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From: don@umd5.UUCP
Newsgroups: net.analog
Subject: Re: filter circuit Q
Message-ID: <821@umd5.UUCP>
Date: Wed, 18-Dec-85 13:34:41 EST
Article-I.D.: umd5.821
Posted: Wed Dec 18 13:34:41 1985
Date-Received: Sat, 21-Dec-85 05:09:13 EST
References: <35500001@ICO.UUCP>
Organization: U of Md, CSC, College Park, Md
Lines: 62

> 
> There is a question that has been puzzling me
> about filter circuits that i was hoping someone
> could explain to me. In physics they talked about the
> "Q" of a mechanical circuit, which was a measure of its
> ability to reject off resonant frequency signals. I see
> how an LC circuit forms a filter, but i don't see how to
> change the Q of the circuit. Anybody have an explanation
> for me?
> 

There's two "flavors" of Q: Loaded and Unloaded.

Unloaded Q is determined by the series resistance present in both
inductors and capacitors. This series resistance dissipates energy
in the circuit, and affects the "sharpness" of the response peak of
a resonant LC circuit.
   "Most diagrams of resonant circuits show only inductance and
capacitance; no resistance is indicated. Nevertheless, resistance is
always present. At frequencies up to about 30 MHz this resistance is
mostly in the wire of the coil. At higher frequencies energy loss in
the capacitor also becomes a factor. *This energy loss is equivalent
to resistance* [emphasis mine]. When maximum sharpness or selectivity
is needed, the objective of design is to reduce the inherent resistance
to the lowest possible value. The value of the reactance of either the
inductor or capacitor at the resonant frequency of a series-resonant
circuit, divided by the series resistance in the circuit, is called
the Q (quality factor) of the circuit, or:  _Q= X/R_  where:
Q= quality factor
X= reactance in ohms of either the inductor or capacitor
R= series resistance in ohms"                                            (1)
   Q can be used to determine the voltage across the LC circuit:
_V= Q*E_  where: E= the voltage being applied to the circuit.
   _Loaded_ Q:
   "However, when the circuit delivers energy to a load (as in the case
of the resonant circuits used in transmitters) the energy consumed in the
circuit itself is negligible compared with that consumed by the load."
"The Q of a parallel resonant circuit loaded by a resistive impedance is:
  _Q= R/X_  where:
R= parallel load resistance in ohms
X= reactance in ohms." "The effective Q of a circuit loaded by a parallel
resistance becomes higher when the reactances are decreased. A circuit
loaded with a relatively low resistance (a few thousand ohms [!!!]) must
have low-reactance elements (large capacitance and small inductance) to
have a reasonably high Q."                                               (2)

(50/75 ohms is relatively low to me !)
I hope this answers your question(s) completely ...
---
(1) & (2) _The ARRL Handbook for the Radio Amateur_ pps. 2-32 & 2-34
reprinted and edited somewhat without permission...

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