Page 321 - Analog and Digital Filter Design

3 1 8 Analog and Digital Filter Design

&R+l +-(1+1212/w)-o.5
ER-1
where seff =-
- 2
3
hlw = 0.064
5.7 3 7
~~,y =~+-(1+12 ~0.064)~'~ =4.24
2
Therefore, Z,,, = 183/[15.625 + 1.393 + 0.677 Ln(15.625 + 1.444)]

2, = 183/[17.018 + 1.92081

2, = 9.663 R.
This impedance will be used a little later on, in the equations for capacitors. The
effective permitivity is 4.24, so the wavelength along the track is 145.7 mm and the
track width of 25 mm is much less than a quarter wavelength, as suggested earlier.

To replace inductors by PCB tracks you need narrow tracks that can be easily
etched. Consider using tracks 0.5mm wide. Since wllz is now less than one, a dif-
ferent equation can be used to find the characteristic impedance.

Z,,, = -Ln(!+0.25:).
60
6
The effective relative permitivity is now given by the expression:
[
121' -'"
E,~ = - - --) %)'I
&R -1
&R +1
+
+
(1
2 2 + 0.04 i( 1 -
The ratio wllz = 0.3125, and hlw = 3.2.
5.7 3.7
[
+
(1
= - - + 38.4)-'.' + 0.041(1- 0.3 125)2]
2 2
= 2.85 + 1.85[0.1593+0.02162]
= 3.185
Z,,, = 33.62Ln(25.6+0.078) = 109.12 SZ
This impedance will be used now in the equations for inductors. The length of
a narrow track used to form an inductor is given by the expression:

Here L is the required inductance, c is the velocity of light (3 x 10Sm/s), Z,, is
the impedance (= 109.12) of a 0.5mm wide line, and is the relative effective
permitivity (= 3.185) of the dielectric for such a line.