0
\$\begingroup\$

In an LC tank circuit, why do XC (XC1 + XC2) and XL need to be roughly similar?

For example, in this Colpitts Oscillator taken from https://www.electronics-tutorials.ws/oscillator/colpitts.html, I calculate XC to be 703ohms and XL to be 677ohms.

Image taken from https://www.electronics-tutorials.ws/oscillator/colpitts.html

Edit: I just realised that in order for a tank circuit to resonate, the reactances need to be equal at that frequency.

\$\endgroup\$
2
  • \$\begingroup\$ At resonance, the imaginary part of the admittance is equal to zero. If you include that fact in your analysis, you should be able to work out why. \$\endgroup\$
    – jonk
    Commented Aug 6, 2018 at 3:29
  • \$\begingroup\$ "Resonance"? This oscillator needs -180 deg phase shift at the desired frequency. We have a 3rd-order lowpass in the feedback path (circuits with "resonance" produce 0 deg) \$\endgroup\$
    – LvW
    Commented Aug 6, 2018 at 7:51

1 Answer 1

Reset to default
1
\$\begingroup\$

Collector feedback direct to the base is usually called negative feedback i.e. there is a natural inversion or a 180 degrees phase shift. This is the opposite of what is needed to make an oscillator. Positive feedback is needed hence, to acquire the extra 180 degrees needed to create positive feedback C1, C2 and L are used.

That extra 180 degrees precisely happens when the series combination of C1 and C2 produce an impedance magnitude equal to the impedance of L.

\$\endgroup\$
1
  • \$\begingroup\$ The function of the Colpitt oscillator is based on a third-order lowpass (r,out-C1-L-C2) in the feedback path which produces the required 180deg phase shift at one single frequency only. The resistance r,out is the dynamic output resistance at the collector node; in addition, the lowpass is loaded by the input resistance at the base node. These resistive terminations do also determine the phase shift of the whole network. Therefore, both imaginary impedances are not "precisely" identical - but the are roughly equal. \$\endgroup\$
    – LvW
    Commented Aug 6, 2018 at 7:47

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.