Capacitance divider of Colpitts Oscillator

[promach]

For the following Colpitts Oscillator, Why does employing the capacitance divider up-converts the source impedance of transistor by a factor of (1+C2/C1)²

[TimFox]

It's the combination of the two capacitors and the inductor that effects the impedance transformation.
Such resonant transformer networks are common in narrow-band RF applications.  The text book you show probably covers that.

[ferdieCX]

Hi, the thing is called "capacitive transformer".
Although the text in the picture is in Spanish, you can see the equations that demonstrates it.
When both X1 an X2 are of the same sign, in our case capacitive, one of the two simplifying conditions automatically implies the other,
and you can go from eq. (19) to (20)

[promach]

Wait, I am asking about source impedance (not just resistance) of transistor (in the case colpitts oscillator) which is equivalent to the common, bottom node for X2 and RL

What is RL'  ?


Edit:

I found http://www.zen22142.zen.co.uk/Analysis/analysis.pdf which is more explanatory in English text version.

[ferdieCX]

Wait, I am asking about source impedance (not just resistance) of transistor (in the case colpitts oscillator) which is equivalent to the common, bottom node for X2 and RL

What is RL'  ?


Edit:

I found http://www.zen22142.zen.co.uk/Analysis/analysis.pdf which is more explanatory in English text version.

The input capacitance Cin of the transistor is part of the tuned circuit, so it is canceled by the inductance L1 at the oscillation frequency and you only have to care about the input resistance.
To make the oscillation frequency independent of Cin, the external capacitors are usually many times bigger than Cin

The "capacitive transformer" admitance Yi looks like the parallel of  RL' (the transformed resistance) an a capacitor of value C = C1 // C2
The explanation in English that you have found is very complete.