# Serpentine resistor - Number of bends effect

I was trying to simulate two serpentine resistors on COMSOL multiphysics with exactly the same length, width and thickness. I drew them on AutoCAD. The only difference between the two resistors is the number of bends. For resistor(1), number of bends is only 2, while for resistor(2), number of bends is 18.

After running simulation on AC/DC module and Electric Currents (ec) sub module, resistance values came as the following:

Resistor(1) = 13.221 ohm (2 number of bends)

Resistor(2)= 12.654 ohm (18 number of bends)

I expected that the number of bends will not affect the resistance value, but it actually does in an inverse manner (though not much, but it does) Is anyone aware why would the number of bends affect the resistance value?

Simulation parameters:

Resistor(1): Length:337mm; Width: 1mm; Thickness: 0.2mm; Number of bends:2

Resistor(2): Length:337mm; Width: 1mm; Thickness: 0.2mm;Number of bends:18

Meshing:

Resolution: Extra fine.

Minimum Element Size: 0.18mm.

Maximum Element Size: 4.2mm.

Maximum Element Growth Rate: 1.35

Curvature Factor:0.3

Resolution of Narrow Regions: 0.8

Effect of meshing resolution on resistance values:

• I don't know about you, but I don't trust any simulation that I didn't screw up myself. :-) No, seriously... one of the problems here is that it might be difficult to distinguish real effects due to the bends from simulation artifacts. Can you set simulation parameters, like resolution of the grid? – CuriousOne Aug 8 '16 at 9:25
• @CuriousOne lol I understand :) I edited my question and added some simulation parameters. Is there anything else I can add? – HaneenSu Aug 8 '16 at 9:55
• Can you change the mesh size? What happens when you plot the two resistance values as a function of minimal mesh size? Does there seem to be a convergence to some final values for an "infinitely fine" mesh? Just a thing to try... not sure it will be more enlightening about the simulation quality, but it's something I would do. There should, of course, be a real difference because the current density changes in the corners... but by how much... that's the real question, right? – CuriousOne Aug 8 '16 at 9:56
• @CuriousOne makes sense. In literature I didn't find anything that mentions the effect of the number of bends (as far as I could look), it seems since the effect is small, it was overlooked. Will try your suggestion and post the outcome. – HaneenSu Aug 8 '16 at 10:08
• @CuriousOne It seems meshing resolution affects the simulation outcome but not to the extent to cover the effect of the number of bends. Regarding your previous comment about current density near the corners, this is what I am not sure about. In what manner should the current density change near the corners and is the inverse relationship between number of bends and resistance explainable? – HaneenSu Aug 8 '16 at 10:45

You are measuring the length through the center, but the current will not follow that path, rather it will crowd into the corners and, on average, take a somewhat shorter path through each bend, so your total resistance will be less for the serpentine pattern with more bends. Below is a field solver simulation showing a current density at the inside corner that is 5x higher than the average in the straightaways.

One rule of thumb is that a corner square is equivalent to 0.56 squares rather than one square. Your first example has 4 corners, the second 36.

If I back-calculate the effect from the ratios in your simulation I get 0.55 as the effect of one square, which is pretty close.

Your results do not surprise me. Actually, to compute the resistance, you need to solve the Laplace problem in your areas. My coauthor and I solved a similar problem in http://arxiv.org/abs/1506.01599 . I believe the reason you have a difference is the width/length ratio is small but finite. I would predict the relative difference would be smaller for a smaller ratio. To understand the difference, you may note that the shortest distance from one end of a resistor to another (within the area of the resistor) is shorter for the resistor with more bends. Imagine for a moment that your resistors are racetracks and try to determine the shortest trajectories drivers would select to win, for example, how they would pass the bends - closer to the "inner radius". In the same way, current tends to choose a trajectory that is the shortest in some respect.

• In case you cannot move diagonally, you could shave off exactly 1 width per bend. As such a nice ballpark estimate for impact of bends on resistance seems to be: length - (width*#of bends) . In this case (337-2)/(337-18) gives a ratio around 1.05 which is close to the 1.44 ratio you would get by doing 13.221/12.654. – Dennis Jaheruddin Aug 8 '16 at 14:24
• @DennisJaheruddin Thanks for the useful contribution. I guess you meant 1.044 – HaneenSu Aug 9 '16 at 3:39

The error you see is a consequence of the effective width of the trace at the corner. Pick a corner, and draw a line from the inner corner point to the outer corner point. Observe that the line makes a 45° angle to both the entering and leaving trace. Further observe that this line is longer than the width of the trace. Specifically, it's sqrt(2)*width, or about 1.41*width. Since the trace in the corner is wider, the cross-sectional area is proportionally higher and the resistance proportionally lower. Changing to two 45° bends or rounded corners will reduce the effect. Carefully constructing a constant cross-section trace will eliminate the effect.

Some circuit board designers use curves instead of angled corners to reduce the impact of the varying resistance. Here's a paper that discusses some corner designs, and how they affect impedance. Their finding is that there's very little difference between corner designs at high frequency, but unfortunately there's no mention of DC parameters, which definitely do change.

So in summary, though you controlled the length to be identical, you missed the area change due to the corner shape.