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Exercise E1 Series Resonant Circuit
(written test, approx. 10 % of a 120-minute written test, SS2022)

A real capacitor behaves like an RLC resonant circuit, with an equivalent series resistance $R$ and an equivalent series inductance $L$.
A capacitor shall be given with the following values:

  • $C=10 ~\rm nF$
  • $R=88 ~\rm m\Omega$
  • $L=60 ~\rm pH$

ee2:7el8zlzjglaazxtw_question1.svg

1. What is the impedance $\underline{Z}_{RLC}$ of this real capacitor for $f_0=100 ~\rm MHz$? (Phase and magnitude)

Path

The impedance $\underline{Z}_{RLC}$ is given by: \begin{align*} \underline{Z}_{RLC} &= R + \underline{X}_{L} + \underline{X}_{C} \\ &= R + {\rm j}\omega L - {{\rm j}\over{\omega C}} \\ &= R + {\rm j}\cdot \left(\omega L - {{1}\over{\omega C}} \right)\\ &= R + {\rm j}\cdot {X}_{LC} \\ \end{align*}

Putting in the numbers, only for the reactive part ${X}_{LC}$: \begin{align*} {X}_{LC} &= 2\pi\cdot \quad \quad f_0 \quad \quad \cdot \quad L \quad \quad \quad \; - {{1}\over{2\pi \cdot \quad \quad f_0 \quad \quad \cdot \quad \quad C \quad\quad}} \\ &= 2\pi \cdot 100 \cdot 10^{6}{~\rm Hz} \cdot 60 \cdot 10^{-12}{~\rm H} - {{1}\over{2\pi \cdot 100 \cdot 10^{6}{~\rm Hz} \cdot 10 \cdot 10^{-9}{~\rm F}}} \\ &= -121.45... ~\rm m\Omega \\ \end{align*}

With the real and imaginary parts, we can derive the magnitude and phase: \begin{align*} X_{RLC} &=\sqrt{R^2 + {X}_{LC}^2} \\ &=\sqrt{(88 ~\rm m\Omega)^2 + (-121.45 ~\rm m\Omega)^2} \\ &= 150.0... ~\rm m\Omega \\ \end{align*}

\begin{align*} \varphi &=\arctan \left( { {{{X}_{LC}}\over{R}}} \right)\\ &=\arctan \left( { {{-121.45 ~\rm m\Omega}\over{88 ~\rm m\Omega}}} \right)\\ &= -0.9437... = -54.07...° \\ \end{align*}

Result

  • $X_{RLC} = 150.0~\rm m\Omega $
  • $\varphi = -54.07° $

2. The impedance magnitude of task 1. can be interpreted as the impedance magnitude of an effective ideal capacity $C_0$.
In this case, the magnitude of the impedance of $C_0$ would be $X_{C0}=X_{RLC}$.
Which value would $C_0$ have for the given $f_0$?

Path

The calculated impedance of $X_{RLC}$ has to be set equal to $X_{C0}$ \begin{align*} X_{RLC} &= X_{C0} \\ &= {{1}\over{2\pi f \cdot C}} \\ \rightarrow C &= {{1}\over{2\pi f \cdot X_{RLC}}} \end{align*}

With values: \begin{align*} C &= {{1}\over{2\pi 100 \cdot 10^{6} {~\rm MHz}\cdot 0.1500... {~\rm \Omega} }} \\ &= 10.6... ~\rm nF \end{align*}

Result

$ C = 10.6 ~\rm nF $

3. What is the resonance frequency $f_{\rm r}$ for the given capacitor? What is the impedance in this case?

Path

The resonance frequency is given as \begin{align*} f_{\rm r} &= {{1}\over{2\pi\sqrt{LC}}} \\ &= {{1}\over{2\pi\sqrt{60 \cdot 10^{-12}{~\rm H} \cdot 10 \cdot 10^{-9}{~\rm F}}}} \\ &= 205.5... ~\rm MHz \end{align*}

At resonance, the impedance is given purely by the resistor.

Result

  • $f_{\rm r} = 205.5 ~\rm MHz$
  • $88~\rm m\Omega$