Adjustment of Q with frequency.

[amdx]

I measure the Q of a 105nH inductor at 50Mhz my equipment max.
The Q is 80.
This is 57 / 80 = .71 ohms
But I'm using the inductor at 100Mhz.
Is there a way to calculate the Q at the higher frequency?

Thanks, Mikek

 

[[email protected]]

I measure the Q of a 105nH inductor at 50Mhz my equipment max.
The Q is 80.
This is 57 / 80 = .71 ohms
But I'm using the inductor at 100Mhz.
Is there a way to calculate the Q at the higher frequency?

If L and R would remain constant, the reactance and Q would double.
They probably won't stay exactly the same.
Anyway, for the values given, the reactance is 33 ohms, so the
resistance would be 0.4125 ohms.

 

[Tim Williams]

Veeeery crudely, air core copper inductors (solenoids, etc.) go as Q ~
sqrt(F).

If it's got a core or funny construction or pixie dust, who knows. Also,
self resonance (although I wouldn't expect that small an inductor to have
SRF that low).

Tim

 

[RobertMacy]

I measure the Q of a 105nH inductor at 50Mhz my equipment max.
The Q is 80.
This is 57 / 80 = .71 ohms
But I'm using the inductor at 100Mhz.
Is there a way to calculate the Q at the higher frequency?

Thanks, Mikek

usually sqrt(f/f0) times the losses.

 

[amdx]

I measure the Q of a 105nH inductor at 50Mhz my equipment max.
The Q is 80.
This is 57 / 80 = .71 ohms
But I'm using the inductor at 100Mhz.
Is there a way to calculate the Q at the higher frequency?

Thanks, Mikek

Thanks for all the answers, but the sticky point was that
I didn't assume R to remain the same. I wondered if there
was some way to calculate how the R would change.
I was looking for an R to put in LTspice for the inductors.
Mikek

 

[Don Lancaster]

If L and R would remain constant, the reactance and Q would double.
They probably won't stay exactly the same.
Anyway, for the values given, the reactance is 33 ohms, so the
resistance would be 0.4125 ohms.

Is this a ferrite core?
If so, the permeability varies with frequency.

--
Many thanks,

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Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
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[John S]

Thanks for all the answers, but the sticky point was that
I didn't assume R to remain the same. I wondered if there
was some way to calculate how the R would change.
I was looking for an R to put in LTspice for the inductors.
Mikek

You had your answer several times. Please read the thread.

 

[Maynard A. Philbrook Jr.]

I measure the Q of a 105nH inductor at 50Mhz my equipment max.
The Q is 80.
This is 57 / 80 = .71 ohms
But I'm using the inductor at 100Mhz.
Is there a way to calculate the Q at the higher frequency?

Thanks, Mikek

hmm
i get Xl 65.94

and if we assume you have lets say a .7 for DCR on the coil
this comes to
~ 94 for Q

Maybe I've over looked something but, if you know what the
BW at 50 mhz is and if you trust your meters, then double it.

Jamie

 

[Maynard A. Philbrook Jr.]

Thanks for all the answers, but the sticky point was that
I didn't assume R to remain the same. I wondered if there
was some way to calculate how the R would change.
I was looking for an R to put in LTspice for the inductors.
Mikek

I am glad you did because last night, late I should say, I was
graphing that with my casio Fx 7400G and the bats died. Glad
I kept a note list of the running programs i had stored in it.


Jamie

 

[amdx]

If L and R would remain constant, the reactance and Q would double.
They probably won't stay exactly the same.
Anyway, for the values given, the reactance is 33 ohms, so the
resistance would be 0.4125 ohms.

Yes, I made a mistake, I used 100Mhz when I should have used 50Mhz.
The problem as I see it, effective L will change slightly and R will
change more. I was looking for a way to calculate that.
My experience is more with 100uh coils than 100nH coils, maybe the R
is more constant with tiny coils. I'm trying to find out.
Thanks, Mikek

 

[amdx]

Is this a ferrite core?
If so, the permeability varies with frequency.

Aircore 1/4" ID, #16 wire.

 

[amdx]

Veeeery crudely, air core copper inductors (solenoids, etc.) go as Q ~
sqrt(F).

If it's got a core or funny construction or pixie dust, who knows. Also,
self resonance (although I wouldn't expect that small an inductor to have
SRF that low).

Tim

Simple coil 1/4" ID #16 wire.

So Q = 80 @ 50Mhz so Q = ? @ 100Mhz
according to the Veeeery crudely rule?
Mikek

 

[RobertMacy]

Yes, I made a mistake, I used 100Mhz when I should have used 50Mhz.
The problem as I see it, effective L will change slightly and R will
change more. I was looking for a way to calculate that.
My experience is more with 100uh coils than 100nH coils, maybe the R
is more constant with tiny coils. I'm trying to find out.
Thanks, Mikek

As I said, get a copy of femm.

You'd see that in an air core coil, the inductance changes slightly with
higher frequency because the carriers 'crunch' down changing their
effective 'pattern' in free space. [Envision a different, smaller
conductor.] The resistance tends to follow skin depth formula of
sqrt(f/f0) Although that is off too in the 1-2% ranges.

 

[amdx]

Is this for that flea power FM broadcast transmitter you were working
on in rec.ham-radio.antennas? If so, the coil is probably for the
output low pass filter for the xmitter. Q will have an effect on the
bandpass ripple, but with a filter intended to only remove the 2nd
harmonic and up, it's not going to be very critical.

Anyways, assuming this is an air inductor, the change in Q is going to
be directly proportional to frequency.
Q = Xl / R
Double the frequency and you double the inductive reactance, while
keeping the resistance constant. So, the Q @ 100 MHz should be 160.

Is there something different about coils at 100Mhz vs 1Mhz?
The R is not constant in a 240uh coil, Q often will start to decrease
in a 240uh coil near 1.5Mhz, and it is not near SRF.
Mikek

 

[amdx]

usually sqrt(f/f0) times the losses.

Which one is F0? 50Mhz or 100Mhz?
what does 0 stand for?
Thanks, Mikek

 

[RobertMacy]

Simple coil 1/4" ID #16 wire.

So Q = 80 @ 50Mhz so Q = ? @ 100Mhz
according to the Veeeery crudely rule?
Mikek

approx Q*(f/f0)/sqrt(f/f0), or Q*sqrt(f/f0) = 113

 

[amdx]

You had your answer several times. Please read the thread.

Ok I reread the thread, here are 4 answers I got.

1) If L and R would remain constant, the reactance and Q would double.
They probably won't stay exactly the same.

2) air core copper inductors (solenoids, etc.) go as Q ~
sqrt(F).

3) Double the frequency and you double the inductive reactance, while
keeping the resistance constant. So, the Q @ 100 MHz should be 160.

4) usually sqrt(f/f0) times the losses.

So instead of a non answer, how about being helpful.
If you knew the answer you could have copied and pasted it for me.

I think the real answer is, it is not an easy calculation, it gets into
the changes in skin effect, proximity effect, interwinding capacitance
and eddy currents when going from 50Mhz to 100Mhz.
Mikek
btw, What is your answer?

 

[RobertMacy]

Which one is F0? 50Mhz or 100Mhz?
what does 0 stand for?
Thanks, Mikek

usually f0 is like a starting point, or data and f is what you are
varying. In this case f0 would be 50MHz where the data was taken; and f,
the operating frequency, has been taken out to 100MHz.

thus losses due to skin effect tend to follow a sqrt(f/f0) function,
translating to the effective losses increasing about 1.4 times, up to
around 1 ohm.

 

[Maynard A. Philbrook Jr.]

Simple coil 1/4" ID #16 wire.

So Q = 80 @ 50Mhz so Q = ? @ 100Mhz
according to the Veeeery crudely rule?
Mikek

I think you are already in trouble..

#16 at 100 Mhz?

I need to dig up the skin formula.

Jamie

 

[amdx]

usually f0 is like a starting point, or data and f is what you are
varying. In this case f0 would be 50MHz where the data was taken; and f,
the operating frequency, has been taken out to 100MHz.

thus losses due to skin effect tend to follow a sqrt(f/f0) function,
translating to the effective losses increasing about 1.4 times, up to
around 1 ohm.

Ok, and sorry you ran with my mistake, I calculated Q with 100Mhz
instead of the correct 50Mhz.
So 33/ 80 = 0.4125. Then 0.4125 x 1.4 = 0.5775 ohms.
Sure would be nice to have equipment to measure such things.

Mikek