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

 

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?

 

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 =^.^=

 

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

 

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

 

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

...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 =^.^=

 

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 =^.^=

 

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
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

 

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.

 

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

 

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

 

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.

 

Thanks for that further info Ninja -- much appreciated!!

Best,

-Bill

 

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.

 

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.

 

"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

 

"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

 

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

Colin =^.^=

 

Thanks Bob -- a great, clarifying explanation!

-Bill