Resonance in RLC Circuits

  • #1

Homework Statement


I did an experiment using an RLC circuit (in series) and an oscilloscope to determine the resonance frequency and amplitude (in volts) of my components (R total = 2300ohm, L = 0.081H, C = 1.5E-8F). I plotted all the data I collected (V vs. f) and got a nice response curve which looks like a bell curve with max amplitude of 21.5V @ 16kHz where my initial input voltage was 8V.

The problem:
I am asked to compare my experimental data to a theoretical calculation of both the resonance frequency and amplitude. But I have no idea what equations to use or how to find the values otherwise. Are there any equations that can help me with that?

Thanks!
 

Answers and Replies

  • #2
gneill
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http://en.wikipedia.org/wiki/RLC_circuit" [Broken]
 
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  • #3
Hey gneill!
Yea I saw this article already, but I did not see anything about finding the amplitude.
Finding the resonance frequency looks easy enough 1/sqrt(LC)...

I found another formula meanwhile: A=(v0/C)/sqrt((R^2)(w^2)+L^2(w^2-(1/LC))^2). I tried differentiating it with respect to w (resonance frequency) after substituting all my variables in and substituting it into the original equation of A to find the amplitude...

However the results I'm getting for the amplitude do no produce anything that looks like a response curve and the values of amplitude range between ~8-10 as oppose to my experimental values which range from 5-23 (I've collected my data by repeating the experiment 5 times so I'm sure that my values are rather accurate). Any ideas?
 
  • #4
gneill
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Are you familiar with the concept of impedance? (Sort of an "up scale" version of resistance).

Which component(s) were you taking the voltage measurements across for plotting?
 
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  • #5
I saw impedance show up in my reading from time to time but I am not too familiar with it.
I attached the circuit diagram that I used... The measurements were taken across the capacitor.
 

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  • #6
gneill
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Impedance is a sort of resistance that takes into account the effects that reactive components (like capacitors and resistors) have on the voltage and current in a circuit. It is measured in Ohms, just like resistance, but it is a complex number rather than a simple real number. This allows the impedance to reflect the fact that in a circuit driven by a sinusoidal source, the current will lead or lag the voltage waveform by some amount (phase shift) -- complex numbers allow the angular offset of the waveforms to be represented.

The impedance of a resistor is just the resistance value -- a straightforward real number. The impedance of capacitors and inductors is purely imaginary:

[tex] Z_C = \frac{1}{j \omega C}[/tex]

[tex] Z_L = j \omega L [/tex]

The impedance values can be treated just like resistances when analyzing a circuit. Of course, the rules of complex arithmetic apply, and often you are interested in the magnitude of a result rather than a complex value.

In your case you have a circuit that is essentially a voltage divider, with a resistor and inductor "on top" of a capacitor across which you're measuring an output voltage.

If your driving voltage is V (a sinusoid) then the voltage across the capacitor will be

[tex] V_o = V_{in} \frac{Z_C}{R + Z_L + Z_C}[/tex]

Of course the result will be complex, reflecting the phase shift being produced by the reactive components.

When you measure with a meter or oscilloscope you're typically looking at the magnitude of the output voltage. So take the absolute value, and obtain:

[tex] |V_o| = V_{in} \frac{1}{\sqrt{R^2 \omega^2 C^2 + \omega^4 L^2 C^2 - 2 \omega^2 L C + 1}}[/tex]

This isn't quite the expression that you found.

Now, you specify that the input voltage was 8V, but presumably that was 8V RMS? That would give a peak-to-peak value of about 22.6V, which is close to your reported maximum. Were you measuring peak-to-peak on the oscilloscope for your results?

I don't see the "bell curve" you report coming out of this expression. A peak, yes, but I'd expect a relatively large flat response at low frequencies (since the inductor has a low impedance at low frequencies and the capacitor a high impedance there), followed by a peak and then a plunge in output as the capacitor "shorts out" higher frequencies and the inductor blocks them.
 
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  • #7
Thanks for the explanation about the impedance!
Yea, I measured peak to peak to get my results. I tried plotting a theoretical curve using the formula for amplitude you provided but I am getting an exponential curve (20Hz-25000Hz) which gives me a much higher peak value than the one you found (22.6). I used excel with the following formula using the variables I listed originally.
8.6/SQRT((R*(w^2)*(C^2))+((w^4)*(L^2)*(C^2))-(2*(w^2)*(L)*(C))+1)
What can be the problem?

Edit: Re: the bell curve, I managed to fix it and the curve now looks like you described it. The theoretical curve is still an exponential though...
 
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  • #8
gneill
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Thanks for the explanation about the impedance!
Yea, I measured peak to peak to get my results. I tried plotting a theoretical curve using the formula for amplitude you provided but I am getting an exponential curve (20Hz-25000Hz) which gives me a much higher peak value than the one you found (22.6). I used excel with the following formula using the variables I listed originally.
8.6/SQRT((R*(w^2)*(C^2))+((w^4)*(L^2)*(C^2))-(2*(w^2)*(L)*(C))+1)
What can be the problem?

Edit: Re: the bell curve, I managed to fix it and the curve now looks like you described it. The theoretical curve is still an exponential though...
I must apologize. When I "LaTexed" the formula I missed a square on the first R in the denominator. The formula should have read:

[tex]
|V_o| = V_{in} \frac{1}{\sqrt{R^2 \omega^2 C^2 + \omega^4 L^2 C^2 - 2 \omega^2 L C + 1}}
[/tex]

I've fixed it in the original post, too.

Again, sorry for the typo.
 
  • #9
I tried the new formula and although the curve starts to resemble my experimental one, my peak value is only 9.9V while the lowest value is 8.6V. For which frequency did you find the peak value to be 22.6V?
 
  • #10
gneill
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I tried the new formula and although the curve starts to resemble my experimental one, my peak value is only 9.9V while the lowest value is 8.6V. For which frequency did you find the peak value to be 22.6V?
Actually, my curve's peak was a bit higher than that, around 26.3V. Since the supply was 8V RMS (how accurate is that number?), that makes the peak-to-peak driving voltage 2√2 times that, or about 23 V.

The peak occurred around f = 3260 Hz, or ω = 20490 rad/sec.
 
  • #11
What program did you use to find the values and plot them?
I tried maple and excel but I can't get the anything above 10V for the peak....
 
  • #12
gneill
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What program did you use to find the values and plot them?
I tried maple and excel but I can't get the anything above 10V for the peak....
I use Mathcad, a commercial program. Yes, I'm spoiled.:biggrin:

But Excel should be able to handle it fine. You can do semilog plots with Excel easily enough.
 
  • #13
I noticed that when I put the equation and calculate the values the term under the square root always equals to ~1 so that most of my terms end up being very close to the numerator term (ie. Vin). looks like all the other values are too small to affect the 1 under the square root in the equation. I think I tried every variation of the equation by now and it still doesn't work.
There is also barely any difference if I use "w" in Hz or rad/s since the number under the square root still tends to ~1...
 
  • #14
gneill
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I noticed that when I put the equation and calculate the values the term under the square root always equals to ~1 so that most of my terms end up being very close to the numerator term (ie. Vin). looks like all the other values are too small to affect the 1 under the square root in the equation. I think I tried every variation of the equation by now and it still doesn't work.
What frequency range are you plotting?
 
  • #15
20Hz - 25000Hz since that's pretty much what I used in my experiment.
 
  • #16
gneill
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20Hz - 25000Hz since that's pretty much what I used in my experiment.
And that's Hz, not radians per second (ω) ?
 
  • #17
I converted the frequency into radians. I attached my excel file, I think it'll be easier to see what I did that way.
 

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  • #18
gneill
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I don't see a problem there. You've plotted RMS values based upon your 8V RMS input. You were recording P-P volts though, right?
 
  • #19
Yea I was recording peak to peak. Am I missing something here?
Edit: I think I see it now... do I have to convert from RMS to p-p (RMS*2*SQRT(2))?
 
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  • #20
gneill
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Yea I was recording peak to peak. Am I missing something here?
8V RMS has a p-p amplitude of about 22.6V.

I'll trade you an excel workbook; attached is one that I put together just before you sent yours!
 

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  • #21
Yea it all makes sense now. I was wondering though, why did my experimental curve started from a much lower point (~5V) while the theoretical starts almost near the peak? Both graphs have the same scale and cover almost the same frequencies. Also the resonance frequency is around 4kHz as oppose to the 15kHz in my experimental curve. Looks like a very large error compared to what it should have been.
 
  • #22
gneill
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Yea it all makes sense now. I was wondering though, why did my experimental curve started from a much lower point (~5V) while the theoretical starts almost near the peak? Both graphs have the same scale and cover almost the same frequencies. Also the resonance frequency is around 4kHz as oppose to the 15kHz in my experimental curve. Looks like a very large error compared to what it should have been.
I can't say for sure why the difference. It could be that one or more part values were not as they seemed, or that the coil had some internal resistance, or the capacitor had some leakage resistance, or the supply voltage didn't remain steady over the frequency range.

You might note also that the peak doesn't occur right at the natural frequency ωo, but at the damped frequency, ωd (you'll find it on the wiki page).

EDIT: Actually, the peak occurs before the damped frequency! Solving for the peak (differentiating the transfer function w.r.t. ω , setting to zero, blah, blah,... yields

[tex] \omega_{peak} = \omega_o \sqrt{1 - \frac{1}{2}\omega_o^2 R^2 C^2} [/tex]

in this case about 3.3 kHz.

P.S. Are you sure that the capacitance was 1.5 x 10-8 F and not 1.5 nF or 1500pF or .0015 µF?
 
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  • #23
Yea I see what you mean. I appreciate the help gneill! Thanks.
 
  • #24
gneill
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theuniverse, I made an edit in my previous post that might be worth looking at.
 
  • #25
If it is 0.0015 µF? then how would it affect the result?
 

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