Maximizing Q in a series resonant L/C circuit?

Hi All,

If I have a resonant L/C series circuit, how do I maximize the Q? Do I: 1. Make the capacitor much larger in comparison to the inductor (IE: 1000pF cap vs. a 1nH inductor); or 2. Make the inductor much higher in value than the capacitor (IE: 1000nH inductor vs. a 1pF cap)? One of my books states to "make the L/C ratio higher to maximize the Q", but does that mean make the L or the C higher in a series circuit? (Logically I would say make the cap larger and the ind smaller, but I hate guessing at such things!)

Thanks,

-Bill

Reply to
billcalley
Loading thread data ...

On a sunny day (Sat, 16 Jun 2007 14:37:55 -0700) it happened billcalley wrote in :

Q = (w x L) / R, so make R small, and L big (w = 2 x pi x f). A big L gives a low value for C for the same frequency.

As R is linear proportional to number of turns, and L to the square of the number of turns, f linear to the number of turns, then

2 x more turns = 4 x L / 2 x R = 2 x more Q. So use many turns.... and a small value of C to keep tuned.

This right?

Reply to
Jan Panteltje

It depends on what frequency range your working in, but its more usefull to think in terms of impedance, ie they L and C will have the same impedance at resonance, id go for something in the range of 10-300 ohms for high frequency, not sure about lower frequency. depends on what is atatched to it as well of course.

Colin =^.^=

Reply to
colin

Hi Jan,

Thanks for the response! What's strange is that when I run a simulation, and I use two perfect (infinite Q) L and C between 50 ohm terminations, this series resonant circuit *always* has a much sharper gain response (higher Q) when the L is high in value and the C is small in value -- when I use a small L and a large C, the series circuit is always flat and has almost no selectivity. I wish I knew the reason for this, but it has me stumped!

Best Regards,

-Bill

Reply to
billcalley

Hi Colin,

Thanks! But when I use two ideal (infinite Q) Ls and Cs between

50 ohm terminations, this series resonant circuit has a much higher Q when the L is high in value and the C is low in value, than when the L and C is visa-versa in values?

Thanks!

-Bill

Reply to
billcalley

Ummmm? Q = omega*L/R for a series resonant circuit.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
 Click to see the full signature
Reply to
Jim Thompson

well yes like I said it depends what the load is, it also depends if the load is in series or parallel.

if you make the impedance of the series L/C much higher than 50ohms termination then you will get much higher overall Q.

you might need to go as high 250 ohms if you want a high Q of say 50. wich is probably somewhat optimisticly high.

conversly if the 50ohms termination was accros a parallel LC then you might want to go as low as 1 ohm for the same Q.

Colin =^.^=

Reply to
colin

its quite easy to understand, as the frequency changes 1 octave from the center, the impedance of the series LC doubles. if impedance of the L/C is low then it will offer little resistance so the effect wil be minimal, if the impedance is large it wil have a sharp response to frequency, of course at resonance the impedance drops to near zero in either case.

you should be able to work out that the parallell case is the opposite.

Colin =^.^=

Reply to
colin

If I understand what you circuit is, the two terminations are part of the series network and total to 100 Ohms. With perfect L's and C's there is no other resistance in the circuit. So the Q = Xl /100. As was stated above, a higher L creates a higher Q . In this case, the resistance is constant.

Reply to
Bob Eld

In either a series or parallel circuit the inductance and capacitance will cancel each other out at resonance. Off resonance you'll either see an LC circuit as a capacitance or an inductance.

In a series circuit, to have much effect the resistance has to see a _big_ reactive impedance, so you need big L (high impedance) and little C (again, high impedance). At resonance these impedances cancel each other out and the resistor sees a short circuit (except for the coil losses). Unfortunately it's hard to wind a big, high Q coil, which is why you don't see a lot of high Q series resonant circuits out there.

In a parallel circuit, to have much effect the resistance needs to see a _small_ reactive impedance. So you need little L (low impedance) and big C (again, low impedance). At resonance the coil and capacitor _admittances_ cancel out, so the resistor sees an _open_ circuit and the signal can get by. It's easy to get a high Q big capacitance, and it's easy to get a (relatively) high Q little coil, so you see lots of parallel resonant circuits out there.

--
Tim Wescott
Control systems and communications consulting
 Click to see the full signature
Reply to
Tim Wescott

There are two varieties of Q, UNLOADED, and LOADED.

UNLOADED means just that, you have an L and a C, and only the R that is a part of the L. C is usually assumed to be perfect. To get a high Q, you want the X(L) to be large with respect to R(L), which means large wire, or other than air core

LOADED Q is the resultant Q when the L and C are actually part of a circuit. Suppose X(L)=300, R(L)=1, and R(LOAD)= 50. You now have an UNLOADED Q of

300/1 = 300, and a LOADED Q of 300/(50 + 1)= ~6. Clearly, if you want a higher Q, you will have to use a larger inductor to get the X(L) up. This is what you are seeing.

Things get more interesting when the R(LOAD) and the inductor loss resistor R(L) are the same order of magnitude

Tam

Reply to
Tam/WB2TT

Thanks guys; Tim and Colin's explanations finally made the light bulb go on! I just had never, for whatever reason (stupidity?), thought of this before on those kinds of clear, concise terms.

Best Regards,

-Bill

Reply to
billcalley

Several posters have already explained this, but maybe explaining it another way would help get the point across?

A series LC circuit with 50 ohm source and 50 ohm load impedance is really an RLC circuit. The resistance is 100 ohms. The quality factor is, by definition, Q = Energy Stored / Energy Dissipated. Or in electronic terms Q = Inductive Reactance / Resistance = 2*pi*f*L/R.

Notice that L is in the numerator? A larger L will result in a larger Q.

You made a point of saying that you were simulating with ideal inductors and capacitors, so I presume you understand that real-life inductors (and to a lesser extent capacitors) have losses of their own, unrelated to the source and load. Inductors also have paracitic capacitance which limits their usefulness at high frequencies. These realities are among the reasons that we don't all specify huge inductors and small capacitors to keep the Q up and the losses down.

Reply to
Ninja

There have been a number of good responses to this already, but I was a bit surprised to see that no one has yet talked about what "Q" actually IS. It may help make things clearer if we take a look at that.

"Q" stands for "quality factor," and it's basically a measure of the the "purity" of a reactive circuit - more specifically, how well it approximates a purely reactive circuit. If a resonant circuit, for instance, were to be made up of purely reactive elements, its impedance would either be infinite (parallel resonant circuit) or zero (series) at the resonant frequency. The "Q" would also be infinite, regardless of the ratio of L and C. Note, though, that any resistive element in the circuit - including the load - takes us out of the "purely reactive" situation.

The definition of "Q" is the ratio of reactive power to resistive power in the circuit. (Of course, a reactance doesn't really dissipate any power, but we can treat reactances as resistances in coming up with a figure for reactive "power".) In a series resonant circuit, since the current is the same in all elements, the simplest formula to use for power is the square of the current times the impedance, so we have

Q = P(reactive)/P(resistive) = (I^2X)/(I^2R) = X/R

Since, at resonance (where Q is defined), Xl and Xc are equal, either could be used here - Xl is typically used, because in most practical circuits it is the resistance of the inductor that will dominate the Q if the load is ignored (i.e., we're looking for the Q of the series resonant elements themselves) - so the Q of the circuit is basically the Q of the inductor itself at the resonant frequency.

(As an aside - why isn't the reactive power the sum of the "power" in the two reactive elements? Think about the relative phasing of the voltages across the L and the C to understand this.)

Similarly, if we're talking about the Q for a parallel resonant circuit, we still are talking about the ratio of reactive to resistive power - but here, since it's the voltage that's the same across all elements, we have

Q = (V^2/X)/(V^2/R) = R/X

The exact same definition, but it results in a formula that is the inverse of the previous case.

Now, you should also from these be able to see why, for the real-world situation (where there's going to be SOME resistance involved, even if it's just the load) you'd want to maximize the L vs. the C (or vice-versa) - are you looking for the biggest, or smallest, value of the reactance (X) at the resonant frequency in order to maximize the Q?

Bob M.

Reply to
Bob Myers

Q = wL/R, and at resonance wL = 1/wC.

Q^2 = (wL)^2/R^2 = (wL/wC)/R^2.

Q = sqrt(L/C)/R.

Q is proportional to the square root of the L/C ratio.

--
Tony Williams.
Reply to
Tony Williams

"Bob Eld"

** Looks like the ONLY correct answer.

The dopey OP failed to include source and load resistances in his " ideal" series resonant circuit.

The inductor determines a tuned circuit's Q factors in practice, as it is normally the weak link.

...... Phil

Reply to
Phil Allison

"Tony Williams"

** For which set of circumstances ??

Naughty of you to leave that crucial bit out.

For fixed R and given frequency, Q is simply proportional to L.

Your formula quantifies how the Q of a tunable resonant circuit behaves as L or C is varied.

...... Phil

Reply to
Phil Allison

[...good explanation snipped]

Im surprised no ones suggested reading "art of electronics" its good at explaining this sort of thing.

Colin =^.^=

Reply to
colin

Thanks Bob -- a great, clarifying explanation!

-Bill

Reply to
billcalley

You are cooking the books by sticking to a 50 ohm generator and load. They are determining Q, in the sense that they are probably the dominant lossy elements in the circuit, not the inherent Q of the L and C.

Q = wL/R, and you are forcing R, external to the L and C themselves. That ain't fair.

John

Reply to
John Larkin

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.