On the page 2 of the article about the surface resistance (link given below), the following is stated:

It is assumed that electric current flows on the surface of the material only.


What does that mean? What's the area through which current is flowing? Relationship between the current and current density is: $$I = \iint_S \vec j \cdot d\vec S$$

Component of the current density perpendicular to the area contributes to the current through that area.

Since we're talking about the surface resistance, current density component parallel to the upper horizontal surface of the sample is relevant to the phenomenon (between two electrodes as shown on the figure 1). However, in this case, it seems that defining area which is perpendicular to the current density makes no sense or better to say, surface is zero and therefore current should be zero.

On the page 1 of the article, surface current, $I_s$, is defined. It's relationship to surface resistance is given by the Ohm's law. Specific surface resistance, $\rho_s$, is also defined as: $$\rho_s = R_s \frac {D}{L}$$ where $D$ is the width of the upper sample surface (figure 1 of the link) and $L$ is the distance between the electrodes.

This definition is different compared to the specific resistance I learned about (it also has different dimension): $$\rho = R \frac {A}{L}$$

In the context of figure 1, $A$ would be cross sectional area of the sample between the electrodes and $L$ remains the distance between them. The area $A$ in this formula should be the same as the area of integration in the first equation. However, it seems that in the definition of the current, $I_s$, given in the article, no area is defined whatsoever. This doesn't make sense. Any ideas?


1 Answer 1


It means the current is confined to a thin region at the top of the sample, "thin" meaning much lower in thickness than the electrode dimensions and spacing. Usually, this happens when you have a thin layer of conductive material on top of an insulating substrate, or a semiconductor substrate whose surface is doped to make conductivity much higher there. To answer your question, in general it may not be possible to define an exact area through which a uniform current density flows, because the current density can vary as a function of depth. Concepts of sheet resistance/conductance and surface current density still remain meaningful, however.

Suppose the conductivity $\sigma=1/\rho$ (in units $\text{S/m}$) is a function of distance $x$ from the top surface. Since the conductive layer is very thin, we assume it's only non-zero for $x<t$. The surface current density, defined as current per unit perpendicular width, is calculated as $$J_s = \int\limits_0^tJ(x)dx=\mathcal E\int\limits_0^t\sigma(x)dx.$$ Here, $J=\sigma \mathcal E$ is the usual current density (current per unit area) you are familiar with, $\mathcal E$ being the lateral E-field. Now we define sheet conductance $\sigma_s$ as the integral on the right hand side, so that we can write $$ J_s = \sigma_s\mathcal E$$ in analogy with the usual "3D" version of Ohm's law. The sheet resistance is just $\rho_s = 1/\sigma_s$. Sheet resistance/conductance are a characteristic of the surface itself, not of a particular device. Note that we have not referred to a particular device width or length in defining it.

Now to see why these definitions can be useful, suppose we want to calculate the resistance of a region of the sample with width $D$ and length $L$. The current is $$I=\int\limits_0^D\int\limits_0^tJ(x)~dx~dy=DJ_s=D\sigma_s\mathcal E=D\sigma_s\frac V L.$$ $$R =\frac{V}{I}=\frac L D \rho_s.$$ Because of this relation, sheet resistance $\rho_s$ is sometimes defined as the resistance of a "square" device (meaning $L=D$), and reported in units $\Omega/\square$.

  • $\begingroup$ So, there is a small area, adjacent to the surface. Current density doesn't need to be the same over the area, equation written is the most general form. One thing I'd like to ask is, why don't we use the more common relationship between resistance and specific resistance (which includes the area rather than $D$ only)? Since, the problem is essentially the same, only here we have a much smaller area. $\endgroup$ Commented Dec 16, 2023 at 13:51
  • $\begingroup$ The problem I have is that $R_s$ is supposed to be independent of the thickness, $t$ to give a measure of how well does the thin film conduct. However, this isn't the case because even though $L/D$ might be the same, $R$ will be smaller for thicker samples (bigger surface area of conduction means less resistance and bigger current). $\endgroup$ Commented Dec 16, 2023 at 14:01
  • $\begingroup$ As I explained, the cross section area may not be well-defined. Current density might decay exponentially with depth for example. What area should we use then to calculate the resistance? If area is well-defined, e.g. in the case of a uniform conductive layer, you CAN use the cross section area, you'll get the same result as above. $R_s$ (which I suggest you call $R$, because it's just a resistance regardless of where the current is flowing) does depend on the thickness of the conductive layer, why do you think it isn't supposed to? A thicker layer will result in lower $\rho_s$ and hence $R$. $\endgroup$
    – Puk
    Commented Dec 16, 2023 at 14:26
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    $\begingroup$ @DarioMirić It is important to distinguish between cross section area through which current flows ($\sim Dt$), and the device "layout" area as seen from above ($LD$). Square means $L=D$, so $R=\rho_s$, independent of square size (i.e. layout area). This is why sheet resistance $\rho_s$ is often reported in units $\Omega/\square$. All this has nothing to do with thickness. Finally, in my experience sheet resistance is more commonly denoted by $R_{\text s}$ or $R_\text{sh}$, hence my advice to not use $R_s$ to refer to the resistance of a particular device. $\endgroup$
    – Puk
    Commented Dec 16, 2023 at 15:35
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    $\begingroup$ Ohh, so different square than I thought. I got it, thanks. $\endgroup$ Commented Dec 16, 2023 at 15:50

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