26.10.2012 Views

Radio Frequency Integrated Circuit Design - Webs

Radio Frequency Integrated Circuit Design - Webs

Radio Frequency Integrated Circuit Design - Webs

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

136 <strong>Radio</strong> <strong>Frequency</strong> <strong>Integrated</strong> <strong>Circuit</strong> <strong>Design</strong><br />

Figure 5.38 An IC in a package (delidded).<br />

where L is expressed as nanohenry per millimeter. For typical bond wires, this<br />

results in about 1 nH/mm.<br />

For two round wires separated by d and a distance h above a ground<br />

plane, the mutual inductance is estimated by<br />

2h<br />

M ≈ 0.1 ln�1 +� d � 2<br />

�<br />

(5.42)<br />

where M is expressed as nanohenry per millimeter. As an example, for a pair<br />

of bond wires separated by 150 �m (a typical spacing of bond pads on a chip),<br />

1 mm from the ground plane, their mutual inductance would be 0.52 nH/mm,<br />

which is a huge number. If the height is dropped to 150 �m above the ground<br />

plane, then the mutual inductance of 0.16 nH/mm is still quite significant.<br />

Note that parallel bond wires are sometimes used deliberately in an attempt to<br />

reduce the inductance. For two inductors in parallel, each of value L s , one<br />

expects the effective inductance to be L s /2. However, with a mutual inductance<br />

of M, the effective inductance is (L s + M )/2. Thus, with the example above,<br />

two bond wires in parallel would be expected to have 0.5 nH/mm. However,<br />

because of the mutual inductance of 0.5 nH/mm, the result is 0.75 nH/mm.<br />

Some solutions are to place the bond wires perpendicular to each other or to<br />

place ground wires between the active bond wires (obviously of little use if we<br />

were trying to reduce the inductance of the ground connection). Another<br />

interesting solution is to couple differential signals where the current is flowing<br />

in opposite directions. This could apply to a differential circuit or to power<br />

and ground. In such a case, the effective inductance is (L s − M )/2, which, in<br />

the above example, results in 0.25 nH/mm.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!