Adjustment of Q with frequency.

Yes, and you should listen to R. Macy.

Q as a function of frequency:

Q(w) = w*L/(R_DC + ((w_0*L/Q_0) - R_DC)*sqrt(w/w_0))

w = 2*pi*f

R-DC := DC resistance

Subscript "_0" means the frequency at which you know the Q.

The presumption is that you are operating below any frequency where the self resonating capacitor has substantive effect. That is, you are in the "sweet range" of an air core inductor.

Q = w*L/R_total

Generally, R_total = R_DC + R_skin + R_core + R_dielectric + R_radiate

For what I gave you, it is only R_DC & R_skin (R_dielectric lumped in)

Generally too, *everything* is a function of frequency except R_DC, by definition. So the formula, again, is bounded to the "sweet range" of an air core inductor.

Gee, there was fifth different answer!

"multiply R by sqrt(2)" Well I know that's wrong, no frequency term. I'm sorry you got involved at all. Let alone a second time, with a no response post. Why did you waste your time? Mikek

btw, What is your answer?

I am.

Thanks for that.

I always thought interwinding capacitance had an effect on Q. The higher the interwinding capacitance the lower the Q. Maybe very low with a 1/4" dia. 5 turn coil, but what about a 200uH coil with 15pf interwinding capacitance. Is there something there or do I have it all wrong? Thanks, Mikek

self resonating capacitor has substantive effect. That is, you are in the "sweet range" of an air core inductor.

definition. So the formula, again, is bounded to the "sweet range" of an a ir core inductor.

Interwinding capacitance has an effect on Q if it is substantial for the fr equency of operation. I mean, at self resonance, the Q is zero. If the va lue "L" (Henrys) of the inductor is *not* "flat-ish" in your range of opera tion, then it is having an effect.

But like I said, you have to figure out how substantive the interwinding ca pacitance is for the frequency of operation. For filter type work, using t he inductor where interwinding capacitance is substantive is a no-no. You should be using inductors where their values are "flat-ish."

200 uH is pretty big for an air core. But maybe you never said it is air c ore. I can't remember.

Sorry, I don't want to confuse things here, the coils I'm discussing for the 110 Mhz filter are are around 100nH, 1/4" id. 5 turns.

I have worked with AM BCB coils that were large, 3" x 3" and had

240uH of inductance and 8pf to 15pf of interwinding capacitance.

I think I've worked through the math, I'll check it and then post it. Thanks, Mikek

Q(w) = 6.28*100Meg*105nH/(.002+((6.28*50Meg*105nH/80)-.002)* sqrt(100/50))

Q(w) = 65.94 /(.002+ (.412125 -.002)*sqrt 2)

Q(w) = 65.94 /(.002+ .410125 * 1.414)

Q(w) = 65.94 /(.002+ .57991675 )

Q(w) = 65.94 / .58191675

Q(w) = 113.3

The measured Q of the 105nH coil was 80 at 50 MHz.

Using this formula, the calculated Q is 113.3 at 100MHz.

I appreciate the exercise, time for a drink!

Mikek

Hope the font works.

Add 50% in your head and it's within about 6%, closer than many caps.

Best regards, Spehro Pefhany

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My answer is that you appear to be a lost cause.

And again you wasted your time with a non helpful response.

btw, what is your answer?

Mikek

No, I didn't waste my time. The object was to get you to waste your time.

btw, what is your answer?

My answer is that you appear to be a lost cause. Still haven't learned to read, have you?

Boys! Please... stop it! Show some restraint. Maybe this coming New Year we can all make a pledge to be nicer. I often write something in response and then look at it.. do I really need to post this.. and then throw it away.

George H.

You're right, George. I failed to restrain myself so I hereby apologize to Mikek and the group.

Cheers to all.

Sounds about right. Look at the Q v f plots for the Coilcraft "Midi Spring" inductors. That graph suggests your calculation is reasonable.

No problem.

Um, I have google. Just terrible.

Thanks, I'm just here to learn enough do a few electronic projects. I didn't learn the math 40 years ago when I should have. I lean on people here to help.

So, I note the sqrt(2) gives me an answer of Q = 113.13

Very close to the long formula,

Q(w) = w*L/(R_DC + ((w_0*L/Q_0) - R_DC)*sqrt(w/w_0))

Using this formula, the calculated Q is 113.3 at 100MHz.

But I understand the long formula has more utility. Mikek

PS. I'm usually the one trying to stop the sparing between Thompson and Larkin.

Say what?

Jamie

Actually Robert Macy said "usually sqrt(f/f0) times the losses". I said sqrt(2) because, IIRC, the R was measured at 50MHz and the estimate was to be for 100MHz. So, R*sqrt(100/50) is the same as R*sqrt(2). Don't use sqrt(2) unless your frequencies are 2:1. The long equation is the correct one to use for most any two arbitrary frequencies. Keep in mind that these are approximations. There are many factors which affect all this.

Cheers.