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I am in the process of investigating a crystal oscillator for a design, and I've been refreshing my understanding of resonant circuits. I've closed most of the gaps in my knowledge, however I am struggling with a solid interpretation for the impedances of an anti-resonant circuit. Anti-resonant circuit

The above image portrays an LC circuit that I was analyzing to understand anti-resonance (this is the circuit I simulated in LTSpice, hence the dotted current source). The reason I exclude the motional capacitance of the crystal is due to the inductive reactance of the motional elements at the anti-resonant frequency. With that in mind, this is my current understanding of the above circuit:

  • Without the resistor: at the resonant frequency \$\omega_{0}=\frac{1}{\sqrt{ LC }}\$ the impedance of the network is \$\lim Z_{ \omega \to \omega_{0} }=j\infty\$, due to \$(1)\$
    • The impedance is purely imaginary $$ Z=j\left( \frac{X_LX_C}{X_C-X_L} \right) \tag{1} $$ $$ Y=j\left( \frac{1}{X_{c}}-\frac{1}{X_{L}} \right) \tag{1} $$
  • With the resistor: The resistor decreases the phase angle of the \$LR\$ branch by adding a real component, thus allowing some small, real current to flow .
    • At some different frequency \$w_{1}\$, \$C\$ and \$L\$ still resonate, creating a very large impedance
    • The impedance at \$\omega_{1}\$ is primarily real $$ \frac{1}{Z}=\frac{R}{R^2+X_{L}^2} +j\left( \frac{1}{X_{C}}-\frac{X_{L}}{R^2+X_{L^2}} \right) $$

My confusion comes from the introduction of the resistor changing the purely imaginary impedance to primarily real. Even if the resistor is some relatively small value, the large impedance is still caused by the resonating reactive components, so why is the impedance still large and almost entirely real?

I understand the math and if assuming reactive impedance goes to zero, then the impedance becomes purely real and can be solved to be \$Z=\frac{L}{CR}\$, but I don't know how to explain the behaviour otherwise.

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  • \$\begingroup\$ Your equation (1) is giving admittance (Y), not impedance. If you think about the whole problem in terms of admittance rather than impedance you will likely get the solution for yourself. \$\endgroup\$
    – The Photon
    Commented Aug 10, 2023 at 17:44
  • \$\begingroup\$ @ThePhoton Do you mind explaining a little bit about how to interpret the problem with admittance? It would help me start (: \$\endgroup\$ Commented Aug 10, 2023 at 17:57
  • \$\begingroup\$ Your image does not correctly portray the equivalent circuit of a crystal. Maybe have a look at my site here. \$\endgroup\$
    – Andy aka
    Commented Aug 10, 2023 at 18:54
  • \$\begingroup\$ Should there be a capacitor in series with the inductor? See Figure 14 of the IEEE Std 176 "IEEE Standard on Piezoelectricity". Without it the impedance of the crystal oscillators near zero at DC, when it should be infinite. \$\endgroup\$
    – C. Dunn
    Commented Aug 10, 2023 at 18:55
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    \$\begingroup\$ Impedance at anti-resonance is infinite: if you add a small resistance, it's still almost infinite, but now is absorbing a small amount of power which you can model as a large eqivalent parallel resistor, or a small equivalent series resistor \$\endgroup\$
    – david
    Commented Aug 11, 2023 at 6:53

2 Answers 2

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First, we need to find the resonance frequency. In this case of a parallel resonance circuit, we need to find the imaginary part of the admittance of the resonant circuit.

The capacitor admittance is

\$B_C = \frac{1}{X_C} = jωC\$

And the admittance L + R

\$B_L = \frac{1}{R + jωL} = \frac{R - jωL}{R^2 + (ωL)^2} = \frac{R}{R^2 + (ωL)^2} - \frac{ jωL}{R^2 + (ωL)^2}\$

At the resonance, the imaginary parts of admittance become zero \$B_C=B_L\$.

$$ ωC = \frac{ ωL}{R^2 + ω^2 L^2} $$

$$ ωC (R^2 + ω^2 L^2) = ωL $$

$$ R^2 + ω^2 L^2 = \frac{ωL}{ωC} $$

$$ ω^2 L^2 = \frac{L}{C} - R^2$$

$$ ω^2 = \frac{\frac{L}{C} - R^2}{L^2}$$

$$ ω^2 = \frac{L - C R^2}{C L^2}$$

$$ ω^2 = \frac{L}{C L^2} - \frac{C R^2}{C L^2}$$

$$ ω^2 = \frac{1}{LC} - \left(\frac{R}{L}\right )^2$$

Now to find the impedance at the resonance we need to take the admittance real part.

In our case, it is:

$$Y = \frac{R}{R^2 + ω^2 L^2} $$

And we know that \$ω^2\$ is equal to:

$$ ω^2 = \frac{1}{LC} - \frac{R^2}{L^2}$$

So we have:

$$Y=\frac{R}{R^2 + \left(\frac{1}{LC} - \frac{R^2}{L^2}\right) L^2} $$

$$Y=\frac{R}{R^2 + \left(\frac{L^2}{LC} - \frac{R^2 L^2}{L^2}\right) } $$

$$Y=\frac{R}{R^2 + \frac{L}{C} - R^2 } $$

$$Y=\frac{R}{\frac{L}{C}} $$

$$Y=\frac{RC}{L} $$

Thus, the impedance at the resonance is equal to:

$$Z = \frac{1}{Y}= \frac{L}{RC}$$

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  • \$\begingroup\$ This derivation is useful, although it doesn't quite answer my question. However, paired the comment from @david it resolves my confusion. Thanks! \$\endgroup\$ Commented Aug 11, 2023 at 21:03
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Well, notice that the impedance of your circuit is given by:

\begin{equation} \begin{split} \displaystyle\underline{\text{Z}}&=\displaystyle\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}\space\text{||}\space\left(\text{R}+\text{j}\omega\text{L}\right)\\ \\ &=\frac{\displaystyle\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}\cdot\left(\text{R}+\text{j}\omega\text{L}\right)}{\displaystyle\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}+\text{R}+\text{j}\omega\text{L}}\\ \\ &=\frac{\displaystyle\frac{\displaystyle\text{j}\omega\text{C}}{\displaystyle\text{j}\omega\text{C}}\cdot\left(\text{R}+\text{j}\omega\text{L}\right)}{\displaystyle\frac{\displaystyle\text{j}\omega\text{C}}{\displaystyle\text{j}\omega\text{C}}+\text{j}\omega\text{C}\text{R}+\text{j}\omega\text{C}\text{j}\omega\text{L}}\\ \\ &=\frac{\displaystyle\text{R}+\text{j}\omega\text{L}}{\displaystyle1+\text{j}\omega\text{CR}-\omega^2\text{CL}}\\ \\ &=\frac{\displaystyle\text{R}+\text{j}\omega\text{L}}{\displaystyle1-\omega^2\text{CL}+\omega\text{CR}\text{j}}\cdot\frac{\displaystyle1-\omega^2\text{CL}-\omega\text{CR}\text{j}}{\displaystyle1-\omega^2\text{CL}-\omega\text{CR}\text{j}}\\ \\ &=\frac{\displaystyle\left(\text{R}+\text{j}\omega\text{L}\right)\left(1-\omega^2\text{CL}-\omega\text{CR}\text{j}\right)}{\displaystyle\left(1-\omega^2\text{CL}\right)^2+\left(\omega\text{CR}\right)^2}\\ \\ &=\frac{\displaystyle\text{R}+\omega\left(\text{L}-\text{C}\left(\text{R}^2+\left(\omega\text{L}\right)^2\right)\right)\text{j}}{\displaystyle\left(1-\omega^2\text{CL}\right)^2+\left(\omega\text{CR}\right)^2} \end{split}\tag1 \end{equation}

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

Solving for the extremum (it is a maximum) of \$\displaystyle\left|\underline{\text{Z}}\right|\$, gives (assuming that \$\displaystyle\text{CR}^2<\left(1+\sqrt{2}\right)\text{L}\$):

$$\frac{\displaystyle\partial\left|\underline{\text{Z}}\right|}{\displaystyle\partial\omega}=0\space\Longrightarrow\space\omega=\frac{\displaystyle1}{\displaystyle\text{L}}\cdot\sqrt{\frac{\displaystyle\sqrt{\text{L}\left(\text{L}+2\text{CR}^2\right)}}{\displaystyle\text{C}}-\text{R}^2}\tag2$$

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  • \$\begingroup\$ I'm not sure if I'm misinterpreting this, but I'm not sure I understand the relevance of this response to the problem. I assume this is the method of finding the resonant frequency of the circuit? Could you provide some more information? \$\endgroup\$ Commented Aug 10, 2023 at 19:34

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