Physical Electronics: Uniform Current Density

[jegues]

Homework Statement

Typical semiconductor fabrication technologies employ copper (ρ ~ 15.7nΩm; μr ~1) as the interconnect material. A process is designed for a clock frequency of 1GHz, has interconnect cross sections ranging between 35nm x 70nm and 0.18um and 90nm. Quantitatively determine whether it is reasonable to expect a uniform current density across these wire cross-sections

Homework Equations

The Attempt at a Solution

I know that,

\vec{J} = \sigma \vec{E}

and that,

J = \frac{I}{A}

I'm not sure how to obtain J though.

With the information given, I can calculate the skin depth,

\delta = \sqrt{\frac{2}{\omega \sigma \mu}} \approx 2\mu m

but I don't know how they helps me determine whether the current density will be uniform across these two wire sections.

The area of these sections can easily be calculated by multiplying the two dimensions.

Any ideas?

 

[rude man]

Homework Statement

Typical semiconductor fabrication technologies employ copper (ρ ~ 15.7nΩm; μr ~1) as the interconnect material. A process is designed for a clock frequency of 1GHz, has interconnect cross sections ranging between 35nm x 70nm and 0.18um and 90nm. Quantitatively determine whether it is reasonable to expect a uniform current density across these wire cross-sections

Homework Equations

The Attempt at a Solution

I know that,

\vec{J} = \sigma \vec{E}

and that,

J = \frac{I}{A}

I'm not sure how to obtain J though.

With the information given, I can calculate the skin depth,

\delta = \sqrt{\frac{2}{\omega \sigma \mu}} \approx 2\mu m

but I don't know how they helps me determine whether the current density will be uniform across these two wire sections.

The area of these sections can easily be calculated by multiplying the two dimensions.

Any ideas?

The skin depth formula tells yu how deeply inside the surface the current will flow.

For example, if the skin depth is d1 and your cross-section is d2 x d3, what fraction of the area d2*d3 do you think will be covered by current?

 

[jegues]

The skin depth formula tells yu how deeply inside the surface the current will flow.

For example, if the skin depth is d1 and your cross-section is d2 x d3, what fraction of the area d2*d3 do you think will be covered by current?

If my surface has dimensions d2 and d3 in the shape of rectangle, and the current will flow within a δ from the edges of the surface, then the current will flow in an area that is,

2δ(d2+d3)

Thus the current will flow in a fraction of the total area,

\frac{2\delta *(d_{2}+d_{3})}{d_{2}*d_{3}}

is this correct?

 

[rude man]

If my surface has dimensions d2 and d3 in the shape of rectangle, and the current will flow within a δ from the edges of the surface, then the current will flow in an area that is,

2δ(d2+d3)

Thus the current will flow in a fraction of the total area,

\frac{2\delta *(d_{2}+d_{3})}{d_{2}*d_{3}}

is this correct?

Close, but I should have started differently, especially in this case:

Look at d2 and d3. You were given four different values, all < 2 μm.

So since δ > d2 and d3 then what can you say about the uniformity of the current density?

 

[jegues]

Close, but I should have started differently, especially in this case:

Look at d2 and d3. You were given four different values, all < 2 μm.

So since δ > d2 and d3 then what can you say about the uniformity of the current density?

It will flow through the entire face of the surface.

 

[rude man]

It will flow through the entire face of the surface.

Exactly. Uniform current density, not affected by skin depth limitations.

 

[jegues]

Exactly. Uniform current density, not affected by skin depth limitations.

So how do I show that quantitatively? Is it as simple as showing that δ is larger than any of the dimensions given and stating the this must be a uniform current density since it is not affected by skin depth limitations?

 

[rude man]

So how do I show that quantitatively? Is it as simple as showing that δ is larger than any of the dimensions given and stating the this must be a uniform current density since it is not affected by skin depth limitations?

Basically, yes.

But realize that "skin depth" does not mean that ALL the current is confined to that depth. And it also doesn't mean the current density is TOTALLY uniform within that depth. Rather, the current density rolls off with distance inside the surface of the conductor in an exponential manner.

Interesting fact: if you took a conductor and calculated the dc resistance of the part covered by the skin depth, that resistance would closely equal the resistance of the entire conductor at the frequency of operation! That statement is more true the thinner the skin depth is compared to the cross-sectional dimensions of the conductor.

 

[jegues]

There is a small extension to this question.

If I were to use the relation,

R = \frac{\rho L}{A}

to estimate the resistance of these interconnects, would I be over-estimating or under-estimating the resistance?

I think I would be over estimating the resistance because since the skin depth is larger than the surface of the interconnect some of the current that I'm trying to push through this interconnect surface will be lost, leading me to believe that the conductor is of larger resistance than it actually is.

Is this correct? What's the right way to think about that?

Thanks again!

 

[rude man]

Look at my post #8, "interesting fact". That says that if the skin depth were a lot smaller than a dimension then the effective resistance at the high frequency would be the same as the dc resistance of a hollow conductor with thickness = skin depth. The dc resistance is of course your R = ρL/A.

In your case however the skin depth is >> dimension so the answer is that the dc and 1 GHz resistance would be about the same.

 

[jegues]

Look at my post #8, "interesting fact". That says that if the skin depth were a lot smaller than a dimension then the effective resistance at the high frequency would be the same as the dc resistance of a hollow conductor with thickness = skin depth. The dc resistance is of course your R = ρL/A.

In your case however the skin depth is >> dimension so the answer is that the dc and 1 GHz resistance would be about the same.

Yes but the questions strictly states, "over-estimating or under-estimating" it must be one of the two. I still don't entirely understand which one it should be and why.

Would it be over-estimating or under-estimating and why?

 

[rude man]

Yes but the questions strictly states, "over-estimating or under-estimating" it must be one of the two. I still don't entirely understand which one it should be and why.

Would it be over-estimating or under-estimating and why?

It actually asked whether the current densiti would be uniform or not.

Well, as I said, it would not be COMPLETELY uniform. At ANY finite frequency there is SOME skin effect. So the ac resistance is always greater than the dc resistance. The inner sections of the conductor will always carry less current than it would at dc.

I think of it as follows: OK, some of the current is forced to the outer sections of a cable. That alone does not seem to change the overall reistance at first blush. But: the power loss will be greater because power loss is proportional to i². So the power loss in the skin depth case is goig to be the same current squared except the R will greater since the inner parts of the conductor are effectively removed. So we get i²R for skin effect > i²R without it.

So a dc measurement always underestimates the ac resistance.