Remembering wire sizes and resistances can be difficult, but this technique makes it a breeze. You only need to remember the number 17. I told you it was simple. It turns out that copper cable with a cross sectional area of 1 square mm has a resistance of approximately 17 milliOhm per meter. So the resistance of a cable is equal to 17milliOhm per meter divided by its cross sectional area in square mm. Now by coincidence it turns out that 17 gauge wire has a cross sectional area of approximately 1 square mm. Combine that with the fact that as you decrease the gauge by 3, the cross sectional area of wire doubles, you can come up with the following rule.

__Rule of Thumb__Copper AWG 17 wire has a cross sectional area of 1 square mm and a resistance of 17 milliOhm per meter. Doubling the cross sectional area of the wire will decrease the gauge by 3 and halve its resistance.Using this rule will get you half way there. You still need to make sure that the cable is capable of carrying the current you specify. As a general rule, the absolute maximum you should pass through a copper cable is 10 Amp per square mm. This would be for a single cable in open air. As soon as you have more than one cable or you enclose them in a conduit, you need to decrease the current in the cable to stop it from overheating. A pretty safe amount of current to run through a cable is 2.5 Amps per square mm, but depending on the size of the cable and where it's placed you may be able to get away with a more or less current. To show how to apply these rules to real world problems I've included an example below.

*I have a sensor at a remote location with 60 meters of copper cabling connecting it back to a base station. The sensor draws a load of 100 mA and is connected by AWG 25 cable. The supply voltage is 12 Volts but the sensor can work with as little as 11 Volts. Is the design feasible? If not, why?*

**Example 1**

*Answer*

*. Start by working out a scaling factor for the cable. AWG 25 is 8 gauges away from AWG 17. Every time you increase the gauge by 3, you double the resistance, so increasing the gauge by 8 is equivalent to doubling the resistance 2.66 times. This gives a scaling factor of 2 raised to 2.66, which at a quick guess is approximately 6.*

**1***Check if the cable can carry 100mA. AWG 17 can safely carry about 2.5 Amps, applying the scaling factor means that AWG 25 can carry about 400 milliAmps. Therefore the cable can carry the current.*

**2.***Check that the voltage drop in the cable isn't too high. Calculate the resistance of the cable, remembering that there are 2 cables, one supplying the current, and one providing a return path for it.*

**3.**AWG 17 cable is 17 milliOhms per meter. Multiply this by 6 to give a cable resistance of about 100 milliOhms per meter for AWG 25

R = 2 x 60 meters x 0.1 Ohms per meter = 12 Ohms

Calculate the voltage drop in the cable

V = I R = 0.1 Amps x 12 Ohms = 1.2 Volts

This leaves only 10.8 Volts to power the sensor, which isn't enough. To solve this problem a larger cable would need to be used.

As this is just an estimation tool, you'll still need to perform calculations to get an accurate set of numbers, but this will however get you in the ball park and allow you to know if your design is heading in the right direction.

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