Wires and Voltage Ratings Calculation
This is a method to help on calculations relative to the electric cables dimensioning in the direct current operating systems.
From the following formulas it is possible to obtain:
- which voltage drop will be in relation to the line length, the current draw, the section of the cable used
- how much current can be drawn in the end of a line depending on its length, the section of the cable used and the voltage drop accepted
- which cable section must be used to have a certain voltage drop in relation to the length and the current draw of the load
Each conductor, depending on its section, gives a certain resistance to the passage of current. The following is a table that summarizes the specific resistance values for the copper cables and the sections more usually used.
|Cable section in mm2||Specific resistance|
- V. drop: Voltage drop on the load (difference between the output tension from the supplier and the minimum operating tension of the connected device)
- Length: Line length ( distance from the supplierx 2 ) in metres
- I. load: Max. Current drawn from the device connected in the end of the line ( in mA )
- Section: Section of the cable used ( which corresponds to the specific resistance as shown in the table above) in mm2
|(Specific resistance x Line length) x I.load|
|V. drop =||= Volt|
A sounder that draws max. 2A must be installed on a 100 metres line. The line is installed using a cable with a section of 1 mm2.
Which voltage drop will be in the end of the line?
Solution: ( 0,018 x 200 ) x 2000 / 1000 = 7,2 Volt
|V. drop max allowed x 1000|
|I.load max =||= mA|
|Specific resistance x Line length|
Is it possible to install a 24 V sounder that draws 2A max. in the end of a line 100 metres long using a cable with the section of 1 mm2 taking into consideration that the sounder operates with a minimum of 13 V ?
( 24 - 13 ) x 1000 / 0,018 x 200 = 3055 mA = 3 A .... so the answer is : YES.
And if the sounder operates with a minimum of 18 V ?
( 24 - 18 ) x 1000 / 0,018 x 200 = 1666 mA = 1,6 A .... so the answe is : NO.
Together with the choice of the battery capacity and the definition of the current suppliable by the supplier, the correct dimensioning of the cables sections allows the devices to have optimal tension values according with the regulations established by the manufacturer.
Supplying a sensor or a call point with a lower tension than the minimum value stated means for the device a situation of instability, low efficiency, poor immunity to the noises.
The following data must be considered:
|Voltage at Source||Vs||For a correct and preferable result consider a critical situation as mains loss||24 V|
|Minimum voltage on sensor||Vc||Pointed out by the manufacturer data||13,5 V|
|Sensor current||Ic||Refer to manufacturer data sheet or measured by a multimeter (in milliampere)|
NOTE: In the case of NOT self-powered devices must be used the higher absorption in the different conditions: Service, Stand-By, Alarm
|Line length||L||Segment of cable between the source and the sensor (in metres)||50 m|
Calculated the data, they can be insert in an easy formula to obtain the measure of the smallest cable section that is able to guarantee the optimal functioning: the minimum section, measured in mm2, must be equal or bigger than:
Section = ( 2L x Ic x 0,038 ) : [ ( VS - Vc ) x 1000 ]
Section = (100 m x 10 mA x 0,038 ) : ( 24 V - 13,5 V) x 1000 ) = 0,0036 mm2
The cable for the installation must have the conductors section equal or bigger than 0,04 mm2
The method described can be used to dimension with a good approximation a more complex tension network, with ramifications.
In this case it is enough to identify all the connection points from the source on, to effect the calculation for each single segment between two points and then summing all the results.
Take always into account the minimum sections given by the bodies of regulations that happen to be bigger than those obtained from the calculations.