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Radio Frequency Integrated Circuit Design - Webs

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The Use and <strong>Design</strong> of Passive <strong>Circuit</strong> Elements in IC Technologies<br />

Since most of the applications lie in the 900-MHz to 5-GHz band, it is<br />

easy to see that making lines much thicker than about 4 �m will lead to<br />

diminishing returns. Going any thicker will yield little advantage at the frequencies<br />

of interest, because the center of the conductor will form a dead zone, where<br />

little current will flow anyway.<br />

Example 5.1 Effect of Skin Depth on Resistance<br />

A rectangular aluminum line has a width of 20 �m, a thickness of 3 �m, and<br />

a length of 100 �m. Compute the resistance of the line at dc and at 5 GHz<br />

assuming that all the current flows in an area one skin depth from the surface.<br />

Assume that aluminum has a resistivity of 3 �� �cm. Note that there are<br />

more complex equations that describe the resistance due to skin effects, especially<br />

for circular conductors [1]; however, the simple estimate used here will illustrate<br />

the nature of the skin effect.<br />

Solution<br />

The dc resistance is given by<br />

R = �L<br />

Wt<br />

= 3 ��cm � 100 �m<br />

20 �m � 3 �m<br />

The skin depth at 5 GHz of aluminum is<br />

= 50 m�<br />

�<br />

� = √�f� = √<br />

3 �� �cm<br />

� � 5 GHz � 4� × 10 −7 N<br />

A 2<br />

= 1.23 �m<br />

We now need to modify the original calculation and divide by the useful<br />

cross-sectional area rather than the actual cross-sectional area.<br />

R =<br />

�L<br />

Wt − (W − 2�)(t − 2�)<br />

3 �� �cm � 100 �m<br />

=<br />

20 �m � 2 �m − 17.5 �m � 0.54 �m<br />

= 98.2 m�<br />

This is almost a 100% increase. Thus, while we may be able to count on<br />

process engineers to give us thicker metal, this may not solve all our problems.<br />

99

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