Equation for Maximum Q in resonant circuits

What do you mean by optimal?

You can't create a filter with Qf > 100 (by measure of overall bandwidth) using components with Qc < 100 (by defintion of reactance/ESR). Narrow bandpass/stop filters for instance. If you try, you end up with Qf < Qc regardless, and insertion loss being huge (roughly, I would guess, equal to the ratio of Q's).

The resulting sharpness or efficiency of a filter (i.e., eta = (transmitted + reflected) / incident power) would then seem to be limited by the Qf/Qc of each complex pole pair (a similar rule should hold for real poles as well), so that if you are limited to components of a certain Q, your only choice for sharper cutoff (up to a limit) is to accept greater absorbed power losses in the transition band. Which ultimately limits the sharpness both in amplitude (components of too-low Q can't be adjusted to peak the response) and frequency (Qf < Qc).

Note the distinction between definitions of Q: for a simple (all-parallel or all-series) RLC network, they are one in the same, when all losses are lumped into the R equivalent. But when that network is placed in a system with other impedances, the overall response need not be characteristic anymore. Another way to put it: this Qc is a local measure, per component (or perhaps per resonator, in a bandpass/stop design), while the Qf is the component(or resonator)-in-system effect, measured by bandwidth and absorption loss.

Tim

-- Seven Transistor Labs Electrical Engineering Consultation Website:

For example if you want to increase loaded Q, you need a much smaller resonance reactance than load resistance, so you can pick a small L and high C, but once you reduce the reactance, you have decreased the unloaded Q which would affect the equivalent resistance parallel to the tank, which would reduce the loaded Q.

So I tried to come up with some general formula to maximize loaded Q. In a previous post here (long thread about how clueless I am building oscillators):

one of the experts here suggested that L and C should really be determined once you have some known loaded Q and tank impedance. However suppose you want to maximize the Q knowing RL and Rs where RL is the load impedance and Rs is the equivalent series resistance of the inductor.

My attempt at this resulted that, assuming Rs is constant, the optimum loaded Q would happen when the reactance at resonance is equal to the square root of RL*Rs.

Assuming Rs is constant may not be a good assumption since it will change with changing inductance, but once you have starting value, some tweaking is possible, especially that Rs changes slower than changes in L.

My question is: is this result a well known result? Would I find this in an Electronics text?

Thanks.

erdependent relations of circuit parameters relating to loaded Q.

onance reactance than load resistance, so you can pick a small L and high C , but once you reduce the reactance, you have decreased the unloaded Q whic h would affect the equivalent resistance parallel to the tank, which would reduce the loaded Q.

a previous post here (long thread about how clueless I am building oscillat ors):

ussion

d once you have some known loaded Q and tank impedance. However suppose you want to maximize the Q knowing RL and Rs where RL is the load impedance an d Rs is the equivalent series resistance of the inductor.

aded Q would happen when the reactance at resonance is equal to the square root of RL*Rs.

with changing inductance, but once you have starting value, some tweaking is possible, especially that Rs changes slower than changes in L.

an Electronics text?

I'm not sure exactly what you are asking. But one way to get feel for LRC circuits is to try and make your own. I assume you have a 'scope, function generator and soldering iron.

Then you can try loading it down.

If you make a series LRC, you can make your 50 ohm function generator a low impedance by throwing away signal and loading it with an ohm or two. (Do you need a picture?)

George H.

Ok this is flying way over my head. Maybe I should have posted this in basi cs. I am only dealing with a simple RLC circuit. I assumed Rs, RL are fixed (ESR and load resistance), and X (reactance) is variable.

By optimal Q I mean the maximum loaded Q I can get from this arrangement, o r in other words, the sharpest resonance peak.

I know assuming Rs is fixed as I change the inductance L (hence change the reactance X) is not a good assumption, but I don't know how bad it is eithe r. I wonder if I can arrive at the optimal L and C values (hence the sharpe st resonance), if I just start with the Rs = DC resistance of a certain c oil and given a specific load RL, I calculate X = sqrt(RL*Rs) then constr uct a new coil, then measure the new DC resistance, and so on until I narro w it down a bit, then start including high frequency effects like skin effe ct.

It's worth noting that after some manipulation, the equation I mentioned is another way of saying: you will get the maximum loaded Q when RL=Rp wher e Rp is the equivalent parallel resistance.

I do!

Wouldn't hurt :)

But what is a "simple RLC circuit"?

If you're looking at the RLC (all series or all parallel) circuit, it's a dynamic system yes, but it doesn't... *go* anywhere. It doesn't filter, it doesn't dissipate power, it just... is.

So what are you doing? Are you looking at what happens when you have a series or parallel RLC network strung between a source and load, or in parallel with?

Tim

I think the circuit in question is the "parallel tuned doubly terminated" type, like this:

and the inductor is modeled with some ESR. Maybe the OP can correct me if I'm wrong.

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Hamed,

Go back to first principles and you can figure it out yourself.

One definition of Q is the energy circulating divided by the energy lost per cycle. So if 2% of the circulating energy is lost per cycle, you have a Q of 50.

In a parallel tuned circuit with no load resistor, energy is only lost in the series resistance. This varies with the RMS current; so at the same frequency and power level a tank with a bigger L and smaller C will have less loss due to Rs, because the voltage swing is higher, current swing is lower.

If Rs is negligible but you have a parallel load resistance, then the loss depends on the current through the load resistance, which varies as the tank's RMS voltage increases. So for two tanks resonating at the same frequency, the same power level and with the same parallel load R, the one with the bigger C and smaller L will have a higher Q, because the voltage swing is less and the current is higher.

Have a play with for more insight.

Ok. Then the exact answer isn't simple, because you have a low frequency zero: Z_L plus Rs causes finite gain at DC. At higher frequencies (assuming sqrt(L/C) >> Rs), you can convert Rs to Rp and determine the Q (notice Rp is in parallel with Rsrc, since the voltage source , but: to maximize Q, normally you would simply shrink L and raise C. But if Rs is constant, that fails, because the Rs-to-Rp conversion depends on L. In which case, your result may well be correct (the geometric average of Rs and R, a handy result).

The second best approach is to use a better impedance, and an impedance matching network. It's easier to make high Q, high impedance resonators than ridiculously low DCR/ESR ones, like you get from a naive filter synthesis. That's why tight filters are usually made with cap dividers or tapped inductors on rather more sane dimensions of inductors.

The best solution is to use something other than Ls and Cs, like mechanical, SAW or crystal filters, of course.

Tim

Basically what you have is a constrained optimization problem. As Tim Williams mentioned, as the loaded Q increases, the insertion loss also increases (since the source and load resistances are fixed, and you need to decrease the equivalent parallel resistance of the inductor to increase the loaded Q, creating a voltage divider with the source that reduces output.)

The insertion loss increases proportionally to -log(1 - Q_l) in terms of the loaded Q (where the unloaded Q is normalized to 1) and the loaded Q increases proportionally to 1 - e^(-I_L) in terms of the insertion loss, also normalized to Q_u = 1. So you want to simultaneously maximize the loaded Q while minimizing the insertion loss, subject to the constraint that the normalized Q_L < 1. Once you know the normalized loaded Q, and the source and load resistances are fixed, you should be able to solve for the required unloaded Q, and hence inductor value for a given Rs.

Hopefully I haven't made a mistake in formulating this, I'll play around with SciPy and see if I can come up with anything.

From another point of view, at resonance you have the following equivalent circuit:

Vin o--Rs---*-----*-----o Vout | | Rp RL | | o-------*-----*-----o

with Rp=QL^2*r.

QL is the inductor Q (ratio of reactance to resistance woL/r) and r is the series resistance.

From this you get

a) the insertion loss at resonance (from the resulting voltage divider) Vout = (Rp//RL)/(Rp//RL+Rs) * Vin

and b) the equivalent resistance R=(Rs//Rp//RL) and from this the filter (loaded) Q, Q=RCwo

Pere

erdependent relations of circuit parameters relating to loaded Q.

onance reactance than load resistance, so you can pick a small L and high C , but once you reduce the reactance, you have decreased the unloaded Q whic h would affect the equivalent resistance parallel to the tank, which would reduce the loaded Q.

a previous post here (long thread about how clueless I am building oscillat ors):

ussion

d once you have some known loaded Q and tank impedance. However suppose you want to maximize the Q knowing RL and Rs where RL is the load impedance an d Rs is the equivalent series resistance of the inductor.

If Qu is the unloaded Q of L then the series combination of L and ESR can b e transformed into an equivalent L in parallel with an R=Qu^2 x ESR. Then Rp equivalent becomes Rp'= Rp || R. Once this Rp' has been fixed, the pa rallel circuit Q= Rp'/(WoL)=WoRp'C. Q therefore increases with decreasi ng L and increasing C. The unloaded Qu of decreasing L improves, but probab ly not the case of increasing C where ESL probably rapidly overtakes L. Wha tever, the product LC must remain 1/Wo^2 at all times, and the reactance is infnite at resonance.

aded Q would happen when the reactance at resonance is equal to the square root of RL*Rs.

The reactance at resonance is oo for the parallel LC.

with changing inductance, but once you have starting value, some tweaking is possible, especially that Rs changes slower than changes in L.

an Electronics text?

No.

Thanks for the insight. Makes sense, without thinking about equations!

Thanks for the insight.

Yes I'm aware it's a bad assumption. Just don't know how bad :) Shrinking L yes can improve the Q, but wouldn't it also decrease Rp. So for a given RL (load resistance), decreasing Rp will decrease the parallel combination of Rp and RL hence decrease the loaded Q.

Their turn will come :)

hmmmm .. I thought decreasing the parallel resistance decreases Q

I'll have to read this slowly

be transformed into an equivalent L in parallel with an R=Qu^2 x ESR. Th en Rp equivalent becomes Rp'= Rp || R. Once this Rp' has been fixed, the parallel circuit Q= Rp'/(WoL)=WoRp'C. Q therefore increases with decrea sing L and increasing C. The unloaded Qu of decreasing L improves, but prob ably not the case of increasing C where ESL probably rapidly overtakes L. W hatever, the product LC must remain 1/Wo^2 at all times, and the reactance is infnite at resonance.

My assumption is that Rs is fixed not Rp. Rp will change as L changes becau se Rp = X^2/Rs

Sorry, you're right. Decreasing the parallel resistance decreases loaded Q, but it increases the _ratio_ of loaded Q / unloaded Q, which _increases_ the insertion loss. So for a given unloaded Q, the insertion loss will increase as the loaded Q approaches the unloaded Q. I believe I have that correct now!

an be transformed into an equivalent L in parallel with an R=Qu^2 x ESR. Then Rp equivalent becomes Rp'= Rp || R. Once this Rp' has been fixed, th e parallel circuit Q= Rp'/(WoL)=WoRp'C. Q therefore increases with decr easing L and increasing C. The unloaded Qu of decreasing L improves, but pr obably not the case of increasing C where ESL probably rapidly overtakes L. Whatever, the product LC must remain 1/Wo^2 at all times, and the reactanc e is infnite at resonance.

ause Rp = X^2/Rs

If Rp is free to change, then obviously making it >> Qu^2 x ESR (by factor of 10 or more) makes overall circuit Q the same as Qu, and that of course i s your "optimum." Circuit Q is always less than constituent component Q.

On Thursday, January 15, 2015 at 6:00:56 AM UTC-7, snipped-for-privacy@gmail.com >

Rp == Qu^2 x ESR for Qu >> 1 so you can not make it larger by a factor of 10 or more.

Ideally you would want to make Rp infinitely large for a given load resistance RL, but that means increasing X which would lead to decreasing loaded Q. Hence I thought my little result of Rp == RL provides a nice compromise.

What given load resistance RL??? Why even discuss Rp if you have an RL??? I'm beginning to think you're a troll.