I would like to build a Colpitts oscillator. The oscillator will have a frequency of \$f = 1/2 \pi\sqrt{C_{tot} L}\$. But what implications does my choice of \$L/C\$ have? Can I arbitrarily choose my tank circuit to be very inductive, or very capacitive? For resonators, the impedance goes as \$Z=\sqrt{L/C}\$, but it is not obvious to me how this impacts on the behavior of the oscillator.
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\$\begingroup\$ In a colpitts oscillator the choice of capacitors determines the feedback ratio. The expression for said feedback ratio depends on the topology in use, see for instance CB, CC & CE colpitts oscillator. If the feedback ratio is too small your amplifier might not be able to sustain the oscillation. As for the inductor, don’t see how it affects the rest of the circuit, except of course the resonant frequency with the formula that you’ve written. \$\endgroup\$– rr1303Commented Dec 6, 2020 at 21:43
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2 Answers
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It's all about being practical:
Choose your inductor so that at the desired circuit oscillation frequency, the inductor's self resonant frequency is several times higher. If you want accuracy choose an inductor that has a SRF that is ten or more times higher than the oscillation frequency. This limits the inductor to a much smaller range of acceptable values vs frequency.
At the other end of the scale, choose capacitor values that are large enough to dwarf parasitic capacitances from tracks and internal transistor capacitance.
This pretty much narrows things down to a few choices.
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You know that \$X_{_\text{CT}}=\frac1{\omega_{_0}\,\cdot \,C_{_\text{T}}}\$ and that \$X_{_\text{L}}=\omega_{_0}\cdot L\$. At resonance, \$X_{_\text{CT}}=X_{_\text{L}}\$. At much lower frequencies, the tank impedance is dominated by the inductor. At much higher frequencies, the tank impedance is dominated by the capacitor. But right at that exact point, their contributions are equal and therefore the combined impedance is at its peak.
As \$X_{_\text{CT}}=X_{_\text{L}}\$, it follows that \$\frac1{\omega_{_0}\,\cdot \,C_{_\text{T}}}=\omega_{_0}\cdot L\$. So, \$\frac1{\omega_{_0}\,\cdot \,C_{_\text{T}}\,\cdot\,\omega_{_0}\cdot L}=1\$ or \$\omega_{_0}^2=\frac1{C_{_\text{T}}\,\cdot\, L}\$. That's how you get the fact that \$\omega_{_0}=\frac1{\sqrt{L\,\cdot\,C_{_\text{T}}}}\$. So, by definition, you are computing the (angular) frequency that causes the two impedances to be equal.
The combined impedance will be \$Z_{_\text{T}}=X_{_\text{CT}}\mid\mid X_{_\text{L}}\$. You can easily arrive at \$Z_{_\text{T}}=\frac12\sqrt{\frac{L}{C_{_\text{T}}}}\$, at resonance.
I don't know if you are thinking of a BJT (which works in CE, CB, and CC arrangements), JFET or opamp solutions, but in all cases of oscillation you must have positive feedback that is sufficient to sustain the oscillation. This is where you need to make sure that your loaded output feedback followed by the attenuation of the capacitive divider and its loading at the input are sufficient at the required phase.
You wrote almost nothing about what you are attempting and I am not interested in attempting to write chapter and verse on the topic, even if I could (which I could not.) So I'm stopping here.
