# Determining trace length for an Aluminum Resistor 1 Micron thick and 10 Microns wide

I put the time into writing this article on calculating the trace length for a 100 ohm resistor mean’t for high frequency usage, so I thought it would make sense to share it with others. Calculating on silicon element values can be frustrating, so I hope this helps you with your own calculations.

*Abstract***— In an effort to determine how best to create a 100 ohm resistor with a 10,000 Angstrom thick layer of Aluminum the following information has been compiled for review. **

Keywords: RESISTOR, ALUMINUM RESISTIVITY, RESISTOR IN SILICON, CROSS SECTIONAL AREA.

INTRODUCTION

In order to create an on device linear aluminum trace resistor using a minimum cross sectional area of 10 Microns and a layer thickness of 10,000 Angstroms (1 Micron) a number of calculations and alternatives are presented in the following sections.

*Objectives*

- Basic Calculations
- Adjustments & Effects
- Cautions

Basic Calculations

The resistivity of aluminum [1] is 8.82 ×10^-8 Ω•m which comes into play when calculating the resistance of a trace line.

To calculate Resistivity ( R ), we apply the following formula for resistance:

R = ρ L / A

Given (in SI units):

ρ = 2.82 ×10^-8 Ω•m

L = 1 μm = 1×10^-6 m

A =height×width

= (1×10^-6 m * 10×10^-6 m) = 1 * 10 ^-11 m^2

This calculation provides us with our cross sectional area. As you can see in the equations above we use our height of 10,000 Angstroms which translates to 1 Micron multiplied by our minimum feature width of 10 microns to get a total cross sectional area of 10^-11 m^2.

By applying this value to the previous equation we are able to calculate the resistance of a one micron line with the following calculations:

R = (2.82×10^-8)×(1×10^-6) / (10^-11)

R = .00282 Ohms/Micron (assuming 10 micron wire width)

Now that we know the resistance of a on Micron line we can simply divide 100 ohms by the resistance of a single micron line to determine how many microns of length are required to achieve 100 ohms of resistance.

100 ohms / 0.00282 = 35460.99291 Microns

35460.99291*10^-6 = .03546 Meters long

As shown above it 35.46 mm of trace length are required to produce 100 Ohms of resistance assuming a trace width of 10 microns.

Adjustments & Effects

35.46 mm is fairly long for a trace length on silicon, so we thought it might be helpful to explore what would occur if the minimum feature requirement (trace width) was reduced to 5 microns.

Please refer to Section I (Basic Calculations) for details on how these calculations were performed:

Assuming minimum feature size 5 micron

ρ = 2.82 ×10^-8 Ω•m

L = 1 μm = 1×10^-6 m

A =height×width = (1×10^-6 m * 5×10^-6 m) = 5 * 10 ^-12 m^2

We get

R = (2.82×10^-8)×(1×10^-6) / (5^-12)

R = .00564 Ohms/Micron (assuming 5 micron wire width)

100 ohms / 0.00564 = 17730.49645 Microns

17730.49645*10^-6 = .01773049645 Meters long

As can be seen above decreasing the minimum feature size from 10 microns to 5 microns has an inverse effect on the length requirement for our line resistor. Using 5 micron wide wire we are able to achieve a 100 ohm resistor using only 17.7 mm of trace length.

If possible reduction of trace width would be an ideal way to limit the length requirement, and luckily the width has a directly proportional effect on the required length.

As such it can be deduced that reduction in trace width to 2.5 microns would yield an 8.85mm length requirement, and further scaling could be performed up to the limit of the materials and technology in use.

Cautions

For low frequency applications the natural solution would be to snake or spiral the 35 mm line, however for operating at high frequencies, and in general for reliability as a resistor only device it is recommended that as direct a route as possible be taken.

In order to avoid high frequency problems, and significant RF generation it is possible to use two layers of aluminum and route lengths across from the underside to minimize the frequency transmission and inductance that is generated by the system.

Inductance is a real problem when snaking or spiraling outputs, so for ideal results a direct line or simple double or triple snake would be best. If at all possible separation of the “snakings” with a ground plane of some kind may improve performance.

Conclusion

If the minimum feature size must be 10 microns than a length of 35mm is required to achieve 100 ohms of resistance.

Any adjustments to the layer thickness or the trace width will have a proportional effect on the required length, so it would be wise to attempt to use lower layer thicknesses, or smaller feature sizes to solve the problem if those avenues are available.

Failing the ability to change either minimum feature size, or layer thickness a multilayer technique would likely provide better RF performance, but in all cases the RF problems are much less likely with a straight line approach.

Other conductive materials with a higher resistivity may provide a better solution to this problem if none of the suggestions are suitable for this application.

References

[1] hypertextbook.com, Resistivity Diagram, http://hypertextbook.com/facts/2004/ValPolyakov.shtml Retrieved 5 June 2014