Connect 2 different wire gauge; is the smaller gauge affecting the capacity of the bigger wire gauge?

Question

Hello, I would like first to thank anyone taking the time to read this or help. I was wondering how to calculate how many amp I can use by connecting 2 different wire gauge. I am on a 12v dc circuit and would ie to connect a 12v led module to a battery for a long distance. The led modules come with a 22 awg gauge cable from the factory. I would like to run it for a long distance to the power source (battery or power supply) I plan to use awg 16 wire connected to the led module so I can make a big distance safely. I will solder the awg 16 to the awg 22 of the led module. I usually know how to calculate the max amp that I can run for a single wire gauge by looking at spec of wire. However I am not sure if the 22 awg wire gauge from the led will limit my 16 awg? Is the 16 awg maximum amp capacity is limited by the 22 awg or will the 16 awg be able to use it's full amp capacity?

So to calculate how far I can get with the awg 16 can I just calculate the 22 awg capacity and add to the 16 awg capacity or I need to calculate the 16 awg capacity as if it was a 22 awg?

I am sorry my english is not great I live in montreal and speak french

Thank again for any help

cheers

Gough Lui · Answer

Your question is an interesting one can be solved by a little bit of thought - but the short answer is that each gauge and type of wire has its own maximum current capacity (ampacity) and connecting the two together does not change the fact that the 16AWG portion has its own maximum current rating and the 22AWG portion has its own maximum current rating. Both maximum current ratings must be respected. The long answer: There's actually two different issues which are being mixed together. The first that you have mentioned is maximum current carrying capacity (or ampacity) and this refers to the maximum current the cable is rated to carry so it doesn't cause damage to its insulation . This is not necessarily a good current level to operate at, because the second issue is cable resistance which causes a voltage drop. Lets say your long run is 20m with 16AWG wire with an 80 degrees C rated insulation. Its maximum current capacity is 19A. But if you tried to pass 19A through the cable, you would lose 10.01V (190W of heat) with only 1.99V remaining to power your device. So in most practical instances, your cable choice should be constrained by resistive losses rather than the cable current carrying capacity. However, you can size your cable protection to respect the maximum ampacity to avoid cable damage during fault conditions (i.e. use a 15A fuse for example). Knowing this, the ampacity of your 16AWG cable may be 19A and the ampacity of your 22AWG cable might be 8A . How does this affect things? Well to put it simply, the ampacity of each size of cable must be respected . Think of a home electrical distribution supply - the main supply from your utility comes through thick cables and may be fused at 100A (cable has an ampacity at least 100A). But this is then connected to your end circuits including power points and appliances through thinner wire - mostly with ampacity around 20-30A. This is why each circuit has its own fuse or circuit breaker protecting that branch from fault current induced damage. This is an instance where one thick cable is powering many appliances through a number of thinner wires. But it also means that say you run one 16AWG run from the battery to a central point (with a fuse rated appropriately), you can also run several 22AWG runs from this point outward (separately fused for the 22AWG wire) depending on the total load as long as the total current on the 16AWG section doesn't exceed its limit, the total current on each 22AWG section doesn't exceed its limit and you have sufficient voltage remaining to power your end device. In practice, if you are just powering one device, or you won't exceed the ampacity of the thinnest wire in the system, you could use a single fuse to protect the whole set - i.e. you use a 5A fuse with the whole system when 16AWG can handle up to 19A and 22AWG can handle up to 8A, this is always going to be safe, but on the downside, it means you might not be able to make the best use of the wire in case you might want to add loads later. Regardless, I doubt that the ampacity will cause a problem - it's more likely that the resistance in your wires will make it hard to high currents for run long distances especially at low voltages. Even to push 5A through a distance of 20m will require 9AWG or thicker according to this Voltage Drop Calculator , so to avoid losses, it's best to use as thick of a cable as possible to run the maximum distance. Otherwise, in specialist scenarios, it is possible to increase voltage and then decrease it again later, to reduce transmission losses but the conversion will itself introduce losses. - Gough

jw0752 · Answer

Hi Mao, It sounds like you have an interesting project. Can you tell us what the current draw of the LED module is? Also how far is the long distance that you refer to? One thing that you can do is to try the experiment of hooking it up. Measure the voltage at your power supply and then measure the voltage at the LED module. If you are still getting enough voltage at the module for proper operation then you are good. The lower resistance of the 16 gauge wire will make it possible to put the power supply further away and still have a good voltage at the LED module. In any case there is no harm in trying. John

jw0752 · Answer

Hi Mao, You can always trust the voltage drop and the current to tell you what is happening. You said that you had a 7% voltage drop at the load end. If you began with 12 volts this would mean a drop of 0.84 Volts. You said that the current in the wire was 2 amps. This means that the power loss in the wire would be Current times Voltage Drop or 2A times .84 Volts which equals 1.6 Watts. 1.6 Watts spread out over 10 meters of wire will not be even noticeable. Based on what you have described this would be a safe hookup with no danger. If you want it to be even safer you can put a 3 Amp fuse in the circuit so that if there was ever a short the fuse will blow before the circuit is subjected to high amperage. John

colporteur · Answer

ML, I have attached a rough drawing to assist in my explanation of what you are trying to analyse. In my example there is a signal source on the left feeding a connection of three pieces of wire (A,B &C). Each piece of wire has physical characteristics that impact it electrical properties (i.e. it ability to pass signal). Of the three electrical properties resistance, capacitance and inductance, I will speak to resistance. For this discussion lets keep all the wires the same type of metal. Again different metal different electrical properties. The smaller the gauge of wire (i.e. AWG 8 vs AWG 10) the larger the conductors, the lower the resistance. The gauge size compared to resistance values being opposite can be confusing. The resistance of the conductor, limits the amount of current the conductor can carry. If you attempt to move more current through the conductor that fessible the metal will heat up (because of resistance) and may melt the insulation. The current flowing through the conductor resistance causes some of the signal (voltage) to be lost. We start with a signal at the source but depending on the electrical properties of the conductor the signal at J1 will be reduced and reduced again further at J2 & J3. Let's examine this scenario. The first conductor A is large, with B being smaller and C being smaller still. The limiting factor for the circuit is the size of the smallest conductor. The initial conductors may handle the current but if the smallest can't it become a fuse and will open from overheating. Let's look at the signal. You start with an supply value at the source, that you want to arrive at the destination. As the signal passes through each conductor it is getting reduced. The amount of reduction is dependent on the resistance. If the conductors are long, and the gauge makes the conductors small, there is a lot of resistance for the signal to pass through. A perfect example of resistance at work is the extension cord. Most people purchase an extension cord based on price. Wow, I got a 50ft extension cord for $15. What people fail to look at is the gauge of the cord. 50ft of AWG 18 has more resistance than 50ft AWG 14. Rarely do 50ft 12AWG extension cords go on sale. The consumer runs their weed wacker on an extension cord of 50ft but a small conductor. That is hard on the motor and the cord. The voltage at the end of the weed wacker is going to be reduced because of the current flowing through the resistance of the smaller 18AWG wire vs 14AWG wire. If each piece of wire was a pipe of varying size, the fluid that flows out the end is limited by the smallest pipe. That is why the feed pipes to the house are larger than the pipes that go to each of the taps. If not when you flush the toilet, the shower pressure drop since water has to come from some where if the feed is the same diameter as the taps. The electrical properties discussed centre around the equation for ohms laws.

michaelkellett · Answer

What matters (mostly) is how much current you intend to pass down the wire.

How can you connect a solar panel to a generator - does it have batteries and a place to connect a solar panel ?

It should have instructions that tell you about current ratings.

If not then post some more details (pictures of the generator and panel) maybe and it should be possible to work it out.

MK

Connect 2 different wire gauge; is the smaller gauge affecting the capacity of the bigger wire gauge?

Top Replies