3
\$\begingroup\$

I'm confused by the Colpitts oscillator shown below:

colpitts

I've marked my understanding of the various components, but I'm having trouble understanding the tank circuit.

My understanding is that C1 and C2 form a voltage divider. However, I'm assuming Vin is coming from the emitter, between the two caps which seems unusual given the standard representation: voltagediv
(source: learningaboutelectronics.com)

But intuitively I can see that the voltage is split between C1 and C2. But what is the purpose of this? From my understanding of tank circuits, L1 and C1 alone in parallel should resonate? Why is a voltage divider required?

\$\endgroup\$

1 Answer 1

3
\$\begingroup\$

Yes - it looks like a capacitive voltage divider - however, it isn`t one because the signal input is NOT at the top of C1 but between both caps). If you try to look into the feedback circuit (starting at the emitter node which has a dynamic output small-signal resistance r,out) we face a third order circuit. This circuit is a first order grounded RC block (r,out-C2) that is connected with a second order grounded CL block (C1-L1). Both blocks together provide something like a bandpass response, however, with a squared s in the numerator and a third order denominator (EDIT: First order RC lowpass in series with a second order highpass).

(That means: Your second diagram is not correct: Vout is not across C2)

This bandpass provides - as desired - a zero deg phase for the resonant frequency wo. The necessary loop gain (>1) at this frequency is possible because of the existing voltage enlargement across the inductor at w=wo (although the common collector stage has a gain below unity). C3 acts simply as a coupling capacitor and does not influence the oscillation condition).

A good description of the circuit (including derivation of formulas) can be found here: http://seit.unsw.adfa.edu.au/staff/sites/hrp/teaching/Electronics4/docs/PLL/colpitts.pdf

\$\endgroup\$
2
  • \$\begingroup\$ Thanks, I'll read your answer more carefully and study the linked pdf but it's helped a lot already to know that I'm not looking at a voltage divider. I was confused by the wikipedia page (en.wikipedia.org/wiki/Colpitts_oscillator) which characterizes it as such. \$\endgroup\$
    – new299
    Dec 14, 2014 at 17:17
  • \$\begingroup\$ Yes - a common base Colpitt oscillator (as in wiki) contains two capacitors that form a capacitive divider. This is because in common base we have a relatively large signal gain - and only a fraction of the voltage provided by the passive part of the circuit (tank) must be delivered to the BJT input (emitter). In your circuit, however, which is common collector the BJT does not provide a signal gain >1 . That`s the difference. \$\endgroup\$
    – LvW
    Dec 14, 2014 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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