# resonance

Discussion in 'General Electronics Chat' started by PG1995, Nov 3, 2017.

1. ### PG1995Active Member

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Hi

For a series RLC circuit, it is said that at resonance frequency, the individual voltage across capacitor and inductor is equal and larger than the source voltage but 180 degrees out of phase with each which results into net voltage across capacitor and inductor to be zero.

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Please have a look here. Don't you think that the voltages founds for R, C, and L are completely wrong?

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2. ### ColinMember

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The mistake is 3.33mA x 2.2k The resistor is 1.5k

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3. ### Diver300Well-Known Member

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Also, that is a resonant circuit with a very low Q. The resistance of the resistor is larger than the in reactances. In a circuit that actually has useful resonance, you would expect the reactances to be large compared to the resistance. A series tuned circuit like that, when at the resonant frequency, would have much more voltage across the capacitor or the inductor than across the resistor.

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5. ### ColinMember

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The whole circuit is quite impractical and obviously devised by a University PROFESSOR who ...
(bile and invective deleted by moderator)

Your point is well made, the circuit is not a practical example of a tuned circuit, but something to provide a theoretical learning experience.

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6. ### MikeMlWell-Known MemberMost Helpful Member

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The circuit is intended to demonstrate that at one specific frequency, there exists a value of inductance and capacitance that makes the reactances equal but opposite, leaving only the resistance, which at that one specific frequency, determines the current in the circuit. Look at the following, where X represents the reactance (specifically 1KΩ, but the simulation works for any value). The frequency is f Hz, and the simulation also works for any value:

Here is the result where X=1KΩ and f=1KHz:

Note that (within the computational accuracy of the simulator), V(y) is essentially 0V, meaning that at the resonant frequency, the entire 5V input appears across R1, which makes the current in all four series elements 5V/1.5K = 3.333mA (at a phase angle of zero degrees). Look at node V(z), which has an intermediate voltage, but is in phase quadrature with the input.

To show that my circuit works for all frequencies, here is a plot of the magnitude and phase of V(x), V(y), and V(z) for all frequencies between 1mHz and 1GHz.

Finally, to refute Colin's stupid, insulting babble, here is the practical use of this circuit: If we vary frequency while keeping the reactances fixed, the output from node y makes the circuit into a highQ, very narrow notch filter. If we take the output from node z, then it makes a low-pass filter. Look at the Bode plot:

btw: This was brought to you by a retired professor of Engineering who learned about ducks long before he started as an electronics technician...

For those of you who might be interested in using LTSpice to do analysis in the frequency domain, I include the .asc file...

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7. ### MikeMlWell-Known MemberMost Helpful Member

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I take issue with "not a practical example". Look what happens to the circuit as we change the Q (ratio of X to R1=1.5KΩ). The plots are for Q = 375/1500 = 0.25 = Green, Q = 750/1500 = 0.5 = Red, Q = 1500/1500 = 1 = Blue, Q = 3000/1500 = 2 = Violet and Q = 6000/1500 = 4 = Yellow.

Who is to say that low Q is not more desirable with respect to ringing and/or transient response than high Q. Both low Q and high Q versions of this circuit could be "better", depending on what the goal is...?

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8. ### PG1995Active Member

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Thank you!

MikeMl: Your reply was really helpful. The frequency of 1000 Hz is a resonant frequency. The magnitude and phase of voltage V(y) is almost negligible. But the magnitude and phase of V(z) is relatively quite large. It should have been ideally zero like V(y). Could you please help me with this? Thank you.

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9. ### MikeMlWell-Known MemberMost Helpful Member

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At resonance, the Magnitude of V(z) is a function of the ratio XL/R1 = (2*Pi*f*L)/R1. Note the lower plot in Post #6.

If we set X to 1kΩ (the original problem), sweep frequency from 100Hz to 10kHz, and then plot both Mag(V(z) and Phase(V(z)), here is what we get:

As near as I can read the plot, at 1kHz, the voltage at V(z) is 3.333V at an angle of 90 degrees.

Look at the table you posted. Note that at resonance, the current everywhere in the circuit is 3.333 mA and the reactance of the capacitor (and the inductor) is 1kΩ. By simple application of Ohm, doesn't it follow that the voltage across the capacitor is 3.333V, but lagging the current by 90 degrees?

Do you understand why at much below resonance, V(z) approaches 5V, and much above resonance, it approaches 0V?

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10. ### MikebitsWell-Known Member

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Now have fun with parallel resonance

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11. ### PG1995Active Member

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Thank you, MikeMl.

I think that I was confusing individual voltage across inductor and capacitor with net voltage across both inductor and capacitor at the resonant frequency. At resonant frequency, the net voltage across both inductor and capacitor should, ideally, be zero. I was under the impression that the individual voltage across inductor and capacitor would always be greater than the source voltage. I was wrong. It might be but not always as your example proves.

Using your example from post #5 capacitor is 0.1592 μF and inductor is0.1592 H. Using equation 8 from this pdf: http://mlg.eng.cam.ac.uk/mchutchon/ResonantCircuits.pdf. It gives individual voltage of 3.33 V across inductor and capacitor with 180 degrees phase difference at the resonant frequency. For some reason, the value of V(y) at the resonant frequency given in the table of post #5 seems really inaccurate at least to me. In my opinion, it should be same as V(z) with +90 degrees phase. In the table it is given as 3.333 e^-006 (3.33 x 10^-6) with phase of -2.4622 e^-007 degrees (-2.4622 x 10^-7). Where am I going wrong? Could you please guide me? Thank you.

12. ### MikeMlWell-Known MemberMost Helpful Member

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V(y) is ~zero. The current is not. Therefore, the SUM of the inductor voltage and the capacitor voltage must be Zero.
Depending on Q, the voltage can be either smaller than 5V (the input) or larger.
If the voltage across the inductor is equal but opposite in sign to the voltage across the capacitor (definition of series resonance), then the theoretical voltage at V(y) should be Zero. The simulator calculates 3uV, which with its default convergence settings, is as close as it can get. This is due to how the simulator does its calculations, and also because even its capacitor model has some "leakage resistance", and its inductor model has a bit of "series" resistance...

The following plot may help you visualize what is happening. I plot the vector voltage across the capacitor V(z) Green and add the vector voltage across the inductor as the expression V(y)-V(z) Red. Note that at 1kHz, the voltages are the same, but their phases (relative to V1) are -90 and +90 degrees, respectively. This is what makes the voltage at V(y) very nearly zero...

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13. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

The way to prove or disprove any of this is to just do a little circuit analysis.

The voltage is 5vac of unspecified angular frequency w, the values of L and C are not given but their reactances are, and the total impedance of the RLC string is:
Z=R+s*L+1/(s*C)

and so the voltage across each element where Vs is |Vs| is:
vR=Vs*R/Z
vL=Vs*s*L/Z
vC=Vs*(1/(s*C))/Z

and if we replace s with j*w we get the complex AC representation of all the voltages.

Because the reactances are given and the circuit must be in resonance, that means L and C must follow a certain relationship:
w*L=1/(w*C)

and so:
C=1/(w^2*L)

and if we force w=1 (we have any choice of w) we have:
C1=1/L1

or simply for here only:
C=1/L

Since we forced w=1 we must then also force xL=1000, so we have:
w*L=1000

and with w=1 we have:
L=1000

and since L=1000 then C must be:
C=1/L=1/1000

and we now have everything we need to completely solve anything that might come up as a question in the AC realm.

For example, what is the voltage across R and the current through R?

Using the above for vR we get (using i=j here):
vR=5*(i*w*C*R)/(i*w*C*R-w^2*C*L+1)

and since C=1/L we can reduce that right away to:
vR=5*(i*w*R)/(i*w*R-w^2*L+L)

and since w=1 we can insert that w and reduce and get:
vR=5+0*i=5 volts

and since that is all real and no imag part there is no phase shift. So the resistor has 5 volts AC across it with no phase shift.
This of course means that the current through R is:
iR=5/R

Gee that was tough huh

So you see we can use basic circuit analysis to prove just about anything that has to do with, well, circuits.
It's a funny thing, that circuit analysis, that it actually works with circuits

Granted though you have to be careful to get everything right. That's another story

One thing you will note here. When we got the current through R, we got the current through R for any value of R as long as L and C are such that we obtain physical resonance. This is different than a simulation because in the simulation we only get the current for one value of R unless we do a bunch of R values. That is also possible however, to prove that any R would do the same thing, except for maybe R=0 which would cause infinite current:
iR=5/R
iR=5/0 (tends to infinity)

and R=infinity (open circuit):
iR=5/inf
iR=0

which causes a flow of zero current as we should expect.

As a side note, driving this circuit with a real sine wave will produce an exponential part as well as the sine part we are restricting ourselves to here. That means it will take some time for the circuit to settle into a purely sinusoidal response. This could be very fast or very slow depending on the values of L and C and R, but that's a subject for another time i guess. Once the circuit reaches what we call the steady state the exponential part will be damped out because of the factor:
e^(-a*t)

and then we see the sine response only which is really all that we are after for now.

In fact the time response is not too hard to obtain for this circuit and is:
vR(t)=sqrt(2)*[5*sin(t)-(20*e^(-(3*t)/4)*sin((sqrt(7)*t)/4))/sqrt(7)]

and for t greater than some long time t1 we are left with:
vR(t>t1)=5*sqrt(2)*sin(t)

which is of course the same as the source voltage.
A numerical analysis shows that after about 21.6 seconds the |vR| exponential response part is down to about 1uv which is very very much lower than the sine voltage part of 5v.

Last edited: Nov 6, 2017
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14. ### PG1995Active Member

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Thank you, MikeMl , MrAl.

I think that I understand it now. The ground and node 'y' are at the same voltage.

I was looking at the picture shown at the beginning of this page: http://www.electronics-tutorials.ws/accircuits/series-resonance.html. It looks like that the polarities shown are misleading; at least this is how it looks to me but I might be wrong. I have redrawn it. Please let me know if it's correct. Thank you.

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15. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Two points can be made here.

First, your interpretation of the polarity is a little flawed although you could do it that way. The main point to be made though is that the ASSUMED polarity does not have to agree with the actual MEASURED polarity. When we assume a polarity at the beginning of a problem it can be virtually anything. You are assuming that the inductor voltage is (graphically) - + while they assume + -. That means if we call the inductor voltage vL then your voltage for one circuit condition might come out to:
vL=+2.65v

while theirs would come out to:
vL=-2.65v

They are both right though because of the assumed direction at the start of the analysis.

The second point to be made is that it was good that they had shown the sharpness as R is varied, but the more important graph would be to see how the sharpness changes as L and C are varied. That's because the load is usually constant. In the bigger picture though it is good to see both.
Attached are two pics with three graphs. One is for the series circuit alone when L and C are varied. The other has two graphs where L and C are varied and has one hand drawn plot for the series circuit and one hand drawn plot for the parallel circuit. The series circuit in that plot is approximate but is still representative of the way the response gets sharper or wider in bandwidth as L and C are varied. That paper is from my notes from a long long time ago
Note that in all of my plots the plot is |Z| vs w not current vs w, which is a little different but if you want the current then just think about what it might be with the given impedance. It is max when |Z| is min and lower everywhere else.
It might also be interesting to note that the curve is symmetrical in log coordinates so when w=0.1 we get the same |Z| response as when w=10.

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16. ### PG1995Active Member

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Thank you, MrAl

It's amazing that you still have your notes.

How is it so? I was thinking more in terms of this picture where capacitor in connected in the middle rather than an inductor. Thank you.

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17. ### MikeMlWell-Known MemberMost Helpful Member

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Just a point of clarification about how LTSpice (and I) report voltage at a node: V(x) means the "voltage at node x with respect to Gnd" (the gnd symbol)
V(x,y) or the expression V(x)-V(y) means "the difference in voltage between node x and node y".

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19. ### PG1995Active Member

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Thank you.

I was reading about parallel RLC circuit. I think that at resonance the energy shifts back and forth between inductor and capacitor twice during each half cycle of Vs. This shows you how I'm thinking about it. Could you please let me know if I'm correct? Thanks.

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20. ### MikeMlWell-Known MemberMost Helpful Member

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A lossless parallel resonant circuit has only two nodes, in my example node p and ground (gnd symbol). You can only talk about the voltage at V(p) because by definition V(gnd) = 0.

Imagine that we initially cut the wire at X, and we pre-charge the capacitor C1 to 10V. There is initially zero current in the inductor L1, and then, at time = 0, we connect the wires at X.

Here is what plays out as a function of time...

This shows how the energy initially stored in the capacitor transfers to the inductor, and then back again... Since this is a lossless circuit, this would theoretically go on forever. Note that there is no voltage source other than what is implied by "precharging" the capacitor...

There is only one voltage to display. The energy exchange is shown by the quadrature relationship between the voltage V(p) and the current I(L1). Note that this is a series circuit, so I(L1) = - I(C1).

Why is the peak current 10mA?

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21. ### MikeMlWell-Known MemberMost Helpful Member

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I forgot to say, this circuit is the electrical analog of a pendulum, where potential energy transitions to kinetic energy and back again...

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