How can inductance related effects change AC resistance?

Hello,

I am well aware of both the proximity effect and skin effect but never really understood how these inductance/magnetic related effects are used to subtract away from DC resistance to get the AC resistance of conductors.

Is this purely a way to model it (easy way out) or am I missing something? Are there more "accurate" models that treat these phenomena for their true inductive/magnetic properties?

[I just can't wrap my mind around the fact that AC inductance is changing AC resistance]

Email+NG responses will be appreciated.

Reply to
G Patel
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Skin effect and proximity effect both steer AC current away from regions of the wire. In the case of skin effect, magnetic fields near the center of the wire oppose current trying to travel down the center. So the current is crowded into the outer regions of the wire, reducing the effective cross-section. Less copper, more resistance, just as if you'd drilled out the wire, turning it into a thin-walled tube.

John

Reply to
John Larkin

never

phenomena

What I'm wondering is whether we are just using the REDUCED RESISTANCE model to simplify the issue (instead of dealing with the inductance head on).

Consider the following:

Say there is a series circuit like this...

---(100K resistor)---(1nH inductor)---(130K resistor)----

Obviously the 1nH inductor is of very little significance until we hit very high frequencies... say we're operating at 10GHz... we can consider the IMPEDANCE of the inductor as +j10 or we can get lazy and just ignore the complex nature of the beast and treat it as a 10 ohm resistor.

Is this similar to the way we treat Proximity and Skin effects or is the "resistance reduction" model a more legitimate case for these inductive phenomena?

Reply to
G Patel

are

of

near

wire,

thin-walled

RESISTANCE

hit

Correction: +j10 ==> +j20pi 10 ohm ==> 20pi ohm

Reply to
G Patel

Well with 10GHz we do not use 100k resistors in the signal path, but maybe a

10 ohm resistor. Now be your j10 in series with it and what would be the combined resistance? (14.14 ohms), and what would it be for 1GHz ? (10.05 ohms). Can you see the difference between a resistor and inductor or capacitor. Try to solve it for j62.83. The combined value is sqrt(X^2 + Y^2).
--
ciao Ban
Bordighera, Italy
Reply to
Ban

the

can

and

ohm

is

maybe a

I gave you a hypothetical circuit with ideal components, so why are you correcting me? There is a reason why I made the resistors very large.

(10.05

Wrong. The combined resistance in both cases is unaffected. Inductive Reactance is orthogonal to Resistance, you can't add it like this. On the other hand, the IMPEDANCE is a vector sum of the two.

Yes. There is a HUGE difference.

Hmmm.......

Thanks for your insights.

Reply to
G Patel

I read in sci.electronics.design that G Patel wrote (in ) about 'How can inductance related effects change AC resistance?', on Fri, 22 Apr 2005:

They don't subtract, they ADD. Both effects act to reduce the cross-sectional area of the wire that is available for carrying current.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
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Reply to
John Woodgate

I think Jack the Giant Killer needs to take a course in Beanstalk Climbing.

--
The Pig Bladder from Uranus
Reply to
Pig Bladder

Skin effect is not a lazy way to treat inductance. The resistance increase due to skin effect is real... "real" in the mathematical sense, not imaginary. It's not reactance, and it causes resonant circuits to lose Q and wires to get hot.

And wire resistance *increases* due to skin effect.

John

Reply to
John Larkin

Well if the resistors are large, they dominate and there is no significance. Why do you place 2 resistors and the inductance in series? Is there a reason not to add the resistors to 230k?

You observed that well, it is Z impedance, and the vector is of the calculated length. the current through it would be inverse to the scalar value. If you have 230k the current is identical with or without the 1nH even at 10GHz, with a 10 Ohms resistor that you wanted to take instead of the j10 your Z will be measurable higher, the increase is several orders higher.

What do you want to proof here, I do not understand what is your point? Why did you make the resistors so large?

--
ciao Ban
Bordighera, Italy
Reply to
Ban

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