Question

3. Consider a two-wire transmission line having length L = 1 m and wire separation s = 1.3 mm, operated at 100 MHz. What is the maximum differential mode current to satisfy the FCC limit on Class B devices? Repeat for a common mode current.

Answer 1

The two wire transmission line is shown below.

1.3mm Transmisen Reception im Side

For satisfying the FCC limit of class B devices, within a radius of 3 m from the transmission line and for a frequency f=100MHz, the electric field strength E_d=150*10^-6 V/m

The differential mode current is given as:-

I_d=E_d*r/(f²*S*1.316*10^-14)

where S=loop area=L*s

Given L=length of the transmission wires=1 m

s=1.33 m

S=1.3*10^-3 m²

For maximum differential current, we take r=3 for this range of frequency given.

Thus,
substituting the values,

I_d=150*10^-6 *3/(10¹⁶*1.3*10^-3*1.316*10^-14)

=2630.35 $\mu$ A

This is the maximum value of differential mode current to satisfy the FCC limit of class B devices.

Similarly, the maximum common mode current,

I_C=E_C*r/(f²*S*1.257*10^-14)

Here, E_c=50*10^-6 V/m

Substituting the values,

I_d=50*10^-6 *3/(10¹⁶*1.3*10^-3*1.257*10^-14)=917.94 $\mu$ A