Question 2.9: We want to implement a family of spiral inductors with exter......

We have, with constant $D_o$ and W = S:

$\frac{D_o-D_i}{2} \approx n W+(n-1) S=(2 n-1)  W,$

i.e., for the internal diameter:

$D_i = D_o − 2 (2n − 1)  W.$

The maximum turn number corresponds to $D_i \approx 0$, i.e.:

$n=\frac{1}{2}+\frac{D_o}{4 W}=5.5 \approx 5.$

We then have:

$\begin{aligned} \bar{D} & =\frac{D_o+D_i}{2}=D_o-(2 n-1) W=1.05-0.1 n \text{ mm} \\ \rho & =\frac{D_o-D_i}{D_o+D_i}=\frac{(2 n-1) W}{D_o-(2 n-1) W}=\frac{2 n-1}{21-2 n}. \end{aligned}$

The substrate correction factor is:

$K_\text{g}=0.57-0.145 \log \left(\frac{W}{h}\right)=0.57-0.145 \log \left(\frac{50}{300}\right)=0.83,$

while the gold surface resistance is:

$R_s=\sqrt{\frac{2 \pi f \mu}{2 \sigma}}=0.093 \sqrt{f_{ \text{GHz} }}.$

We therefore obtain for the inductor parameters, with W = 0.05 mm:

$\begin{aligned} L_{ \text{sp} } & =0.521\> 51 n^2  (1.05-0.1 n)\left[\log \frac{21-2 n}{2 n-1}+0.9+0.2 \left(\frac{2 n-1}{21-2 n}\right)^2\right] \text{ nH} \\ R_{ \text{sp} } & =1.2268 n  (1.05-0.1 n) \sqrt{f_{ \text{GHz} }} \\ C_3 & =0.095 \text{ pF} . \end{aligned}$

Neglecting dielectric losses the quality factor $Q_L = 2πfL_{\text{sp}}/R_{\text{sp}}$ increases like the square root of frequency, while (neglecting the capacitance towards ground) the inductor resonant frequency can be approximated as $f_0 = 1/ ( 2\pi \sqrt{LC_3})$. The inductance, quality factor at $f_0/2$ and resonance frequency for n = 1 … 5 are reported in Table 2.3. Notice that the maximum inductance is of the order of 10 nH and that the resonance frequency decreases with increasing number of turns. The quality factor evaluated at half the resonance frequency can be assumed to be an approximation of the maximum quality factor; the parameter is almost independent of the number of turns. The expression for $C_3$ is somewhat approximate, since it does not account for the number of turns.