# RLC cir - Resonant Freq

Discussion in 'Homework Help' started by PsySc0rpi0n, Aug 28, 2014.

Mar 4, 2014
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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Hmm,

I = Vs/Xc = Vs/XL = 100V/40.8Ω = 2.45A

3. ### MikeML AAC Fanatic!

Oct 2, 2009
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Whoever wrote the tutorial needs to go back to school. There is no resonance in the circuit as drawn. LTSpice is telling you the truth!

If a student can find the error and fix it, they would get an A from me...

Let's see if the real engineers can find it?

Last edited: Aug 28, 2014
4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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There are several circuits drawn on that web page--which one are you referring to?

5. ### MikeML AAC Fanatic!

Oct 2, 2009
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The one linked to by the OP. All circuits on that page have the same problem.

Last edited: Aug 28, 2014
6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The right hand circuit in this image is different from the rest, but except for that it looks to me like the circuit exhibits resonant behavior (or perhaps it should be called anti-resonance):

I guess you'll have to explain the problem--I don't see it.

7. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
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I was talking about the circuit with 60Ω resistor, 200mH inductor and 200μF capacitor from 1st Paralel Resonance example around half way to the end of the page...

They calculate the resonant frequency, quality factor and some more values.
One of those values is the inductor current of 2.45A.

I am trying to confirm that value with LTpice...

8. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
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I can't understand why you say that no circuit has resonance. The point is to find the resonant frequency for different conditions like when we change the values for R, L and/or C...

9. ### MikeML AAC Fanatic!

Oct 2, 2009
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Ok, I'll post two LTSpice simulations: Do you see any resonance in a? How about b?

Now you explain the difference between the two circuits, and why the tutorial is screwed up...

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10. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You are assuming that because LTSpice doesn't show what you expected when you drive the circuit with a voltage source rather than a current source, proves that there is no resonance.

Simulators don't work well when you connect an inductor in parallel with a voltage source. Microcap gives an error message (matrix is singular) when you try this, and I'm surprised that LTSpice doesn't.

Try putting a 1 milliohm resistor in series with the voltage source. You will also probably have to let the simulation run for a long time so the initial large transient can die out.

11. ### MikeML AAC Fanatic!

Oct 2, 2009
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No. I was just using LTSpice to show that the circuit is screwed up; not to prove that it is screwed up.

Why would it give an error message? This is perfectly valid in .AC analysis.

You are making the same mistake as the guy that posted the tutorial. What is the output impedance of a voltage source? What effect does that have on the Q of a resonant circuit. Hint: it totally kills resonance...

LTSpice perfectly shows the results of the screwed up thinking that went into the posted tutorial, and predicts accurately what would happen if you connect a very low-impedance (zero in the tutorial schematic) across a parallel circuit consisting of inductor and capacitor.

Note that the voltage V(a) is forced to be exactly 100V in circuit a. That means the current in the 60Ω resistor is defined only by E(a)/R, independent of what is happening in the inductor and capacitor. Likewise the current through the inductor is independent of the current in the capacitor, and the capacitor current is independent of the current in the inductor (can you say no resonance?)

By allowing V(a) to seek its own level (by driving it from a current source which has an infinite output impedance), the Q of the circuit is defined solely by the 60Ω resistor, and everything shows the correct resonant behavior.

It could be that the circuit that the Tutorial developer intended to portray is this one: Note the peak current through the inductor...

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12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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What do you get if you redo the first simulation in post #9, but show I(V1), the current out of the source? Don't show any other currents on the plot so the scale won't be too large due to the currents in the inductor at low frequency and in the capacitor at high frequency.

If I do a transient analysis of this circuit:

with a frequency sweep from somewhat below to somewhat above fr, I get this result. The top plot is the voltage across the inductor, and on the bottom is the current out of the volltage source:

I can't make the horizontal axis logarithmic during a transient analysis which is why the minimum current is not in the middle of the sweep.

Last edited: Aug 31, 2014
13. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Many years ago, parallel resonant circuits were called anti-resonant circuits because they did not exhibit the same characteristics of the series resonant circuit.

Of course the voltage source will not show a voltage resonance in the parallel circuit because of the fact that the voltage is the same across all parallel connected components. And like every parallel circuit, the current flow is the key. In the anti-resonant parallel circuit, the current is minimum, while the resonant series circuit has just the opposite, maximum current flow.

Also from the NEETs series:

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14. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Here is the impedance of a parallel resonant circuit.

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15. ### MikeML AAC Fanatic!

Oct 2, 2009
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One more scenario to convince the Electrician that connecting a voltage source directly across a parallel resonant RLC circuit kills its resonance, and this is physical fact, nothing to do with simulation...

Here are the Bode plots showing the "tuning curves" as a function of the source resistance R2. R2 is stepped 1m 1 10 100 1000 10K Ohms, resulting in the Green, Red, Lt.Blue, Dk. Blue, Violet, and Gray traces, respectively.

When connected this way, R2 is effectively in parallel with R1, so the Q of the circuit is determined by the parallel combination of R2//R1. As R2 approaches zero (I show 1mOhm), the Q is so low that no resonance is evident.

I am sure that the tutorial web page really intended the circuit to be what I show in post #11

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16. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Mike,

Expand your sweep from 1 mHz to 100 kHz. Lowering the R2 resistance increased the bandwidth of that "anti-resonant" circuit.

If we looked at the Norton equivalent of your source circuit, it becomes a 100 Ampere Current source.

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17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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We've got semantic difficulties here. The tutorial referenced:

http://www.electronics-tutorials.ws/accircuits/parallel-resonance.html

doesn't speak of "voltage resonance". Looking at some of the headings we find things like this:

"Current in a parallel resonance circuit"

"Parallel circuit current at resonance"

Further down where he does all the calculations, he's not calculating the voltages in the circuit. He specifically calculates currents, and shows how the total circuit current, Is, shows a resonance.

I posted a simulation showing that the current out of a voltage source applied to the parallel RLC circuit exhibits a resonance.

I suggested this: "What do you get if you redo the first simulation in post #9, but show I(V1), the current out of the source? Don't show any other currents on the plot so the scale won't be too large due to the currents in the inductor at low frequency and in the capacitor at high frequency."

Why don't you try it and see if you get a "current resonance". You're not denying that there is such a thing as a "current resonance", are you?

You have not been precise in what you say here:

You don't say "voltage resonance"; you just say "resonance". The tutorial never claimed that there was any "voltage resonance", nor have I. They never said "current resonance" either; perhaps they should have. If you want to consider "voltage resonance" only, ruling out "current resonance", then you should say "voltage resonance", not just "resonance".

If you are comfortable using the unqualified word "resonance" when what you're talking about is more specifically "voltage resonance", then you surely would not deny that other people could use "resonance" when they specifically mean "current resonance". Context will distinguish the two cases.

I don't need to be convinced that there is no "voltage resonance" in the parallel RLC circuit under consideration; I never said there was. I only assert that there is a "resonance", where it should be clear from the context that it's the current out of the voltage source that exhibits the "resonance".

Again, you've been imprecise. Agreed, there is no "voltage resonance", but there is a pronounced "current resonance", which is what the referenced tutorial described, even if they didn't call it that.

The NEETS glossary quote:

There's nothing there to rule out a parallel RLC circuit. They use the word "resonance" in its general sense, not limiting it to "voltage resonance" or "current resonance". They refer to a "condition", a quite general term not limited to any particular circuit.

If you apply a voltage source across a parallel RLC circuit, you won't see a "voltage resonance", but you will see a "current resonance".

If you apply a current source to a series RLC circuit, you won't see a "current resonance", but you will see a "voltage resonance".

The word "resonance" standing alone can describe can describe any kind of resonance. I don't think I've ever seen the terms "voltage resonance" or "current resonance". It's usually clear from the context what is relevant.

18. ### MrAl Distinguished Member

Jun 17, 2014
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Hi,

I am not sure what the big deal is here. Resonance does not have to show a 'peak' to be resonance, it can show up as a valley too. There are other errors in that paper but i consider them minor which i'll try to explain here.

Also, to see the waveform stabilize faster (exponentials die out) in a circuit simulator do not drive the input with a sine wave, instead drive the input with a cosine wave. Thus you do not have to 'wait' for the sine wave to center itself about the time axis. When using a sine you can also set initial values, but using the cosine wave is easier without having to set initial values.

First, lets get the values straight. The values i see in that paper are:
R=60 Ohms
L=200mH
C=120uf
and they quote the resonant frequency at 32.5 Hz which is close to the more precise 32.487 Hz. No big deal though.

And for the bandwidth they quote at 22 Hz, which is very close to the more accurate 22.1 Hz, but then when they go to calculate the upper and lower frequency they divide the bandwidth by 2 and then subtract and add to the resonant frequency of 32.5 and get a result that is a little more off than we might want to see. This is still an 'estimate', but i'll quote the more accurate upper and lower frequency for your convenience so you can see the difference yourself:
Paper: 32.5-22/2=32.5-11=21.5 Hz, Actual: 23.26Hz, Error: -1.76 Hz.
Paper: 32.5+22/2=32.5+11=43.5 Hz, Actual: 45.37 Hz, Error: -1.87 Hz

So in each case they were off by about 2Hz, but i might add that the method they used is a typical shortcut method just for getting an idea what the upper and lower frequency would be close to. The more accurate method using general circuit analysis yields the slightly better results.

Last edited: Aug 29, 2014
19. ### MikeML AAC Fanatic!

Oct 2, 2009
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Ok, but in your sim, you have (my) R2 set to 1Ohm, which exhibits a lowQ resonance, which I have already shown.

Now redo your sim with R2=0, which is the circuit purported to be analyzed on the Tutorial Web Page. Show me where there are any characteristics of "resonance", be it voltage or current...

20. ### MrAl Distinguished Member

Jun 17, 2014
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Hi,

You should not have to sweep that wide. Say from 32.5-20 to 32.5+20 which is about 12 to 50 Hz, and that should clearly show the dip near 32.5 Hz.
To show the current use a small resistor in series and plot i(Rs).