistors together act like a potential divider; an upper resistor and lower resistor.
This is the linear operating point of this digital circuit.
Next it should be noted that if Vin were taken slightly higher then Vout would fall to Vss much more rapidly than for the small change in Vin - this indicates that there is gain in the configuration and that it is an inverting type of gain.
Next, apply a high value resistor between output and input and the ouput will stabilize and some "halfway" voltage due to negative feedback just like an op-amp would.
Next apply an ac input to Vin via a small capacitor and resistor in series and you get an analogue AC amplifier.
Are you convinced that this type of digital gate can be made into an analogue amplifier? Here's and article showing how a CD4049 can be made into an op-amp. Below is a picture of a CD4049 being used as an AC amplifier: -
Here is the article.
All that remains to make a crystal oscillator is to provide the feedback resistor (maybe 1Mohm to 10Mohm), add a couple of capacitors and the crystal and you have an oscillator. Note that the combination of output impedance and capacitor connected to the output produce a signal that is more sinusoidal that square hence you can make an approximate sine wave oscillator.
Note also that it is this network (together with the input cap to ground) AND the crystal that produces 180 degrees phase shift.
I think, I can agree to everything you have explained. That means: Yes, the Barkhausen condition may be applied to steady-state oscillations only. With other words - if the amplitude condition is met (|T(jωo)|=1 or even >1) and the phase condition is NOT met, we can have decaying or even rising oscillations immediately after power switch-on. But - at which frequency? In reality we have the following situation:
The system will always try to oscillate at a frequency where the phase condition is met. This is the only possibility to allow self-excitement of a system with positive feedback. And this frequency very often will be NOT exactly the frequency (wo) where the amplitude condition is fulfilled per design (or "over-fulfilled"). Hence, the system will try to oscillate at another frequency (let`s say w1). If the amplitude condition is still fulfilled at this new frequency w1 - we will have rising oscillation amplitudes (until any limiting mechanism will come into play). However, if the loop gain magnitude will be <1 at the frequency w1 we will observe damped oscillations (or even no oscillations).
Answer to your second question: If Barkhausens condition (amplitude and phase) is fulfilled for one single frequency only, there is no chance that oscillations at other frequencies are build up.
EDIT 1(example)
Here is a very illustrative example which exactly meets the subject of your question:
There is a harmonic oscillator consisting of two active opamp integrators in a closed loop (one is inverting and the other one is non-inverting). Hence, under ideal conditions we have a loop phase of 0 deg over a rather broad frequency range. However, for real opamps the loop phase will cross the 0 deg line at one single frequency fo only. We will observe that the loop gain magnitude will cross the 0 dB line at another frequency f1 which is always larger than fo (f1>fo). This is due to the opamps finite open-loop gain and corresponding phase shift.
That means: The loop gain magnitude at the frequency fo will be > 0 dB and the Barkhausen condition is (over-)fulfilled. The circuit oscillates at a frequency fo (where the phase condition is fulfilled). Note that for all frequencies below fo the loop gain is even larger than at f=fo. Nevertheless, the frequency for steady-state oscillations is f=fo.
Final remark: In particular, this oscillator topology is most suitable for explaining why and at which frequency a linear oscillator will oscillate.
EDIT 2 (Comment to the example cicrcuit):
Perhaps the following considerations may further help to answer your (very intelligent) question. If you try to use one of the available simulation programs you will find that the two-integrator oscillator will NOT continuously oscillate in case of simplified first-order models for the opamps. Why not? Because the loop phase will meet the 0 deg line for infinite frequencies only. For all finite frequencies, the loop phase is very close to 0 deg but not exactly zero (second-order system only). For this reason, after power switch-on we will have slowly decaying oscillations.
To allow continuous oscillations, a simple resistor across the feedback capacitor will turn one of the integrators into a two-pole system - and the loop gain now is of 3rd order with zero loop phase at a finite frequency.
Best Answer
That's because Barkhausen only applies to linear circuits. Ring oscillators are made from inverter stages that have so much gain that they have to be treated as nonlinear (saturating) elements.
Two inverters (or any even number of inverters) form a bistable system.
A single inverter generally does not have enough delay to oscillate; instead, it drives itself to an intermediate analog state.
Larger odd numbers of inverters have enough delay to oscillate, since the output of each gate reaches saturation before its input changes again.