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Sheet resistance expression

  1. Jul 15, 2015 #1
    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
     
  2. jcsd
  3. Jul 17, 2015 #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.
     
  4. Jul 17, 2015 #3

    jasonRF

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    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
     
  5. Jul 17, 2015 #4

    jasonRF

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    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.
     
  6. Jul 18, 2015 #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.
     
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