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I have been tasked with writing a software to calculate the maximum cable length of the fire detection loop, I am not a electrical engineer myself and therefore I am struggling with some of the calculation aspects, please help.

So having data:

  • Wire type i.e. wire with cross sectional area 1.5 mm square has the resistance of 12.1 Ohms/km or 3.688 Ohms/1000 ft. Or for the USA folks 12 AWG.
  • Vopmin (Minimum operating voltage) 17 V
  • Voutmin (Minimum output voltage) 24 V
  • Total devices current consumption i.e. 400 mA

We calculate the worst case scenario, i.e. all of the detection devices are based at the end of the line.

Having this data, how to calculate the Vdrop (voltage drop across the length of the wire). And the maximum cable length of this circuit?

Please if possible provide the formulas with the example calculation, thank you in advance.

Regards

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  • \$\begingroup\$ You need Ohm's law: en.wikipedia.org/wiki/Ohm's_law so: V = I * R = 400 mA * 12.1 Ohms = 4.84 V. So 4.84 V is dropped across a 1 km cable when 400 mA is flowing. Some pitfalls: Is that 12.1 Ohms one way (one end to the other end) or does that also include the return path ? If 12.1 Ohms is one way then the 4.84 V doubles to 9.68 V. \$\endgroup\$ – Bimpelrekkie Dec 19 '16 at 16:15
  • \$\begingroup\$ Yes there is a return. i.e there are two wires in the cable which goes out of the fire panel + & -, the cable makes circle and returns back to the panel. \$\endgroup\$ – Kugarek Dec 19 '16 at 16:22
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Your question comes down to a basic application of Ohm's Law. It says that the voltage drop across a resistor is the current thru the resistor times the resistance.

You apparently have wire with a resistance of 3.688 mΩ/foot. To get the total actual resistance, you need to know the length of the wire. Remember that current has to run out one wire and back on another, so the wire length the current sees is twice the distance from your power supply to your device.

Since you didn't give the distance, I'll work it backwards to find the maximum distance your setup can support.

It seems the power supply puts out 24 V, and the device at the other end of the cable can operate down to 17 V. That means the cable can drop 7 V. You also say the current is 400 mA. Using Ohm's law, (7 V)/(400 mA) = 17.5 Ω.

Now we can find the wire length that results in that much resistance.

  (17.5 Ω)/(3.688 mΩ/foot) = 4745 ft

Again, it takes two wires, so the distance from the power supply to the device can be at most half that, which is 2373 feet or 723 meters.

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