chris@ICO.UUCP (12/17/85)
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? Thanks in advance chris Chris Kostanick hao!ico!chris ucbvax!ucla-cs!ism780!ico!chris
sgcpal@watdcsu.UUCP (P.A. Layman [EE-Silcon Devices Research]) (12/18/85)
In article <35500001@ICO.UUCP> chris@ICO.UUCP writes: > >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? > Essentially and L-C resonant circuit will be driven by a circuit with some finite output resistance. Thus, Z = R + j( wL - 1/wC ) , or 2 Z = R + j( wL - w L/w ), o o and Q is defined as, Q = w L/R o w is the resonant frequency, and R is the effective resistance at o o resonance. Note this definition also applies for an inductance which includes a parasitic resistance R. Note also that for parallel resonance the w term appears in the denominator. Z = R ( R/R + jQ ( w/w - w /w ) ) o o o o Clearly if the half power points occur at w1 and w2, Q = w /(w1-w2) o Paul A. Layman University of Waterloo, Electrical Engineering, Silicon Devices and Integrated Circuits Research Group (SiDIC) UUCP: {decvax|utzoo|ihnp4|allegra|clyde}!watmath!watdcsu!sgcpal
don@umd5.UUCP (12/18/85)
> > 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... -- --==---==---==-- "What happened ?" "It seems the occipital area of my head impacted with the arm of the chair." "No, I mean, what happened to us ?" "That has yet to be surmised." ARPA: umd5!don@maryland.ARPA, don%umd5@umd2.ARPA BITNET: don%umd5@umd2 UUCP: ..!{ seismo!umcp-cs, ihnp4!rlgvax }!cvl!umd5!don
sandler@utcsri.UUCP (Howard Sandler) (12/18/85)
The Q of a simple LC filter (2nd order) is adjustable independent of the
natural frequency by varying the damping resistor. Without the resistor,
the Q is (theoretically) infinite.
Howard M. Sandler
Dept. of EE
University of Torontorickk@hpvcla.UUCP (rickk) (12/20/85)
The "Q" (quality) of a filter is defined as:
omega * L 1 1
Q = --------- = --------- ; omega= resonant freq = ---------
R omega*R*C sqrt(L*C)
It is a measure of the bandwidth of the filter in that the higher the "Q",
the smaller the bandwidth. To vary the "Q" of a circuit, you just need to
manipulate the terms in the above expression. The most obvious way to increase
"Q" is to reduce the circuit resistance. However it can also be varied by
increasing L and decreasing C to give the proper omega and at the same time
increasing "Q". Note however, that larger inductors usually have larger
resistances, so "Q" in a real circuit will not necessarily increase in the
second instance.
Rick "I really are a digidul n-gineer" Klaus @ hpvcla!rickk