In a particular radio frequency (RF) application, you determine there is a need for a small inductor
Question:
In a particular radio frequency (RF) application, you determine there is a need for a small inductor of \(125 \mathrm{uH}\) and rather than trying to order one and wait for it to arrive, you decide to wind it yourself. The applicable equation is
\[
L=\frac{r^{2} N^{2}}{9 r+10 l}
\]
where \(L\) is the inductance in \(\mu \mathrm{H}, r\) is the radius of the coil in inches, \(l\) is the length of the coil in inches, and \(N\) is the number of turns. A maximum length of 1 inch and a maximum coil diameter of 0.25 inches are required in order to fit in the space available. You have both 45-gauge and 50-gauge wires that are 0.0028 and 0.001 inches in diameter, respectively, but the thinner wire is difficult to wind without breaking. Design your coil.
Step by Step Answer:
The Analysis And Design Of Linear Circuits
ISBN: 9781119913023
10th Edition
Authors: Roland E. Thomas, Albert J. Rosa, Gregory J. Toussaint