Series to Parallel inductance question

Reviewing some material I thought I understood, but it turns out there's some subtlety I'm not getting. The usual formula for calculating the Q of a non-ideal inductor is Xs/Rs, where Xs is the series reactance and Rs is the series resistance. For a parallel circuit, it's the inverse, Rs/Xs. The two are related by the equation Rp=(Q2+1)Rs, where Q is the unloaded Q of the element. Consider a non ideal parallel LC circuit. You have an R and L in series, and this series combination is itself in parallel with a capacitor. Using the above transformations, one can convert the series resistance into a value of parallel resistance, and then using the value of parallel resistance solve for a value of parallel inductance using the second formula above.

But then, what is the resonant frequency of the tuned circuit? Is it given by the original value of series inductance, or the new value of parallel inductance? The texts I've read make the assumption that the reactance of the parallel inductance is the same as the series inductance, and working out the equations with the Laplace transform of the original circuit shows that the resonant frequency is solely determined by the value of the original capacitance and non-lossy inductance. Indeed, it would hardly make sense to try to calculate a new resonant frequency based on this equivalent parallel inductance value, because the transformation is only valid at one frequency! So what's the significance of this new value of parallel inductance, then? For low Qs it can be significantly larger than the original series value. Does any of this make sense?

Reply to
Bitrex
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Did you mean to write Rp/Xs here? So you've got a circuit with R, L and C all in parallel? (A picture would help.)

A little bit, A picture or link to some equations would help. As you add damping (reduce the Q) to a resonant system you decrease the resonant frequency slightly... sqrt(1-damping^2) or somethng like that. Most of the time you ignore the slight shift. But you can see that things 'get weird' if you let the damping get close to one. You can have the damping greater than one too.

George H.

Reply to
George Herold

Sorry, I meant Rp/Xp.

Thanks for the reply. Does the resonant frequency actually shift, or is it the frequency of the peak? IIRC the resonant frequency is constant and set by the ideal component values, but the peak of maximum gain shifts. Perhaps the resonant frequency of the "new" inductance value and the capacitor is the new center frequency?

Reply to
Bitrex

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Hmm, that's mostly just a question of nomenclature. It=92s the peak frequency that shifts, but that=92s what I=92d call the resonant frequency. (When you drive the system at that frequency then the drive and response are in phase.) The undamped frequency appears in the equaitons, and for many practical purposes is equal to the resonant frequency.

George H.

George H.

Reply to
George Herold

Ahh, it makes sense now. At low Q the frequency where the impedance of the "parallel transformed" lossy inductor and capacitor is at maximum shifts off the undamped frequency, and this appears in the transfomration as an increase in the equivalent parallel inductor value. Thanks!

Reply to
Bitrex

It is Rp/Xp

Rp is as you note below. Lp is:

Lp = Ls(Rs^2/Xs^2 + 1)

This is not a "parallel" circuit. It is sometimes called a "series- parallel" circuit, as it is has elements of both.

Which is not strictly true at this single frequency transform.

You note the problem well and ask the right questions. To answer your first question, you need to define "resonance."

Strictly speaking, "resonance" means the power factor is 1 -- the frequency at which the 2nd order circuit provides a purely real impedance. For a "pure" series resonance and a "pure" parallel resonance, that is the usual

w = 1/sqrt(LC)

I have seen authors call "series" resonance XL = XC (same as w = 1/ sqrt(LC)) and parallel resonance as PF=1. But as you see, they are coincident in pure parallel or pure series circuits.

For the "series-parallel" circuit you mention, XL=XC and PF=1 diverge from each other. There is also a third point of interest: It is when I_total is at a minimum, meaning the impedance is at a maximum. This is sometimes called "anti-resonance." Again, these three points are not coincident for the "series-parallel" circuit. (They all converge in the "pure" parallel circuit.)

In my notes, I have:

series res: XL=XC w=1/sqrt(L*C)

Unity PF (par res): w = sqrt( 1/(L*C) - R^2/L^2 )

Max Z (anti-res): w = sqrt( 1/(L*C) - C*R^4/(2*L^3) )

It's been a long time since I derived them, so it is your exercise to double check. I also had a couple "exchange" arrows for the expressions of Unity PF and Max Z, so I might have them swapped. I admit the second two eq's look mysterious to me all these years later. lol!

In any case, a consistant definition is necessary and I think only the PF=1 definition makes sense, since at that frequency the circuit Z (a "dipole") appears completely real, regardless of the particular "internal" circuit of the second order circuit.

hth.

Reply to
Simon S Aysdie

I can't do the math, but if the inductor R and capacitance R are not equal the resonance point will shift and the greater the difference the greater the shift. It was a surprise to me, I thought the R had no affect on the frequency. Mikek

Reply to
amdx

Thanks very kindly for this informative post! It definitely helps clear up my confusion.

Reply to
Bitrex

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No problem. I may have not been completely clear, but the "in my notes" section was for the series-parallel circuit. "Series res" looks the same as for the pure series circuit, as it is a matter of definition. The series-parallel expression for "parallel res" (PF=3D1) is different than for a pure parallel circuit, perhaps obviously. PF=3D1 is the resonance.

Reply to
Simon S Aysdie

It should be easy enough to derive the above equations for PF=1 and maximum impedance. For PF = 1, take the Laplace transform of the series-parallel circuit, substitute j omega and separate into real and imaginary parts, then set the imaginary part equal to zero and solve for omega. For the frequency of maximum impedance, you'd take the magnitude of the transfer function and differentiate, setting the result to zero and solving for the local maximum. To make the differentiation easier one can remove the square root from the magnitude function because I think the result when solving for the maximum will be the same either way.

Reply to
Bitrex

Here's some formula that I picked from a webpage a while back. It gives magnitude of the impedance, The frequency of maximum impedance, and the frequency of zero phase angle of the impedance. I can't vouch for it, but someone went to a lot of trouble. Hope they are helpful.

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Mikek

Reply to
amdx

Here's some formula that I picked from a webpage a while back. It gives magnitude of the impedance, The frequency of maximum impedance, and the frequency of zero phase angle of the impedance. I can't vouch for it, but someone went to a lot of trouble. Hope they are helpful.

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Mikek

Reply to
amdx

Here's some formula that I picked from a webpage a while back. It gives magnitude of the impedance, The frequency of maximum impedance, and the frequency of zero phase angle of the impedance. I can't vouch for it, but someone went to a lot of trouble. Hope they are helpful.

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Mikek

Reply to
amdx

Computer locked up and, well you see what happened.

Reply to
amdx

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kek

That one includes a series resistor in the cap branch also. If one zeros the Rc, then the PF=3D1 (zero phase angle) expression is the same as the one I have in my notes.

I didn't do the same reduction for the anti-resonance expression, as it is pretty messy. It's a nice result page, but like my notes, it doesn't have the derivation work. :(

Reply to
Simon S Aysdie

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If you work it out, post it to see if it agrees with my notes. I don't have the derivation -- or at least I don't know where it is.

Reply to
Simon S Aysdie

The image amdx has posted is my work. Since I posted that, I've done a little more work, added another expression for the frequency at which the real part of the impedance reaches a maximum.

I've also simplified the various expressions, and put them in a form that shows them as a factor multiplying the series resonance frequency.

I haven't shown the derivations because they're mind bogglingly messy. I did the algebra using Mathematica so I'm quite certain the results are correct, but I would certainly welcome verification or falsification by someone else who wants to do the work.

Note that in every case, if RL = RC, the pertinent frequency reduces to

1/(2*pi*sqrt(L*C))

See:

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for the results.

Reply to
The Phantom

Thanks Phantom, I saved that image a while back when I was mulling over a high Q LC for a crystal radio. If I had known I would be posting it, I would have saved the source and given you credit. Thanks, Mikek

Reply to
amdx

Yes, I remember. It was shortly after that original posting I made that I revised the formulas.

Not a problem. I'm not bothered by anybody's use of the formulas.

Reply to
The Phantom

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That's just a big mess of algebra...

Reply to
Fred Bloggs

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