Does the barkhausen criteria requires the magnitude of the loop gain AB for a oscillator exactly to be unity or can a value greater than that will be fine. What will be the consequences if the value is greater than 1.
Electronic – Barkhausen Criteria. Is unity loop gain required
oscillator
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For an oscillator to be stable, it is necessary that the net loop gain over the course of a cycle be exactly one. If one is using a circuit where increases in instantaneous amplitude, at least beyond a certain point, will cause a reduction in gain, then the circuit will oscillate at a level where the gain is sometimes above one and sometimes below one, with the "greater" and "less" parts balancing out over the course of a cycle to yield a net gain of exactly one. Since most amplifier circuits have a gain which naturally falls off as instantaneous amplitude increases beyond a certain point, setting the low-amplitude gain to be slightly greater than one will usually achieve the conditions necessary for oscillation.
Note that if there is much difference between the minimum and maximum gain during a cycle, that difference will cause distortion in the output wave. If the purpose of the circuit is to produce a "wibble wave", and one doesn't care whether the wave is a particularly clean sinusoid, one may use a circuit whose low-amplitude gain is quite high, and whose gain falls off rapidly as the signal approaches the desired amplitude; this will yield a reasonably-quickly-starting oscillator which with predictable output amplitude, but the wave will be pretty distorted. Having gain fall off more gradually with amplitude, and having the zero-amplitude gain be just above unity, may reduce distortion, but such an oscillator will slower to start, and have a less predictable output amplitude. Controlling gain based upon a multi-cycle average amplitude may allow the circuit to start quickly while yielding very low distortion, but adds complexity.
One rather clever approach for controlling gain based upon multi-cycle amplitude is to use a light bulb in the feedback circuit. The resistance of a lightbulb will increase with temperature, which will be a function of the average power dissipated by it, which will be roughly proportional to the square of the average current (not exactly proportional because of the resistance change). If the thermal time constant of the filament is long relative to the oscillation period, the gain won't change much during the course of a cycle, thus allowing the circuit to generate a relatively clean sine wave.
The Pierce oscillator has no noise immunity right at the transitions, and the amplifier likely has a pretty good high frequency performance.
In both cases, the input of the amplifier is connected to ground through C1, and the slew rate through the zero crossing should be fairly rapid, so it would take a lot of energy at high frequency at just the right time to disrupt the waveform. There is, however, no hysteresis so the noise immunity measured in volts is low. The current at the transition is high due to the capacitor, so the noise immunity measured in mA is not so low.
So you could say that the arrangement with hysteresis should have higher noise immunity.
If you define "noise immunity" to be the voltage applied in series with the EXTAL input that would disrupt the waveform that's probably not so useful in terms quantifying immunity to externally generated EMI. However, if you had a non-ideal PCB layout so that there was actually such a voltage due to the ground bouncing around, this could be important.
Early on-chip crystal oscillator circuits used hysteresis (eg. Intel MCS48) but that method was widely abandoned in favor of the now-standard Pierce circuit (fewer problems). It's nice to see that on-chip oscillator evolution continues.
Best Answer
If the loop gain of an oscillator would be greater than one, then unless the circuit is in perfect balance such that the amplitude is precisely zero, oscillations will grow until either they have gotten so big as to swallow up the entire universe (unlikely), or until the circuits in the loop are unable to maintain a gain greater than one for an entire cycle (much more likely). If the gain of the system drops with average amplitude but remains constant throughout each cycle, the oscillator will produce a clean output. If, as is somewhat more typical, the gain drops as a function of instantaneous amplitude, the variation in gain throughout each cycle will cause distortion. Generally, the closer the "maximum" gain is to unity, the less variation there will be in gain during each cycle and the cleaner the output will be.
Note that a relaxation oscillator represents an extreme case of variable loop gain; in an ideal relaxation oscillator, the loop gain will be zero for much of the cycle, and infinite during an infinitesimal portion of the cycle. The "unity gain" requirement applies for relaxation oscillators just as for other oscillators, but the fact that the gain is very high for a very small portion of the cycle makes them numerically far less useful than for resonant oscillators.