Bulk Resistivity & Skin Effect

Thread Starter

Management

Joined Sep 18, 2007
306
Can someone help me understand bulk resistivity and how it changes with increased frequency over a conductor, i.e. the introduction of skin effect?

I know that in a stripline at DC you don't have an loss due to a dielectric but you do have IR loss. And as you increase in frequency you start getting loss due to the loss tangent of the dielectric material.

But what about the the loss over the conductor as you increase in frequency? That is what I am confused about. How skin affects play a roll and how to figure out its value or whatever.

Thanks for the help.
 

Thread Starter

Management

Joined Sep 18, 2007
306
Thanks for the link.

The thing is I can calculate skin depth at any particular frequency for Copper for example. But I do I practically understand that value? How do I know if that depth is bad or not good enough? Any rule of thumbs that would be nice to follow, especially if you looking at a transmission line and trying to figure out loss.

Update: After reading Microwaves101 I understand a little bit more. I biggest question goes back to my original post.

How does the skin effect/depth issue affect the conductor resistance equation below?

R = pL/A where p: resistivity, L: length, and A: area

Thanks.
 

SawabyPlus

Joined Feb 27, 2009
14
one approach is to think of the charge distribution in the cross sectional area of a wire.
the magnetic time-varying field resulting from the time-varying current, with its rotating direction inside the wire itself, will push moving charges outwards, and thus decrease the effective area for the charges to move across. this means an increase in the ac resistance of the wire, as R is inversely proportional to A.
 

Thread Starter

Management

Joined Sep 18, 2007
306
one approach is to think of the charge distribution in the cross sectional area of a wire.
the magnetic time-varying field resulting from the time-varying current, with its rotating direction inside the wire itself, will push moving charges outwards, and thus decrease the effective area for the charges to move across. this means an increase in the ac resistance of the wire, as R is inversely proportional to A.
So when you say inversely proportional to A, in terms of AC, is that A (area) the area where the current is. Since the current covers a smaller area then that resistance is increased?

Sorry if I sound lost. :confused:
 

thingmaker3

Joined May 16, 2005
5,083
Did you mean to use "outside" & "inside" or "above" & "below"? I just got a little confused.
I apologize for my ambiguity. Roughly 2/3 of the current will be outside/above the skin depth, the remainder will be inside/below the skin depth. So, as frequency increases, your effective area decreases, and AC resistance goes up.
 

mik3

Joined Feb 4, 2008
4,843
As the skin depth is greater than the conductor's radius there are not any extra losses, the AC resistance is equal to the DC resistance. The problem arises when the skin depth becomes less than the conductor's radius because the charge does not flow through the center of the conductor but tends to flow near the surface of the conductor. This reduces the effective area of the conductor and thus the its resistance increases.
 
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