Maximum current through 30AWG Kynar wire - Page 1
Forum section: electronics / beginners

author: mohf
author_rank: Contributor
timestamp: March 22, 2012, 12:16:01 am
content: Hello guys!I've got a simple question. I want to know what the maximum current is that can go through 30AWG wire. Searching on google does not give me a clear answer, and it's unclear whether they are talking about currentdissipatedby the wire OR current that goesthroughthe wire. I hope someone can explain/clear this up.Suppose for example, I have phone that wants to charge the Li-ion battery. (it's actually an iPhone with a broken battery connector, but... whatever) The phone puts 4.3V 1A out for about 1 hour. R=U/I so Rbat=4.3 ohms. (if we neglect contact/wire resistance)Lets say we connect 2x 0.25m (+ to + and - to -) Kynar 30AWG wire between the charger and battery, so total lenght =0.5m.http://en.wikipedia.org/wiki/American_wire_gaugeAccording to wikipedia, the wire resistance is aprox 0.34 ohms p/m, so Rwire = 0.17ohms.The current 1A that is dissipated by the wire should thus be: 0.17/4.3*1 = 40mA which SHOULD be safe for the wire. But the 1A that is going THROUGH it is not safe.Please note, I am not a student at the university or like that, I am just a 18-year scholar who likes electronics, and just recently with lots and lots of trial and error and efforts got a (nothing more than a functional) calculator from a PIC16f876A and LCD (that calculates from 0 to 255).

author: Psi
author_rank: Super Contributor
timestamp: March 22, 2012, 12:40:54 am
content: Here's a good site for the maximum current before insulation starts to meltIt even lists your Kynar wire type.http://www.prototypemachining.ca/Articles/WireCapacityCharts.htmlIn 30deg C ambient air, your AWG30 cable with Kynar insulation will melt (125C) at 3.6APretty much all 'wire size vs current' tables will calculated for cable temperature. (energy burnt off in the cable as heat)The energy you get feed into a cable which makes it out the other end doesn't really matter except for the voltage drop which may or may not effect your circuit.Obviously a higher input voltage means less percentage of the voltage is lost.

author: IanB
author_rank: Super Contributor
timestamp: March 22, 2012, 12:54:14 am
content: Quote from: Psi on March 22, 2012, 12:40:54 amI think this is when the insulation will melt. If you are careful, you can probably pass a lot more current than that before the wire itself melts. But it might be a bit messy and odorous.To the OP, what you usually care about is voltage drop. Voltage drop equals current times resistance, so if you pass 1 A through a wire with total resistance of 0.17 ohms the voltage drop will be 1 x 0.17 = 0.17 V. The voltage at the end of the wire will be 0.17 V lower than the voltage at the supply. For most purposes this probably won't matter too much.

author: Psi
author_rank: Super Contributor
timestamp: March 22, 2012, 12:58:24 am
content: Yeah, that's what i meant to say, it's when the insulation will melt.

author: vk6zgo
author_rank: Super Contributor
timestamp: March 22, 2012, 01:20:04 am
content: You are a bit confused!The current of 1amp & the resistance of 0.17 Ohms are all you have to work with.What will cook your wire is Power in Watts.(Dissipationis measured in Watts,not amps!!)The formula for this (given I & R) is P= I^2*ROr in your case: P=1^2*0.17But,1 squared =1,so we have: P= 1*0.17 ,so thedissipationis 0.17 Watt,or 170 mWFuse,& wire makers do not specify their wire in power ratings however,but,rather,in current ratings,so to answer your original question,Kynar 30AWG wire,should handle 1A without fusing,& without even getting very warm--170mW isn't a lot of power.The formula you used in the following sentence does not make sense:The current 1A that is dissipated by the wire should thus be: 0.17/4.3*1 = 40mA which SHOULD be safe for the wire. But the 1A that is going THROUGH it is not safeFirstly,you have used the reciprocal of the Current form of Ohm's Law, which should be I=V/RSecondly,it does not apply anyway,as the 0.17 Ohms is only part of a series circuit which is being fed by the 4.3 volt source.AGAIN,you have already specified the current which will flow in the circuit as 1 amp,that's the only current!

author: amyk
author_rank: Super Contributor
timestamp: March 22, 2012, 11:20:32 am
content: The main limiting factor is actually the thermal resistance of the wire to its environment, which is influenced by its surface area. The dissipated power, through resistance, contributes to heating the wire, and its environment, and a temperature equilibrium is reached once the wire attains a temperature where the power entering the environment is equal to the resistive power dissipation.You can see that the tables assume a particular (fixed!) air temperature. This is because theoretically an infinite amount of current could pass through the thinnest wire (ignoring effects such as electromigration) if it could be maintained at a sufficiently low temperature.

author: Black_Moons
author_rank: Newbie
timestamp: March 24, 2012, 04:00:22 pm
content: mohf: in series all current passes through all elements of a circuit, Hence if your load uses 1 amp, your wiring also conducts 1 amp. There is no 'dissipated current' in a series circuit.If your wire is 0.17ohms, it drops (1 amp / 0.17ohms in voltage or) 0.17v. Its the voltage times that 1 amp that makes the power that is dissipated in the wire.http://www.prototypemachining.ca/Articles/WireCapacityCharts.htmlGives you the actual melting point of the insulation in amps, assuming unbundled, free air, 30c ambient temperature. Bundling, insulation and higher ambient temperatures will decrease the allowable current before the insulation on the wire melts.For short runs it is often the wire insulation temp that is critical and for longer runs it is often the voltage drop that is critical. Both should always be considered however.