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Radio Frequency Integrated Circuit Design - Webs

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

Figure 5.14 Inductor with dimensions.<br />

Solution<br />

We can start by estimating the inductance of the structure by using the formulas<br />

in Section 5.11 and referring to Figure 5.9.<br />

davg = 1<br />

2 (D out + D in) = 1<br />

(270 �m + 161 �m) = 215.5 �m<br />

2<br />

� = (D out − D in) (270 �m − 161 �m)<br />

= = 0.253<br />

(D out + D in) (270 �m + 161 �m)<br />

L = 2.34� o<br />

n 2 d avg<br />

1 + 2.75�<br />

= 2.34 � 4� × 10 −7 N<br />

A 2<br />

3 2 � 215.5 �m<br />

1 + 2.75 � 0.253<br />

= 3.36 nH<br />

Next, let us estimate the oxide capacitance. First, the total length of the<br />

inductor metal is 2.3 mm. Thus, the total capacitance through the oxide is

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