Why is the expression for sheet resistance given without demonstration?

[EmilyRuck]

Hello!
I have in my notes an expression for the sheet resistance of a uniform conductor with length L, width W = L and thickness t. It is

R_{\square} = \displaystyle \frac{\sqrt{\displaystyle \frac{\pi f \mu}{\sigma}}}{1 - e^{-t/\delta}} = \displaystyle \frac{1}{\sigma \delta} \frac{1}{1 - e^{-t/\delta}}

where f is the frequency of the signal, \mu is the magnetic permeability, \sigma is conductivity of the conductor and \delta is its penetration depth.

This is given without any demonstration and it seems a standard expression. Do you know it or something similar? How can it be obtained?
If you don't have an answer, but you have some links or reference books, they will be useful as well!
Thank you anyway,

Emily

 

[EmilyRuck]

I will try to add some details, hoping that it will be useful.

Suppose that the thickness is along the x direction and the width is along y. The current density across a section can be expressed as

J(x) = J_0 e^{-(1 + j)x/\delta}

so it is supposed to be uniform along y.
The sheet resistance or R_{\square} should be obtained through the total current flowing through a section of the conductor and the voltage across a length L:

I = \displaystyle \int_{0}^{W = L} \int_{0}^{t} J(x)dxdy = L \int_{0}^{t} J(x)dx = - L J_0 \frac{\delta}{1 + j} \left[ e^{-(1 + j)t/\delta} - 1 \right]

Knowing that J_0 = - \sigma E_0 and that E_0 = V_0 / L we have

I = \sigma V_0 \frac{\delta}{1 + j} \left[ 1 - e^{-(1 + j)t/\delta} \right]

The impedance is

Z = \displaystyle \frac{V_0}{I} = \frac{1 + j}{\sigma \delta \left[ 1 - e^{-(1 + j)t/\delta} \right]}

The R_{\square} must be the real part of Z.

Even noting that

\frac{1}{\sigma \delta} = \sqrt{\displaystyle \frac{\pi f \mu}{\sigma}}

the real part of Z does not coincide with the R_{\square} of the original post. It is a more complicated expression.

 

[jasonRF]

Emily,

My hunch is that this is defined in terms of power dissipation. From Poynting's theorem, the power dissipated would be
<br /> P = \frac{1}{2} \Re \int dv \, \mathbf{E}\cdot\mathbf{J}^\ast<br />
For power per unit area you would only integrate along your x direction. Using
<br /> E(x) = E_0 e^{-(1 + j)x/\delta}<br />
<br /> J(x) = \sigma E_0 e^{-(1 + j)x/\delta}<br />
doing the integral and setting that expression equal to E_0^2/2 R may get you the expression you want, but it will have an extra factor of 2 in the exponent when compared to your original formula.

jason

 

[jasonRF]

Just one more thought. To apply such a concept it may make more sense to write E_0 in terms of H_0, then set the integral equal to |H_0|^2 R_{\square}. I'm thinking this may be more practical since we usually use PEC boundary conditions on good conductors, then use the surface H to estimate the surface current and hence losses.

 

[EmilyRuck]

Thank you for your very useful observations.
I tried to compute R_{\square} from the power, as you suggested: with L = W = 1. The result, as you predicted, is

R_{\square} = \displaystyle \frac{2}{\sigma \delta (1 - e^{-2t/\delta})}

It is definitely better than mine, even if not exactly what I expected.

A question: why the R_{\square} obtained from power is different from the R_{\square} obtained as \Re(Z)? Should not they be equal?

The curious thing is that the asymptotical behaviour of both

R_{\square} = \displaystyle \frac{1}{\sigma \delta (1 - e^{-t/\delta})} (1)

and

R_{\square} = \Re(Z) = \Re \left \{ \displaystyle \frac{1 + j}{\sigma \delta \left[ 1 - e^{-(1 + j)t/\delta} \right] } \right \} (2)

is the same.

For t \ll \delta, that is \omega \to 0, (1) becomes

R_{\square} \simeq \displaystyle \frac{1}{\sigma \delta (1 - (1 - t/\delta))} = \frac{1}{\sigma t}

and (2) is

R_{\square} = \Re \left \{ \displaystyle \frac{1 + j}{\sigma \delta \left[ 1 - e^{-(1 + j)t/\delta} \right] } \right \} \simeq \Re \left \{ \displaystyle \frac{1 + j}{\sigma \delta \left[ 1 - (1 -(1 + j)t/\delta ) \right]} \right \} = \frac{1}{\sigma t}

the same!

For t \gg \delta, that is \omega \to \infty, (1) becomes

R_{\square} \simeq \displaystyle \frac{1}{\sigma \delta}

and (2) is

R_{\square} \simeq \Re \left \{ \displaystyle \frac{1 + j}{\sigma \delta } \right \} = \displaystyle \frac{1}{\sigma \delta}

Again, the same.

This is the second doubt arousen: why is such a behaviour possible?P. S.
I didn't thought about H field, but up to now I preferred to mantain the E-point-of-view.