Adjustment of Q with 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.

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John

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

-- Seven Transistor Labs Electrical Engineering Consultation Website:

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.

"Notice that Q increases linearly with frequency if L and R are constant."

Add a Q-meter to your MFJ-259B:

When you blow it up, how to fix your MFJ-269 (the MFJ-259B is similar):

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usually sqrt(f/f0) times the losses.

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

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

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Many thanks, 

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

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

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

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

Aircore 1/4" ID, #16 wire.

Simple coil 1/4" ID #16 wire.

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

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

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

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

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?

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.

I think you are already in trouble..

#16 at 100 Mhz?

I need to dig up the skin formula.

Jamie