Why is the expression for sheet resistance given without demonstration?

In summary, the conversation discusses an expression for the sheet resistance of a uniform conductor, given without a demonstration, and the process of obtaining it. The expression involves variables such as frequency, magnetic permeability, conductivity, and penetration depth. There is a discussion about using power dissipation to calculate the sheet resistance, with one participant noting that the result is better than the original expression but not exactly what was expected. The conversation also touches on the behavior of the expression at different values and the possibility of using H-field instead of E-field.
  • #1
EmilyRuck
136
6
Hello!
I have in my notes an expression for the sheet resistance of a uniform conductor with length [itex]L[/itex], width [itex]W = L[/itex] and thickness [itex]t[/itex]. It is

[itex]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}}[/itex]

where [itex]f[/itex] is the frequency of the signal, [itex]\mu[/itex] is the magnetic permeability, [itex]\sigma[/itex] is conductivity of the conductor and [itex]\delta[/itex] 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
 
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  • #2
I will try to add some details, hoping that it will be useful.

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

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

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

[itex]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][/itex]

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

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

The impedance is

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

The [itex]R_{\square}[/itex] must be the real part of [itex]Z[/itex].

Even noting that

[itex]\frac{1}{\sigma \delta} = \sqrt{\displaystyle \frac{\pi f \mu}{\sigma}}[/itex]

the real part of [itex]Z[/itex] does not coincide with the [itex]R_{\square}[/itex] of the original post. It is a more complicated expression.
 
  • #3
Emily,

My hunch is that this is defined in terms of power dissipation. From Poynting's theorem, the power dissipated would be
[tex]
P = \frac{1}{2} \Re \int dv \, \mathbf{E}\cdot\mathbf{J}^\ast
[/tex]
For power per unit area you would only integrate along your x direction. Using
[tex]
E(x) = E_0 e^{-(1 + j)x/\delta}
[/tex]
[tex]
J(x) = \sigma E_0 e^{-(1 + j)x/\delta}
[/tex]
doing the integral and setting that expression equal to [itex]E_0^2/2 R[/itex] 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
 
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Likes EmilyRuck
  • #4
Just one more thought. To apply such a concept it may make more sense to write [itex]E_0[/itex] in terms of [itex]H_0[/itex], then set the integral equal to [itex]|H_0|^2 R_{\square}[/itex]. 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.
 
  • #5
Thank you for your very useful observations.
I tried to compute [itex]R_{\square}[/itex] from the power, as you suggested: with [itex]L = W = 1[/itex]. The result, as you predicted, is

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

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

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

The curious thing is that the asymptotical behaviour of both

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

and

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

is the same.

For [itex]t \ll \delta[/itex], that is [itex]\omega \to 0[/itex], (1) becomes

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

and (2) is

[itex]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}[/itex]

the same!

For [itex]t \gg \delta[/itex], that is [itex]\omega \to \infty[/itex], (1) becomes

[itex]R_{\square} \simeq \displaystyle \frac{1}{\sigma \delta}[/itex]

and (2) is

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

Again, the same.

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

What is sheet resistance expression?

Sheet resistance expression is a measurement used to describe the electrical resistance of thin, flat materials such as semiconductor wafers or thin films. It is expressed in units of ohms per square (Ω/square) and is used to characterize the conductivity of a material.

How is sheet resistance expression calculated?

Sheet resistance expression is calculated by measuring the resistance of a material with known dimensions and then dividing by the surface area of the material. This can be done using specialized equipment called a four-point probe, which applies a known current and measures the voltage drop across the material.

What factors can affect sheet resistance expression?

Sheet resistance expression can be affected by a variety of factors including temperature, impurities or defects in the material, and the presence of a magnetic field. Additionally, the type of material and its thickness can also impact sheet resistance.

Why is sheet resistance expression important in semiconductor manufacturing?

Sheet resistance expression is important in semiconductor manufacturing because it allows for the characterization and quality control of thin films and other materials used in electronic devices. It can also provide valuable information about the electrical properties of a material, helping to optimize its performance in different applications.

How is sheet resistance expression used in other industries?

While sheet resistance expression is most commonly used in the semiconductor industry, it is also used in other industries such as solar cell manufacturing, display technology, and thin film coatings. It is a useful measurement for any industry that works with thin, flat materials and needs to understand their electrical conductivity.

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