Resonant Circuit problem

[h3rroin]

Hello!

I have some problems when calculating on a special RLC-circuit. The circuit looks like this:

1 inductor (50nH) in series with a RLC-circuit that consists of 1 inductor (50nH), 1 capacitor (variable) and 1 resistor (139 ohms) all paralell-connected.

The resonant frequency is at 90 MHz, and the question is how much the inimpedance is then?

I found that the circuit has two resonances, one series resonant and one parallel resonant. Continued then with defining the resonance circuit from Im(Y) = 0, where Y is the parallel resonant circuit Y = jwL1 + 1/(jwL2) + jwC + 1/R. Am I on the right way?

I can draw a picture of the circuit if it is needed!

 

[Dean Huster]

At resonance, the RCL parallel circuit should appear as pure resistance with a value of 139 ohms (the inductive and capacitive reactances cancel each other at resonance). At this same frequency, the inductive reactance of the series inductor is 28.274 ohms. Slapping those two into the "square root of the squares" equation gets you 141.875 ohms impedance without all the engineering "j" factors.

Lessee if there's any disagreement out there. Odds are, there is!

Dean

 

[RadioRon]

Hello!

I have some problems when calculating on a special RLC-circuit. The circuit looks like this:

1 inductor (50nH) in series with a RLC-circuit that consists of 1 inductor (50nH), 1 capacitor (variable) and 1 resistor (139 ohms) all paralell-connected.

The resonant frequency is at 90 MHz, and the question is how much the inimpedance is then?

I found that the circuit has two resonances, one series resonant and one parallel resonant. Continued then with defining the resonance circuit from Im(Y) = 0, where Y is the parallel resonant circuit Y = jwL1 + 1/(jwL2) + jwC + 1/R. Am I on the right way?

I can draw a picture of the circuit if it is needed!

Yes, please draw a picture of the circuit. Do I understand correctly that you want to calculate the input impedance?


edit: oops, did it wrong. Defining resonance as the point the input impedance is purely resistive, I get Zin = 133 ohms

 

[RadioRon]

At resonance, the RCL parallel circuit should appear as pure resistance with a value of 139 ohms (the inductive and capacitive reactances cancel each other at resonance). At this same frequency, the inductive reactance of the series inductor is 28.274 ohms. Slapping those two into the "square root of the squares" equation gets you 141.875 ohms impedance without all the engineering "j" factors.

Lessee if there's any disagreement out there. Odds are, there is!

Dean

If you resonate the parallel section only, as you do, then the entire circuit boils down to a 50nH inductor in series with a 139 ohm resistor. If you define a resonant circuit as one which has a purely resistive input impedance, then this is not resonant. To get a purely resistive Zin, you need to tweak the capacitor value a little bit so that the net reactance of the parallel circuit cancels out the reactance of the inductor.

 

[h3rroin]

                |--^^R^^--|
A x-----~~L1~~--|--| C |--|-----x B
                |-~~L2~~--|

L1=L2 = 50 nH
R = 139 Ohms
C = Variable

Yes, it's the input impedance i want in the circuit at resonant frequency.

But i think i'm getting it now, the circuit becomes:

A x-----~~L1~~-----^^R^^-----x B

At resonance and then you just calculate it with: Z = R + jwL1?

Where omega is the resonant angular frequency?

And then.. just pythagoras on that, simple .

If you define a resonant circuit as one which has a purely resistive input impedance, then this is not resonant.

But I think he meant purely resistive @ resonance, or?

Thank you everyone! Great

 

[Jamjoom(EE)]

Hello, Wat Dean said is right, at resonance the circuit is pure resistive where the impeadances tend to cancel each other out. I need to correct something h3rroin, that resonance means IM(Z)=0 (impeadance=0) which means IM(Y)=infinity (admittance=infinity).

Regards,
Ahmad Jamjoom

 

[RadioRon]

                |--^^R^^--|
A x-----~~L1~~--|--| C |--|-----x B
                |-~~L2~~--|

L1=L2 = 50 nH
R = 139 Ohms
C = Variable

Yes, it's the input impedance i want in the circuit at resonant frequency.

Thank you for drawing a circuit. However, would you be so kind as to define exactly what is the "input". I ask this because I am not sure if you are showing a one port network or a two port network. Your diagram does not define where the input is. If it is a one-port network, then the input would be across A and B (that is, A relative to B). If it is a two-port network, then I can assume the input is at terminal A with respect to an implied ground reference. But then I would want to know what is attached to B, and I would want to confirm that there is an implied ground reference for both ports. I hope you can see my meaning.

 

[h3rroin]

RadioRon:

Oh I maybe forgot to say, but it is a one-port network, input is across A and B. I get your point

I think that the impedance 141.8 ohms is the parallel resonance. But how do I get the series resonance?

Thanks!

 

[RadioRon]

RadioRon:

Oh I maybe forgot to say, but it is a one-port network, input is across A and B. I get your point

I think that the impedance 141.8 ohms is the parallel resonance. But how do I get the series resonance?

Thanks!

You are misunderstanding the way this circuit will work. There is no separate parallel and series resonances, there is only one frequency at which the network can be reduced to a pure resistance. ***Edit: THIS IS WRONG. SEE LATER POST*** We can model the circuit as a series RLC for ease of analysis. At this one frequency, the +j reactance of the series inductor will be cancelled out by some -j reactance contributed by the parallel network. So, the reactances of the parallel inductor and capacitor are not equal in magnitude, as this sub-network has to look like a capacitor in order to contribute some capacitive reactance to cancel out that series inductive reactance.

I confirmed this by simulation, but it is also intuitively apparent that you can't just choose a capacitor value that provides the same inductive reactance as the 50nH inductor at 90 MHz. If you do this, you are left with an inductive reactance in the overall circuit at 90MHz and so the circuit would clearly not be resonant.

Start by calculating the inductive reactance of the series inductor, and then you can deduce that you need the same magnitude of capacitive reactance from the parallel network at resonance. Convert this to a susceptance. Then calculate what amount of parallel capacitive susceptance is necessary to negate the parallel inductive susceptance and leave you with the necessary excess capacitive susceptance. Now you can calculate the value of the capacitor.

When doing your parallel to series conversions to calculate the overall impedance of the circuit be sure to include the resistor.

 

[h3rroin]

Okay, but what I meant was not that it has two resonances, I was a little bit unclear. I meant that there must be two different impedances, right? One with low impedance (series) and one with high impedance (parallel). I am almost sure about that.

I'll try to calculate a little and then return with some conclusions, but thanks for the help so far! It's helping me to understand this

 

[RadioRon]

Okay, but what I meant was not that it has two resonances, I was a little bit unclear. I meant that there must be two different impedances, right? One with low impedance (series) and one with high impedance (parallel). I am almost sure about that.

I'll try to calculate a little and then return with some conclusions, but thanks for the help so far! It's helping me to understand this

Ah, yes you are absolutely correct, I made a mistake in my explanation. I redid my calculations more carefully. To better understand the circuit I started by setting the resistor to infinity rather than 139 ohms. It then becomes clear that there is one frequency where the Xind = Xcap and no current can flow through the parallel network, so the Z of this parallel network goes to +jinfinity and then goes through 0 to get to -jinfinity. Obviously when you put this parallel network reactance in series with an inductive reactance of about +j28 ohms, the frequency at which the sum still equals 0 shifts over only an infinitely small amount. Then, when you calculate as I suggested in my previous post, only setting R=inf, then we find that there is another frequency where the imaginary part of the circuit Z goes through zero, up at about 124.66 MHz. This is the frequency at which the parallel network has a capacitive reactance that matches (except for sign) the reactance of the series inductor.

Now, when I put the resistor back in to the network, with value 139 ohms, I find by calculation and also by simulation that the two frequencies at which the imaginary part =0 are 90MHz and 122 MHz. At 90 MHz, the Zin is 133 ohms, where at 122 Mhz the Zin is 11.5 ohms.

Is this the same as the result that you calculate?

 

[h3rroin]

I feel I suffer from mental exhaustion, but I'll try anyway. (and I haven't had the time for some calculation)

Lets see.

L1 reactance: +j28 ohms
L1 susceptance: 1/j28 ohms (?)

C susceptance (in order to cancel the two inductive susceptances that come to existence): +j2/28 ohms (or -2/(j28) ohms if you like) (?)

Am I thinking right?

Then I'm not really sure how you have calculated the series resonance. Is it by this formula:

Im(Z)=0

1/(jwL1)+jwL2+1/(jwC) = 0

?

Hmm... I did not really get Zin=133 ohms, rather 141.8 ohms (as mentioned in a earlier post.)

Isn't it: Z=R+jwL = 141.8 ohms (@ 11.5 degrees)

As I said, slight risk of mental exhaustion!

Cheers!

EDIT:
To reply Janjoom's post:

Doesn't that depend if you have a RLC-circuit in series or parallel?

 

[Roff]

I know you are trying to get a handle on the mathematical analysis, but a simulation will do wonders for visualizing what's going on.

 

[h3rroin]

Yep, I tried to analyze and simulate it in multisim. But I can't get it working

It's the variable cap that bothers me (among others).

Is it the network analyzer you have to use btw?

 

[Roff]

Yep, I tried to analyze and simulate it in multisim. But I can't get it working

It's the variable cap that bothers me (among others).

Is it the network analyzer you have to use btw?

I used LTspice, and just tried some different values of C, as well as running a parameter sweep of C.

 

[RadioRon]

My head is starting to hurt too. I approached this by first simulating it to get an idea of what to expect. This was misleading because the simulation output a graph of magnitude, real and imaginary parts of Z vs frequency from 50 to 150MHz. When the graph auto-scaled it put the Y axis with a very high max value since the magnitude of Z goes very high at 90MHz, and so I completely missed the other zero crossing of the imaginary part at 122 MHz, where the impedance is quite small.

Next, instead of solving it all analytically like a mathematician might do, I spreadsheeted the problem with Excel, with one column for frequency, and several others for various admittance and impedance terms. It was in this spreadsheet that I saw the imaginary part crossing zero at 122 Mhz.

I feel I suffer from mental exhaustion, but I'll try anyway. (and I haven't had the time for some calculation)

Lets see.

L1 reactance: +j28 ohms
L1 susceptance: 1/j28 ohms (?)

Yes, that is correct.

C susceptance (in order to cancel the two inductive susceptances that come to existence): +j2/28 ohms (or -2/(j28) ohms if you like) (?)

Am I thinking right?

No. First, you have to name which resonance you are solving for. The original problem only states that the capacitor was tuned for resonance at 90 MHz, but there are two resonances. So I chose for the high impedance parallel resonance to be at 90 MHz.

Its too hard to type a whole explanation, so I've pm'd my spreadsheet to you. It shows the calculation steps to find Z given C. Then I just varied C.

Then I'm not really sure how you have calculated the series resonance. Is it by this formula:

Im(Z)=0

1/(jwL1)+jwL2+1/(jwC) = 0

?

Hmm... I did not really get Zin=133 ohms, rather 141.8 ohms (as mentioned in a earlier post.)

Isn't it: Z=R+jwL = 141.8 ohms (@ 11.5 degrees)

As I said, slight risk of mental exhaustion!

How can Z =141.8 @ 11.5 degrees at resonance, when we define resonance to be the frequency at which the angle =0?

 

[MrAl]

Hello there,


The resonant impedance partly depends on how you want to 'tune' the capacitance, C.
Tuning it for LC resonance we get 141.8, but tuning it for circuit resonance
we get 144.5 or so. We can also look for other resonant points (later).
If we tune it for LC resonance, we end up with a resonant frequency that is
lower than what we really wanted, so we have to find another way to solve
for C.
BTW, the imaginary part only goes away in some circuits, not all, and
not this one in particular.


Following the LC resonance rule:

C=1/(4*pi^2*F^2*L2)

Z then becomes:

Z=1/(1/R-j/(w*L2)+j*w*C)+j*w*L1

At F=90MHz this is:

Z=28.2743338823*j+139.0

This is the complex impedance, so the magnitude of Z is therefore:

141.8465

However, although 90MHz is the resonate point of L2 and C, it is
not the resonate point of the circuit.

Exciting the circuit at 88MHz for example get get:

|Z|=144.2

which is higher than at 90MHz.

A quick look at some other frequencies close to 90MHz and |Z|:

86MHz 140.1087
87MHz 142.8814
88MHz 144.2395 (peak)
89Mhz 143.9309
90MHz 141.8465
91MHz 138.0561
92MHz 132.7975


This tells us that the circuit resonate frequency is closer to 88MHz.

Thus, to get the circuit itself to resonate at 90MHz we would have
to lower the value of C (by about 2pf), but to get a better idea we
could perhaps start with this:

Z(jw)=1/(1/R-j/(w*L2)+j*w*C)+j*w*L1

and take the abs value we get:

|Z|=sqrt((0.11111111111111/(pi*(1.8*10^+8*pi*C-0.11111111111111/pi)^2+pi/19321)-(1.8*10^+8*pi*C)/((1.8*10^+8*pi*C-0.11111111111111/pi)^2+1/19321)+9.0*pi)^2+1/(19321*((1.8*10^+8*pi*C-0.11111111111111/pi)^2+1/19321)^2))

now take d|Z|/dC and we get:

dZ/dC=9*pi*sqrt((50706032400000000000000*pi^4*C^2-125200080000000*pi^2*C+81*pi^2+77284)/(50706032400000000000000*pi^4*C^2-62600040000000*pi^2*C+81*pi^2+19321))

and set that equal to zero and solve for C we get:

C=-(sqrt(324*pi^2+19321)-417)/(450360000000*pi^2)
C=(sqrt(324*pi^2+19321)+417)/(450360000000*pi^2)

the first one we want:

C=-(sqrt(324*pi^2+19321)-417)/(450360000000*pi^2)

so

C=6.005512e-011

and this will give us a circuit resonance at 90MHz or very close to it.

Z will then be approximately 144.53 or so. Note that the original cap value
was about 6.25e-11 so this value for C is about 2.5pf lower.



You can find all the resonances by finding d|Z|/dF and
setting it equal to zero:

dZ/dF=0

and then find all the solutions. This should show all the resonate
points, but it isnt that easy to do. Another way is to simply
sweep |Z(jw)| through a wide range of frequencies like 10MHz to 1000MHz,
looking for peaks and dips. Once you find a peak or dip, you
can home in on it by narrowing the frequency range, to say
80MHz to 100MHz for the one near 90MHz for example.

 

[Roff]

BTW, the imaginary part only goes away in some circuits, not all, and not this one in particular.

MrAl, if you are saying that the phase of the impedance in this circuit is never zero, that's wrong.

You can find all the resonances by finding d|Z|/dF and
setting it equal to zero:

dZ/dF=0

You appear to be defining resonance as the minima and maxima of impedance. H3rroin and RadioRon are defining it as the frequencies where the impedance is purely resistive (phase angle is zero).

 

[MrAl]

Hello Roff,

I appreciate your response and will try to address these issues
as time permits. I have to disagree on several issues though and
will explain thoroughly.

MrAl, if you are saying that the phase of the impedance in this circuit is never zero, that's wrong.

In the context of the discussion, saying that the impedance in this circuit
never goes to zero means it does not go to zero at resonance. See below
for more info on this.

You appear to be defining resonance as the minima and maxima of impedance. H3rroin and RadioRon are defining it as the frequencies where the impedance is purely resistive (phase angle is zero).

Well, i assure you that i not only appear to be defining resonance in this
way, i actually am.
Unfortunately H3rroin and RadioRon's definition of resonance is not correct.

Why?
If we define resonance as the point where the response is purely
resistive, we find that many 'tuned' circuits are no longer tuned
in that they dont have a resonant point, and this simply cant be true.
This means that resonance must be defined in terms of the amplitude,
not the phase.
What has happened in the past is that many people carry over the
definition of resonance in a circuit with only
*one inductor and one capacitor*
which has a point where the impedance becomes resistive and that
point coincides with the resonant point. Thus, it becomes common
to say that the resonant point is the point where the imag part
goes to zero, but this is not true in many circuits with more than
one cap and one inductor.

To show these points clearly i've included a diagram below
of the current response of this circuit (Circuit 1).

If you or anyone else still disagrees for some reason, keep the original
LC tuning of this circuit (C=62pf approx) and change L1 to a 150nH
inductor instead of 50nH and then try to tell me the circuit no longer
has any resonant points. Alternately, try to tell me the response
shown in Circuit 2 below has no resonant points. Good luck

Dont get me wrong, if we have a good reason for looking for the
point where the response becomes resistive that's just fine, but
we cant confuse this with the resonant point because this is
only going to be true SOMETIMES, for SOME circuits, and not
for every circuit.



**broken link removed**

**broken link removed**

 

[RadioRon]

One of the reasons I read this board is that I always learn new things or I get reminded of lessons learned once but long since forgotten. This is a very good case in point. Here is an example where the answer to the simple and most usual problem (or the "rule of thumb") does not necessarily extend to the general or more complex case. I looked up the more general definition of resonance, in both mechanics and the electrical world, and indeed it is about maximums of amplitude. So my previous responses to the OP were incorrect.

With this new information, I find the most obvious resonance, (and by this I mean a circuit resonance where the impedance between A and B is a maximum magnitude), when the capacitor tunes to 60.16 pF and the net Z of the circuit is 144.5267 ohms with a phase angle of -24 degrees. I'm not going to recalculate for any other local maximums.

If there is disagreement amongst us practicing engineering types imagine how confused the poor OP is. Sorry about the confusion and extra work h3rroin. ( and I sure hope we got it right this time!)

PS: to h3rroin; my spreadsheet is still correct in how it calculates Z, only my interpretation is wrong and the value of C is wrong.