2
\$\begingroup\$

I was reading on a book (High-Frequency Oscillator Design for Integrated Transceivers, J. van der Tang, Dieter Kasperkovitz, Arthur H.M. van Roermund) this paragraph about the start-up condition for an oscillator:

enter image description here enter image description here Let's consider for instance a common base Colpitts Oscillator (biasing network for the base is not shown):

enter image description here

Now, I may do a small signal analysis of this circuit at open loop, find the expression of the loop gain, and size the components in order to make its absolute value greater than one. But, which is the mechanism that will reduce it once the oscillator has started? Is it automatic? How can I size the components in order to make it exactly equal to 1?

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

Non linearities in the transistor will inevitable make the loop gain equal to 1 at some peak to peak amplitude and, the natural impact of this is that the amplitude is difficult to precisely predict (not normally a big deal). The second impact is that the sine wave output will be naturally flattened on one or both peaks and, again, this is often not regarded as a big problem.

So, in short, the sine wave amplitude builds (because loop gain is greater than 1) and, as clipping starts to impact the sine wave shape, loop gain falls to exactly unity.

\$\endgroup\$
  • \$\begingroup\$ But why will it become exactly 1? It seems too perfect to be real \$\endgroup\$ – Kinka-Byo Jan 23 at 17:40
  • \$\begingroup\$ If it were an amplifier and you put a small input signal in, the gain might be 2 but, as you increased the input signal amplitude, due to clipping, the ratio of output to input would fall from 2. As clipping becomes more apparent at higher input levels there will be a point when the gain exactly equals one. If you made the input signal even bigger, then the gain would be lower than 1 and get lower as the input signal rose even more. So, an oscillator finds its own level and that level is when gain equals unity. \$\endgroup\$ – Andy aka Jan 23 at 18:02
  • \$\begingroup\$ Perfect, thank you very much \$\endgroup\$ – Kinka-Byo Jan 23 at 18:11
  • \$\begingroup\$ To be exact: Due to the time constant within the loop (not avoidable) the circuit can react upon amplitude changes with a certain delay only - and the result is that the loop gain slightly swings around the nominal gain of unity . This is equal to a closed-loop pole that swings between the right and the left half of the s-plane. Hence, we have kind of small amplitude modulation - for a good design so small that it is hard to see. \$\endgroup\$ – LvW Jan 24 at 8:47
  • \$\begingroup\$ @LvW as always, you make an interesting addition. \$\endgroup\$ – Andy aka Jan 24 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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