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Pretty new to this stuff so apologies if I get some words wrong.

In the book I'm reading they described a parallel circuit with a resistor capacitor and an inductor powered by a current source. They say the capacitor has impedance \$\frac{1}{j\omega C}\$ and the inductor has impedance \$j\omega L\$ and the resistor has resistance \$R_{p}\$. The resonant frequency is when the magnitudes of the two impedances of the capacitor and the inductor are equal i.e. \$\omega_0^2 = \frac{1}{LC}\$. But when I use the parallel resistance formula I get the impedance of the circuit should be

$$\left(\frac{1}{R}_p + j\omega C + \frac{1}{j\omega L}\right)^{-1}$$

But at resonant frequency where \$\omega C = \frac{1}{\omega L}\$ this comes out to \$R_{p}\$, not infinity. So I'm struggling to understand what they're trying to say here.

To clarify some things here's the page from the book I'm reading. (Electronics with Digital and Analog Integrated Circuits by Higgins)

enter image description here

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  • \$\begingroup\$ No @mkeith, the book is correct, parallel R with a current source has 2 dimensional impedance \$\endgroup\$ Jan 10, 2021 at 9:27
  • \$\begingroup\$ Where precisely does the book say this erroneous thing? Copy and paste please, verbatim. \$\endgroup\$
    – Andy aka
    Jan 10, 2021 at 11:31
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 what do you mean by a "2 dimensional impedance"? \$\endgroup\$
    – Andy aka
    Jan 10, 2021 at 11:32
  • \$\begingroup\$ We would need to see the schematic they are referring to here. Assuming Rp, L and C are all in parallel, then the only basis for saying Z "approaches infinity" is that Rp is typically chosen as a very high value (itself approaching infinity) and that at resonance, Z=Rp. Otherwise the book is at best sloppy. \$\endgroup\$ Jan 10, 2021 at 12:19
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    \$\begingroup\$ @TonyStewartSunnyskyguyEE75 I said nothing about current source vs voltage source. I said that if they meant LC in parallel and R in series with LC, then the formula for Q would be wrong. What is hard to understand about that? \$\endgroup\$ Jan 10, 2021 at 13:57

3 Answers 3

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What book? Do they actually say the impedance of the RLC circuit approaches infinity or just that the impedance of the LC section approaches infinity. Your calculation is correct but you may be misinterpreting the book.

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  • \$\begingroup\$ That's what I was thinking but I'm trying to figure out what they were really saying. They just have a table (i now posted) that says "Impedance at resonance: Approaches Infinity" but they don't really say much to explain that result. \$\endgroup\$
    – Onye
    Jan 10, 2021 at 2:00
  • \$\begingroup\$ Onye , if you insert for increasing Q or for getting closer to resonance , it will make more sense to you, but the book is 100% correct.. Going to 10 to 100 is going towards infinity ohms in // and towards 0 in series \$\endgroup\$ Jan 10, 2021 at 9:31
  • \$\begingroup\$ This is the correct thinking. Some users are misinterpreting the book. \$\endgroup\$ Jan 10, 2021 at 14:07
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R acts a damping factor to parallel LC and determines the bandwidth, BW, and risetime and R/Zc(fo)=Q=fo/BW

Thus if R were ideally infinite ! Q is infinite and any fo oscillation remains constant forever with Zc(fo)=-ZL(fo) in the complex (reactive) domain, and \$f_o 2\pi=\omega_o\$

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  • \$\begingroup\$ This diode has leakage R so it is pumped for 10ns every cycle to build up the sine wave at almost the resonant frequency, but very close. pull the diode off and it goes forever tinyurl.com/yy4tyvjf \$\endgroup\$ Jan 10, 2021 at 2:37
  • \$\begingroup\$ Thanks for the input. I'm not sure if I understand everything you've said but I'll think more about it. \$\endgroup\$
    – Onye
    Jan 10, 2021 at 6:51
  • \$\begingroup\$ The Book is 100% correct \$\endgroup\$ Jan 10, 2021 at 9:28
  • \$\begingroup\$ look at my simulation pul the gen , R or diode from the circuit. what happens? tinyurl.com/yyw7zz6z \$\endgroup\$ Jan 10, 2021 at 9:39
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You are correct, and the book appears to be wrong. In a parallel RLC circuit, when the driving frequency is the resonant frequency, the impedance is the resistance of R, not \$\infty\$. In a pure LC circuit, with ideal components, the impedance approaches \$\infty\$.

The impedance at any frequency is $$Z= R || X_C || X_L$$

(where \$X_C\$ and \$X_L\$ have imaginary values.)

At resonance (and assuming ideal components)

$$X_C = -X_L$$

so

$$Z_0 = R || X_C ||( -X_C) = \frac{\displaystyle 1}{\displaystyle \frac{\displaystyle 1}{\displaystyle R}+\frac{\displaystyle 1}{\displaystyle X_C} + \frac{\displaystyle 1}{\displaystyle -X_C}} = R$$

If instead, X_L and X_C are given a real, positive values, the formulas become

$$X_C = X_L$$

and

$$Z_0 = R || (jX_C) ||( -jX_C) = \frac{\displaystyle 1}{\displaystyle \frac{\displaystyle 1}{\displaystyle R}+\frac{\displaystyle 1}{\displaystyle jX_C} + \frac{\displaystyle 1}{\displaystyle -jX_C}} = R$$

Both ways of solving for \$Z_0\$ obviously give the same answer,

$$Z_0=R$$.

For clarity, what I assume to be a parallel RLC circuit looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ This is incorrect. The impedance is 2 dimensional not 1. The book may confuse some , but it is correct \$\endgroup\$ Jan 10, 2021 at 9:26
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75. You are correct that Impedance is complex (2 dimensional) but you are mistaken that the impedance of a parallel RLC circuit at resonance is not R. The impedance at resonance is R. I have added the math to my answer. \$\endgroup\$ Jan 10, 2021 at 13:12
  • \$\begingroup\$ The Impedance of X(f)= 0 . I ignored R as it is irrelevant to LC resonance if Q>1. I am not wrong, The book is not wrong. This answer is wrong in the initial assumption. It's just a silly misunderstanding. And this is like explaining how to add 1+1 to me.. which is 0. Proof XOR gate \$\endgroup\$ Jan 10, 2021 at 13:55
  • \$\begingroup\$ And what exactly is "the initial assumption" you speak of? \$\endgroup\$ Jan 10, 2021 at 13:59
  • \$\begingroup\$ Your opening statement “the book is wrong” \$\endgroup\$ Jan 10, 2021 at 14:04

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