58
2) Finding the maximum cable length using voltage drop
The current flowing from the Distributor into each coaxial cable connected to the TS-905 Infrared transmitter/
receiver unit is 0.1 A, since the number of Infrared transmitter/receiver units connected to each coaxial cable
is 1.
Assuming that RG-59/U coaxial cable is used, the voltage drop between the Distributor and the Infrared
transmitter/receiver unit is calculated by the following equation:
Voltage drop = 16.82 Ω x (50 m / 100 m) x 0.1 A
= 0.841 V
A remaining voltage of 4.159 V (5 V – 0.841 V) is the maximum allowable voltage drop between the Central
unit and the Distributor. The current that flows between the two is 0.4 A.
When RG-6/U coaxial cable is used between the Central unit and the Distributor, the cable length L1 between
the two is,
L1 = { (Voltage drop 1 / Current 1) / Coaxial cable loop resistance 1 per 100 m }
= { (4.159 V / 0.4 A) / 12.82 Ω } x 100 m
= 81 m
Maximum cable length L between the Central unit and the Infrared transmitter/receiver unit is calculated
by the following equation:
L = L1 + 50 m
= 81 m + 50 m
= 131 m
Similarly calculated for other types of coaxial cables, the maximum cable length between the Central unit
and the Infrared transmitter/receiver unit is found in the following table.
The table below shows the required maximum cable length for the example 2, the shorter length of the
calculation results (1) and (2) above.
RG-59/U
111 m
RG-6/U
131 m
RG-11/U
483 m
RG-59/U
111 m
RG-6/U
131 m
RG-11/U
483 m
