How does impendance affect circuit Q?

I have been reading two different tutorials that talk about impedance ver sus circuit Q in RF circuits. One tutorials says raising the resistance in a resonant circuit decreases circuit Q and visa versa, but the other sourc e says a higher impedance in a tuned circuit increases circuit Q. I am awar e that both reactance and resistance make up the total impedance equation, but the two statements appear to contradict each other. Which source is co rrect?

Thanks,

Michael

Reply to
Michael
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w0 ~= 1/(2 pi sqrt(LC)) (exact for Q -> infinity)

0---*----*----* Big R -> high Q | | | Q = R/(w0 L) = w0 R C R L CCC R L CCC R L | | | | 0---*----*----*

0---RRRR---LLLL--*--0 Small R -> high Q | Q = w0 L/R = 1/(w0 R C) CCC CCC | |

0----------------*

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

** Sorry, should have been in radians/s (no 2 pi)

w0 ~= 1/sqrt(LC) (exact for Q -> infinity)

0---*----*----* Big R -> high Q | | | Q = R/(w0 L) = w0 R C R L CCC R L CCC R L | | | | 0---*----*----*

0---RRRR---LLLL--*--0 Small R -> high Q | Q = w0 L/R = 1/(w0 R C) CCC CCC | |

0----------------*

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
Principal Consultant 
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Reply to
Phil Hobbs

Michael wrote: > > I have been reading two different tutorials that talk about impedance versus circuit Q in RF circuits. One tutorials says raising the resistance in a resonant circuit decreases circuit Q and visa versa, but the other source says a higher impedance in a tuned circuit increases circuit Q. I am aware that both reactance and resistance make up the total impedance equation, but the two statements appear to contradict each other. Which source is correct? > > Thanks, > > Michael >

they're both correct, if you separate reactance and resistance and treat them separately.

remember impedance is not pure resistance. A large coil (inductor) will resist the flow of AC, but will not heat up like a plain resistor of the same ohm value.

A high Q circuit has low losses, adding pure resistance (which is a loss) will decrease the Q factor.

Reply to
Cydrome Leader

Q is about the actual coil. The circuit will load down the Q, but if you need a high Q coil, you start with that since the circuit will load it down.

Lots has been done to get good Q. At UHF, silver plating the coils (or cavities) helps a lot. At low frequencies, Litz wire helps a lot.

YOu may see very light coupling to the tuned circuit in an oscilator.

Michael

Reply to
Michael Black

To OP, If you are using google groups click on the "show original" menu option (triangle in Upper RH corner) to see Phil's ASCII art.

As an aside do you know of any practical circuits that have a parallel resistance that limits the Q?

All the ones I can think of are series... (mostly inductor resistance.)

George H.

Reply to
George Herold

Colpitts oscillators.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

At the resonant frequency, if you excite a parallel LC and then watch the thing oscillate (without any external connections to confuse things) the sinewave ringing voltage will decay and eventually die out, below measurable levels. The rate of decay defines the Q of the LC resonator.

You can then blame the losses on an equivalent parallel resistor (R-L-C all in parallel) or an equivalent series resistor (all in series, in a ring.) High Q (low energy loss) corresponds to a big parallel resistor, or a small series resistor.

The actual loss, and oscillation decay, can be from a number of physical mechanisms, which we can lump into a single equivalent resistor, if we want to.

--
John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
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Reply to
John Larkin

I believe the first one was referring to using non- reactive resistance in the circuit which lowers Q..

The second one is most likely talking about the reactance resistance of L when not resonance.

Depending on how you think it's like 8 to 12 times on the Xl of an L around the resonant freq. This can give you the effect of a sharp skirt.

Jamie

Reply to
Maynard A. Philbrook Jr.

The usual way to temperature-compensate a crystal oscillator starts with a parallel resistor across the rock. That lowers the Q, and makes the oscillator easier to pull (not easy, just easier) off its temperature-dependent resonance.

Reply to
whit3rd

"Michael"

------------------------------------- I have been reading two different tutorials that talk about impedance versus circuit Q in RF circuits. One tutorials says raising the resistance in a resonant circuit decreases circuit Q and visa versa, but the other source says a higher impedance in a tuned circuit increases circuit Q. I am aware that both reactance and resistance make up the total impedance equation, but the two statements appear to contradict each other. Which source is correct?

--------------------------------------

** The statements are the same.

The Q of a tuned circuit is the ratio of the impedance Z of the inductor at the resonant frequency, to its resistance R.

So Q = Z/R

For more Q you can:

  1. Increase the resonant frequency by reducing the capacitance.

  1. Increase the value of the inductor while keeping the same resistance.

  2. Reduce the value of R in the inductor.

.... Phil

Reply to
Phil Allison

Hmm. How does lowering the Q cause it to be easier to pull? The required delta-X would be just the same, ISTM, if you're doing LC things to the resonator. Reducing the Q does make the phase slope shallower--are you intending to use a phase shifter in series with the active device?

I'd expect something like a series LC in series with the crystal, the C being a varactor padded with an N750 or something like that.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

It is my understanding that the load capacitance is replaced with a varactor, and one uses the inductor-like part of the resonance curve. The intent is to make the inductor-like region span a larger frequency range. The effect is also, of course, to raise the phase noise (so a true ovenized oscillator works better). An external LC would be useful against the phase noise problem, I suppose.

Reply to
whit3rd

Sure, but when you do that you have to shift the resonant frequency of the crystal by adding reactance to the tank. Reducing the Q doesn't help there. It would help if you're just adding phase shift, because a lower-Q tank has a shallower slope of phase vs detuning.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

I'm not sure what the question is. Altering the capacitance DOES phase-shift. The resonant frequency of the crystal doesn't change, really, just the operating point on the resonance peak. Using a 'stock' crystal, the peak is inconveniently narrow, you might start to skip beats. Either a custom crystal, or a Q-killer resistor, treats that problem. I'm remembering most of this from an old conversation, might have it wrong: seems like a useful treatment.

Reply to
whit3rd

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