resistor resistance change over frequency

I'm making high Q measurements. My Q meter only goes to 625, so I have added a ~ 2.15 ohm resistor in series with the coil. I'm testing a reference inductor with and without the resistor to find the value of the resistor. The resistor value comes out as follows.

500kHz 2.19 ohms 600kHz 2.20 ohms 700kHz 2.23 ohms 800kHz 2.26 ohms 900kHz 2.28 ohms 1000kHz 2.29 ohms 1100kHz 2.30 ohms 1200kHz 2.30 ohms 1300kHz 2.44 ohms 1400kHz 2.45 ohms 1500kHz 2.52 ohms 1600kHz 2.64 ohms

I understand that the resistor mounting will also have an effect, so here is a picture of the banana plug, pcb, and two parallel 4.3 ohm carbon composition resistors. When I measure without the resistor the adapter assembly is removed and the inductor wire is connected directly under the knurled nut.

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OK, I know there will be capacitance and inductance effecting the R, also my pcb maybe causing some of this difference. I don't know what I'm seeing, is it L or C? My tuning cap change is only 0.1pf different but it is both way and often I see no difference at all.

I note the transition of the resistance is not smooth, I'm sure that is my measurements, is there a way to smooth this or, is this discontinuous transition possible. I suspect it should fit a smooth curve.

Thanks for your input, Mikek

Reply to
amdx
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Hi

I am unsure about your test methodology, it sounds like you may be introducing errors

At least as far as the resistors goes, the leaded ones, they should be good way over 100MHz and your figures suggest problems all ready at 1MHz

Cheers

Klaus

Reply to
klaus.kragelund

Please remember that the skin depth in a conductor is inversely proportional to the square root of frequency.

Thus the AC resistance measured at a higher frequency should be higher (this is a real resistance, not to be confused with inductive reactance).

Reply to
upsidedown

Yes, and I probably need to add more information. The above resistances above are, Rseries+RLoss* Minus Rloss equals series resistor resistance listed above. So, I don't think all that change is caused by skin effect in just the series resistor.

  • Rloss is the inductor AC resistance, which in this case varies from 2 ohms to 6.5 ohms from 500kHz to 1600kHz. Thanks, Mikek
Reply to
amdx

A good way for averaging or smoothing is to take an odd number of measurements at each point, sort them and use the middle measurement.

Reply to
Robert Baer

I'm using a Boonton Q meter, it only measures to 625. So I put a resistor in series to lower Q.

I measure Q and record the C used to resonate. I calculate the Xc at the frequency tested, the divide Xc by Q to get R. (since Xc = XL) I call this Rloss.

To test the resistor I check Q with and with out the resistor, on a coil with a Q of about 400. I subtract the Rloss from (Rloss + R series) to get Rseries.

I think* the resistor change over frequency is real and not a measurement error. I just want to know what it is R, and/or L or C and how to incorporate it into my calculations.

I have made the test several time and always have approximately the same curve.

Certainly introducing additional characteristics, don't think they are errors, other than slight measurement errors.

Thanks, Mikek

That's what I thought!

and your figures suggest problems all ready at 1MHz

Reply to
amdx

Skin effect Like DownUnder said as well as other in-phase losses.

Other parts of the resistor could also play a part in resistive losses like the material it is made from.... Kinda like EMI filter cores that dissipate at high frequencies. I think they try to keep that to a minimum though. Try an SMT resistor instead of the leaded parts in your picture. I am not surprised at your measurments. Also try shorting out the resistors right at the body and see how the measurement setup plays into your findings.

It's just the in-phase (zero phase shift) part of the impedance measurement.

boB K7IQ

Reply to
boB

My ancient statistics textbook, which I thought had died a long time ago, just jumped off the bookshelf and threw itself out the window :-)

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Regards, 

Adrian Jansen
Reply to
Adrian Jansen

Also try shorting out the resistors right at the body and see how the measurement setup plays into your findings.

I like that last idea. I'm on that tonight!

I need a stronger argument to convince me this is normal resistor behavior. It is a 20% change from 500kHz to 1600Khz. This is around

1MHZ not 100MHz. I have another person using the same series resistor method, he found his resistor decreased in value 35%. Although he knew how to break out the small capacitance 0.275pf. This is not enough to cause a 35% change.
Reply to
amdx

That's a shame, you could have used it to look-up the definition of "median" and "average".

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Reply to
Jasen Betts

So those two resistors look like carbon film. Do you have any caron composition types to try as well ?

boB

Reply to
boB

OK, I shorted across the resistors with a #16 wire. I ran a scan from 500kHz to 1600kHz, and compared it to three averaged scans >without< the Banana plug/resistor/PCB.

The Rloss runs from 2.17 to 6.54 depending on frequency.

Here is the result of subtracting Rloss from Rloss+shorted resistor PCB.

Khz

500 0.07 ohms 600 0.07 ohms 700 0.04 ohms 800 0.06 ohms 900 0.05 ohms 1000 0.06 ohms 1100 0.05 ohms 1200 0.04 ohms 1300 0.05 ohms 1400 0.03 ohms 1500 0.05 ohms 1600 0.05 ohms

This additional R could be the foil on the PCB and my jumper resistance. So, it comes back to, "Why does the resistance of my 2.15 ohm resistor increase when I use it in series with a coil at frequencies from 500kHz to 1600kHz?"

Thanks for thinking, Mikek

Reply to
amdx

What an eye! My thinking all along was that these were Carbon Composition because I went through a dozen packages of carbon Composition to find these low value resistors. But the 4.3 ohm resistors were from a package labeled Carbon Film. I don't have any Carbon Film composition in this low value. I'll do a better search later. I have one 1 ohm SMD resistor I might try. Thanks, Mikek

Reply to
amdx

No, just old !

Hmmmm..... Good point. I am not sure if you can find a true old-style carbon comp resistor with that low of resistance ? Not sure I've ever seen one now that you mention it.

boB

Reply to
boB

Well, given Kramers-Koenig. It's a two terminal impedance, therefore if the real part is rising with respect to frequency, there must be an inductive component. If it were 100% real, the transform of that impedance (its waveform) would have an analytic (complex) component, which is impossible.

It bothers me, because Coilcraft's SPICE models are formulated by frequency only: they use FDRs to model diffusion terms (wire and core losses). Which are nonphysical, and which most engines can't handle in Transient Analysis (and the ones that can, still suck at it -- because they need to do a convolution across all time, with a huge array approximating the sqrt(1/t) behavior!).

Trivia: a pure diffusion term, has equal parts resistance and inductance, at any frequency. Indeed, it's the perfect damping for any arbitrary capacitance: the capacitor and the inductance resonate at some frequency, with an impedance equal to the resistance. (I think?!)

Tim

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Electrical Engineering Consultation and Contract Design 
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Reply to
Tim Williams

At such high Q's wouldn't it be easier to drive the circuit with a high impedance signal source and measure the effective gain at resonance?

Compromising the system Q to fit the measurement method available doesn't seem like a good approach to me.

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Regards, 
Martin Brown
Reply to
Martin Brown

If the resistor itself is wirewound would it have an effect from frequency?

Reply to
Kevin Glover

Of course. Non-inductively wirewound resistors (depending on the resistance value and the angle of the counter-winding) can be nearly constant (certainly in the frequency range you've been using).

Reply to
Frank Miles

And it would show an inductance increase not a resistance increase. However slight, compared to my 240uh coil. Mikek

Reply to
amdx

Mike, I'd do a second measurement on the resistors R(f) (I/V?) Maybe the error is somewhere else.

George H.

Reply to
George Herold

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