How does impendance affect circuit Q?

M

Michael

Guest
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
 
Michael <google@shadowstorm.com> 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.
 
On Tuesday, July 1, 2014 11:33:36 AM UTC-4, Phil Hobbs wrote:
On 07/01/2014 11:24 AM, 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



w0 ~= 1/(2 pi sqrt(LC)) (exact for Q -> infinity)
Click to expand...
<snup>
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.
Phil Hobbs



--

Dr Philip C D Hobbs

Principal Consultant

ElectroOptical Innovations LLC

Optics, Electro-optics, Photonics, Analog Electronics



160 North State Road #203

Briarcliff Manor NY 10510



hobbs at electrooptical dot net

http://electrooptical.net
Click to expand...
 
On 07/01/2014 11:24 AM, 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
Click to expand...

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

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
 
On 07/01/2014 11:24 AM, 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
Click to expand...

** 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

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
 
On Tue, 1 Jul 2014, 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?
Click to expand...
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
 
On 07/01/2014 02:33 PM, George Herold wrote:
On Tuesday, July 1, 2014 11:33:36 AM UTC-4, Phil Hobbs wrote:
On 07/01/2014 11:24 AM, 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



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

snup
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.
Click to expand...

Colpitts oscillators.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
 
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.)
Click to expand...

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.
 
On Tue, 1 Jul 2014 08:24:05 -0700 (PDT), Michael
<google@shadowstorm.com> 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
Click to expand...

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
http://www.highlandtechnology.com
 
In article <592b2875-a167-445e-9a05-9694a123b41a@googlegroups.com>,
google@shadowstorm.com says...
Subject: How does impendance affect circuit Q?
From: Michael <google@shadowstorm.com
Newsgroups: sci.electronics.basics


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
Click to expand...

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
 
"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.

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

3. Reduce the value of R in the inductor.


..... Phil
 
On 07/01/2014 08:18 PM, whit3rd wrote:
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.)

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.
Click to expand...

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

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
 
On Wednesday, July 2, 2014 7:42:16 AM UTC-7, Phil Hobbs wrote:
On 07/01/2014 08:18 PM, whit3rd wrote:



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

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.
Click to expand...

Hmm. How does lowering the Q cause it to be easier to pull?

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.
Click to expand...

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.
 
On 7/2/2014 4:46 PM, whit3rd wrote:
On Wednesday, July 2, 2014 7:42:16 AM UTC-7, Phil Hobbs wrote:
On 07/01/2014 08:18 PM, whit3rd wrote:



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

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.


Hmm. How does lowering the Q cause it to be easier to pull?

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.

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.
Click to expand...

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

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
 
On Wednesday, July 2, 2014 6:52:35 PM UTC-7, Phil Hobbs wrote:
On 7/2/2014 4:46 PM, whit3rd wrote:

On Wednesday, July 2, 2014 7:42:16 AM UTC-7, Phil Hobbs wrote:

On 07/01/2014 08:18 PM, whit3rd wrote:

The usual way to temperature-compensate a crystal oscillator starts with a
parallel resistor across the rock. That lowers the Q...

Hmm. How does lowering the Q cause it to be easier to pull?



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.

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.

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.
Click to expand...

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:
<http://www.crystek.com/documents/appnotes/VCXOarticle.pdf>
seems like a useful treatment.
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top