Q Factor

Discussion in 'General Electronics Chat' started by mohotmoz, Aug 18, 2009.

  1. mohotmoz

    Thread Starter New Member

    Aug 18, 2009

    I am having some difficulty understanding the meaning of the "Q factor." One definition is that it is the ratio of the reactance to the resistance. This perplexes me, because for an RLC circuit, this is zero at resonance!

    Let's take the example of a non-ideal inductor (in my application I am actually attaching this to a MEMS pressure sensor to create a wireless[​IMG] pressure sensor). It is an inductor in series with a resistor, all in parallel with a cap. What is the "quality factor" of this circuit? Is this even meaningfull without a "load capacitance" also added in? The definition I have found involves a formula like 1/R * sqrt(L/C).. but this does not make sense given the earlier definition of reactance to resistance, which is zero at resonance.

    Thanks in advance and I apologize for the open-endedness of my question. Really been banging my head over this one for a while.
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    For a resonant circuit at resonance, the reactance is that of the inductor (or the capacitor). Q can also refer to the "quality factor" of an inductor, independent of resonance. For series R-L, Q=X(L)/R. For parallel R-L, Q=R/X(L).
  3. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    Hi Mohotmoz!

    Great minds think alike. :) This was always a puzzler for me too...not intuitively, but more philosophically. You are correct: at precise resonance the reactive components go away. It's what happens OFF resonance that Q becomes important.

    However...we ham radio operators being the practical, pragmatic sort that we are.... we generally define Q as XL/R....for a series circut. This REMINDS us that even though during resonance, XL is cancelled out, it really is still there.

    ALSO....in practical life....the quality of an inductor has a much greater effect on Q than the capacitor. It's quite easy (at least at radio frequencies) to build a capacitor with a Q of nearly infinity....whereas the Q of an inductor depends on the wire resistance.

    So....the bottom line is...your thinking is correct.

    Here's another aspect however, where Q can be very important AT resonance. Let's say we drive a very high Q PARALLEL circuit (tank) through a signal generator and a high value of series resistance. We will find that the VOLTAGE across the tank circuit is MUCH higher than the signal generator voltage, because we are storing energy over many cycles. In this case the Q is very directly related to the voltage step-up...and yet, as far as what we "see" at the driving point, it's still a "purely resistive" load.

    Hope this helps some!

  4. Ratch

    New Member

    Mar 20, 2007

    You should not, if you know the definition of Q.

    That is a result, not a definition. For instance, time is not defined by distance divided by speed or velocity. Time can exist without spacial movement.

    Wrong, circuit Q is not zero at either series or parallel resonance. Neither is resistance or individual reactance.

    Let's define what Q really is. Q = 2*Pi*(maximum stored energy)/(energy dissipated per cycle) .

    For a series resistance and inductance, Q = w*L/R = 1/w*C*R, but for a parallel circuit, Q=R/w*L = w*C*R. Those are results, not definitions.

    For a more extensive discussion, see http://en.wikipedia.org/wiki/Q_factor . Google is your friend.

    screen1988 likes this.