billcalley said:
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!)
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.