Why do we get only one frequency as output in oscillators?

Why do we get only one frequency as output in oscillators?

Oscillators work at one frequency by ensuring two things: -

• The signal fed back to sustain oscillations is exactly in phase with the signal it is trying to sustain. Think about lightly tapping a swinging pendulum at exactly the right place and, in the right direction.
• The loop-gain is slightly more than unity. This ensures that a sinewave is produced without too much distortion and it is "sustained". If the loop-gain were less than 1, then it can't "sustain" an oscillation.

So, if we design a phase-shifting network that has a unique phase-shift for each frequency it handles, we will get an oscillator but, only if the signal fed back is sufficient in amplitude to sustain oscillation.

However, some phase shifting networks can produce a phase shift that is a multiple of the basic oscillation frequency. In other words if 1 MHz produces a phase shift of 360 degrees, maybe some higher-frequency might produce 720 degrees (2 x 360). This could potentially give rise to a sustained oscillation at two frequencies (usually deemed undesireable).

So, we design the phase-shifting network to ensure that the higher-frequency "in-phase" candidate is much lower in amplitude than the "basic" candidate and, given that we only allow the gain to be unity or slightly higher (to accommodate losses in the phase shift network) for the frequency we want, the higher-frequency candidate will not cause oscillation.

The above is also referred to as the Barkhausen criteria.

So what happens to those frequencies which have AB>1??

Saturation.

Let's say there are several frequencies with loop gain \$AB \ge 1\$ and \$n2\pi\$ phase shift, but let's call the one with the highest loop gain \$f_x\$. For \$f_x\$, \$AB > 1\$ and you might expect it to produce an oscillation with amplitude increasing in time. But no real circuit can have its output increase in amplitude indefinitely. There is usually some saturation behavior that limits the output amplitude.

And when this happens, it tends to reduce the gain for all frequencies, not just the one that had the super-unity loop gain. So accounting for saturation, this frequency \$f_x\$ will end up with \$AB=1\$ and all the other frequencies that linear analysis told you had \$AB \ge 1\$ but less than at \$f_x\$, now have \$AB < 1\$, so they no longer oscillate indefinitely.

A short answer from my side:

You must not think in magnitude terms only. Don`t forget the phase. The product AB must be a REAL one . A frequency-selective circuitry has a magnitude as well as a phase which is a function of frequency. And - for a correct design - there will be only one single frequency which can fulfill both conditions at the same time (Barkhausens oscillation criterion with loop gain AB=1):

• |A*B|=1 (for practical reasons somewhat larger than "1", for example "1.2") and

• phaseshift exp(j*phi)=1 (phi=0).

For this purpose, most known oscillators use lowpass, highpass or bandpass filters as feedback elements. But there are also other (more advanced) topologies.