IFL Cable Ohmic Resistance Limitations

This article intends to share knowledge about the considerations in deciding on IFL coax cable length and resistance.

In most low power BUCs, power is supplied to the BUC from the modem via the IFL cable. When selecting the proper type of cable for the IF, attention should be given not only to the impedance of the cable (75 Ohm or 50 Ohm), but also to the voltage drop along the cable due to its DC (Ohmic) resistance.  

Two independent parameters limit the total resistance allowed in the IFL cable:

1. The input voltage to the BUC should be above a specified value.

2. The total current draw from the power supply or the modem should not exceed the specification of that power supply.

The voltage drop along the cable and the voltage at the output of the IFL cable can be calculated using the following formulae:

                              The current drawn by the BUC as a function of Vx

Input voltage to the BUC


Vo = output voltage of the power supply or the modem

Po = power consumption of the BUC

R = total resistance of the cable

Vx = input voltage to the BUC

Ib = current consumption of the BUC at a given input voltage (Vx).

By solving equation (3) we receive:

This outcome provides the actual input voltage to the BUC (Vx), given the voltage of the power supply, total resistance of the cable and the power consumption of the BUC.

Using (1), the current consumption of the BUC at Vx input voltage can be calculated.

Example 1

Assume 5W C-Band BUC operated by a modem supplying 24V and IFL cable with total resistance of 2Ω.

The minimum allowed input voltage to the BUC is 15V.

The maximum current from the modem is 3A.

The relevant parameters are:

Po (power consumption) – 50W

Rs (cable total resistance) - 2Ω.

Vo (output voltage of the modem) – 24V.

Using (4) we receive:

This voltage is higher than the minimum voltage which is required for the BUC. Thus the BUC will operate as needed.

Using (1), the current Ib is

The current is lower than the maximum allowed.

Example 2

Using the same modem and BUC, but this time the cable will be 40% longer and thus Rs will be 2.8Ω (instead of 2Ω).

Using (4) again we receive:

This current is higher than the maximum allowed. Therefore the protection circuit in the modem will shut the power supply down.

Also, the input voltage Vx is below the minimum required by the BUC.

In this case, the total resistance of the IFL cable should be decreased. This can be done either by using a shorter cable (which may not be possible in many cases due to the distance between the modem and the antenna) or by using a higher quality cable with lower ohmic resistance.


Q. What happens if the input voltage to the BUC (Vx) is lower than the minimum allowed?

A. The answer depends on the type of the BUC. Some BUCs (like Belcom's BUCs) have minimum input voltage protection that shuts down the BUC if the input voltage is below a predefined limit.

In BUCs that don't have such protection, the BUC may still operate but some abnormalities may be expected, such as low output power, frequency instability and higher harmonics and spurious emissions.

Q.  How can I calculate the total resistance of the IFL cable?

A. The total resistance of the cable can be calculated by multiplying the

length of the cable, by the resistance per length unit of the cable (provided by the cable manufacturer).

Note that the total resistance per length unit is the sum of the resistance of the inner conductor and the outside shield.  Q. What can be done if the current limit of the modem is too low?

A. It is possible to add power inserters (Belcom P.N A1062R-150) and an external power supply between the modem and the BUC. If there is also concern that the input voltage to the BUC will be too low, the power inserter can be connected closer to the BUC using one port of the cable between the modem and the power inserters, and the other part between the power inserter and the BUC. For purpose of calculating the total resistance, only the second cable should be considered

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