As I was playing around with a small AM oscillator, I thought it would be time to understand how they actually work.(adsbygoogle = window.adsbygoogle || []).push({});

I am referring to the 3rd picture on this page (base mode):

http://elektroniktutor.oszkim.de/signalkunde/colpitts.html

Given the working point of the transistor, I describe its small signal behaviour by two relations: ##U_{BE}=rI_B## and ##I_{CE}=B\; I_B## which should be the easiest model one can think off.

Now using the schematic, I find

## I_{CE}/I_B=r[(1/r+1/R_E+1/(i\omega L))(1+C_1/C_2) +i\omega C_1)]##.

This ratio should equal B, which represents two conditions:

a) From the vanishing of the imaginary part, I find Thompsons formula for the frequency.

b) The real part doesn't depend on ##\omega##. Hence I suppose we fulfill it varying the ratio of ##C_1## to ##C_2##.

Now my questions:

c) What happens, if the real part condition isn't fulfilled? Physical oscillators seem to work over a wide range of ratios of the capacities. I suppose one has to consider a more realistic model of a transistor. What would be the most important generalization?

d) As the small signal analysis is linear in all the currents, one can't see why an arbitrary fluctuation gets amplified. I suppose this is mainly an effect of the nonlinear behaviour of ##I_B## on ##U_{BE}##?

What would be the easiest way to take this into account?

Thank you!

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# Understanding a Colpitts oscillator

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