I am trying to determine the resonance frequency of an RLC resonator with added capacitor, as shown below:


simulate this circuit – Schematic created using CircuitLab

Two cases are easy to determine:

$$Rp=0 \rightarrow \frac{1}{\sqrt{LsCs}}$$ $$Rp=\infty \rightarrow \frac{1}{\sqrt{Ls\left(\frac{1}{1/Cs+1/Cp}\right)}}$$

Now I'd like to derive a formula that includes Rp. How should I approach this?


When Rp is infinite you have your equation shown incorrectly. The two capacitors are NOT in parallel; they are in series and as such add up like this: -

\$C_T = \dfrac{1}{\frac{1}{C_S}+ \frac{1}{C_P}}\$

When Rp is zero ohms there is no resonance so your first equation is meaningless.

So, here's the next problem - what do you mean by resonance? Do you mean: -

  1. The frequency at which the impedance is maximum (parallel tuning)?
  2. The frequency at which impedance is minimum (series tuning)?
  3. The frequency that produces the right phase shift to make a pierce oscillator oscillate?

So, when you have decided what to do with the formula in your question and then decided what it is you are actually looking for you might see the wood from the trees.

  • \$\begingroup\$ Thank you! You are completely right about the series capacitance. About the case where Rs = 0, is forgot to draw the series resistance Rs, which makes it bascially a simple RLC series circuit for that case. - I've edited this into my question. \$\endgroup\$ – Douwe66 Apr 20 '18 at 21:06
  • \$\begingroup\$ You still need to reveal what you mean by resonance. Think about it. \$\endgroup\$ – Andy aka Apr 20 '18 at 21:54
  • \$\begingroup\$ I would say the frequency at which the loop impedance is minimum, thus the case where it is at series resonance \$\endgroup\$ – Douwe66 Apr 21 '18 at 6:06
  • \$\begingroup\$ Between which two terminals? \$\endgroup\$ – Andy aka Apr 21 '18 at 8:01
  • \$\begingroup\$ That shouldn't matter in a loop right? \$\endgroup\$ – Douwe66 Apr 21 '18 at 8:40

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