Conductor current rating - Page 1
Forum section: electronics / projects

author: akis
author_rank: Frequent Contributor
timestamp: October 13, 2014, 11:21:17 am
content: How is the current rating of a conductor defined? For example if the current rating of a conductor is 10A, what exactly does that mean, how was it calculated ?The way I understand it, a wire has some resistance per length, passing current through a wire will generate heat on that resistance, depending on the length of the wire, the heat generation and the heat dissipation to ambient without a heatsink, there will come a point where the wire will get hot, and at some temperature it will melt its insulator jacket, and in extreme cases it will melt the solder joints, and maybe even melt itself and snap (like a fuse).A short segment of wire has a smaller resistance than a longer segment, however the longer segment will exchange more heat with the environment, so is the relationship between wire length, wire heat generation and wire heat emission linear?

author: rs20
author_rank: Super Contributor
timestamp: October 13, 2014, 11:22:44 am
content: http://en.wikipedia.org/wiki/Ampacity

author: Richard Head
author_rank: Frequent Contributor
timestamp: October 13, 2014, 01:41:43 pm
content: There are a lot of factors that determine the current carrying capacity of a cable.They are:1. Ambient air temperature2. Type of insulation (PVC/silicone/Tefzel etc)3. Thickness of insulation.4. Cross sectional area of cable.5. Geometry (shape) of conductor.6. Where the cable is located. (ie In a cable tray with many other cables/ Underground/ paired with several other current carrying cables.)As the conductor becomes larger in cross section its ablity to conduct heat away from it diminishes. This is due to the surface area to cross sectional area ratio decreasing.It has a huge affect and that's why one cannot simply use a current density as a rule-of-thumb.

author: tarun172
author_rank: Regular Contributor
timestamp: October 13, 2014, 01:58:29 pm
content: I suggest you look at Advanced Circuits website for more information on this subject. See link here:http://www.4pcb.com/trace-width-calculator.html

author: akis
author_rank: Frequent Contributor
timestamp: October 15, 2014, 08:34:23 am
content: Thanks for good suggestions.

author: rs20
author_rank: Super Contributor
timestamp: October 15, 2014, 10:58:17 pm
content: Quote from: 3roomlab on October 15, 2014, 10:41:16 amYou can easily pass an amp through a trace and measure the voltage, calculate the resistance, and compare that to the resistance you're expecting for the specified thickness of copper. No-one cares what thickness the copper actually is, they only care what the resistance of a trace is (which is a direct consequence of the thickness).

author: lpc32
author_rank: Frequent Contributor
timestamp: October 15, 2014, 11:06:27 pm
content: How does voltage fit into the current rating?

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 15, 2014, 11:38:49 pm
content: Quote from: lpc32 on October 15, 2014, 11:06:27 pmIt doesn't. Unless you consider the rating of the insulation, or spacing between conductors, which isn't exactly part of the "current rating" posed in your question.A given conductor has no way of knowing or being influenced by the voltage in other conductors, at least as long as the voltage of nearby conductors is low enough, and/or the insulation is robust enough, to avoid arcing. All it knows is how much current is being pushed through it, and/or how much voltage difference there is between one end of the conductor and the other end. Of course, those two quantities are related by the conductor's resistance, using Ohm's law. A wire is just a very long, very low value resistor. Resistors have power ratings (equivalently, they could have voltage ratings as measured from one terminal to the other); this is effectively equivalent to a wire's current rating and/or the voltage drop from one end of the wire to the other. But good luck finding a rating that says what the voltage is allowed to be between a resistor and some other component of the circuit which isn't connected directly to the resistor.If you have a pair of conductors carrying power from a source to a load, they're at different potential with respect to each other. This difference doesn't enter into the current rating of the wire. The insulation of the wire, and/or the spacing between the conductors, has to be able to withstand the voltage, though. You may find a voltage rating stamped on the insulation. If the conductors are PCB traces, there may be applicable guidelines for creepage distances which might become important at higher voltages.So you can push a lot more power through a given pair of conductors at high voltage than you can at low voltage, at least until things start to arc.

author: DanielS
author_rank: Frequent Contributor
timestamp: October 16, 2014, 02:27:21 am
content: Quote from: 3roomlab on October 15, 2014, 10:41:16 amIf the PCB's conductivity is critical for your application, you can integrate a test track in your design to do four-wire/Kelvin resistance measurement. If the resistance is lower than required for the test track's width, then the copper thickness should meet or exceed the requirement. If you have doubts about uniformity, you can add more than one test track around the board or on the panelization.Code:V+ O--+ +--O V-|----------|I+ O--+ +--O I-You pass a known current from I+ to I-, measure the voltage across the sample track, calculate the resistance and compare against the expected value given the track width. You would use the thinnest track width you need the measurement for since the thinnest tracks are usually the most vulnerable to over-etch.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 16, 2014, 05:37:55 pm
content: AG6QR:If the limitation is just power, why is it rated as current?

author: SeanB
author_rank: Super Contributor
timestamp: October 16, 2014, 06:26:39 pm
content: Heating in a cable is solely a function of the current flowing in it, and the ability to withstand arcing over or leakage currents to adjacent structures or ground is strictly voltage.Take your typical soldering gun, which uses a very low voltage, typically under 2V, but with a very high current, in the hundreds of amps, to heat up a piece of steel or copper wire that you use as a soldering tip. There it is only current. Then consider a neon transformer, where it has very low current ( typically 60mA) but a very high voltage ( 14 to 22kV) that breaks down the insulating properties of a gas in a tube to give light.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 16, 2014, 09:34:54 pm
content: Still, why are cables rated for amps while resistors are rated for watts?

author: eneuro
author_rank: Super Contributor
timestamp: October 16, 2014, 10:11:07 pm
content: Quote from: Richard Head on October 13, 2014, 01:41:43 pmThis is not true in the case of... copper pipeThere we have 15mm in diameter 1mm thick copper pipe, so its cross section is about pi*14.5mm=45.5mm2, but is empty inside, so we have huge surface.Additionally, heatsinks can be easy mounted by making 3mm holes and attaching aluminium heatsinks simply riveted to this copper pipe-not shown on this photoThat is why I used such copper pipes in my spot welder as last stage to connect working electrodes.Additionally using pipe conductor helps to overcome transients skin efectsFinite Element Analysis of Voltage and Current Driven Transient Skin Effect Problems

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 16, 2014, 10:46:48 pm
content: Quote from: lpc32 on October 16, 2014, 05:37:55 pmQuote from: lpc32 on October 16, 2014, 09:34:54 pmIn the case of a cable, the limitation isn't just power, it's power per unit length. Longer cables can dissipate more power, because their surface area scales linearly with their length. Note that their resistance scales linearly with length, as well. These two factors cancel out.As an example, imagine cable A is 100 meters long, with a resistance of 1 ohm, capable of 5 amps. The power dissipated is I^2*R, or 25 watts.Cable B is made of the same kind of wire, but it's 200 meters long. Its resistance is double the resistance of cable A, and its power dissipation capability is also double. It has a resistance of 2 ohms. Again, at 5 amps, the power dissipated is I^2*R, or 50 watts. Twice the power in twice the length produces the same amount of heating above ambient. The same current was delivered in each case.Because the wire makers don't know how long you're going to cut their wires when you make a cable, it's easier for everyone if they rate the wires in amps, rather than in power per unit length.Obviously, a given resistor is of fixed size, as opposed to a wire which is cut to length by the installer. That's why resistors can be rated in watts. In fact, the power dissipation capability of a resistor is largely due to the package size and thermal characteristics, so resistors of widely differing resistance values, but the same size package and type of construction, will have nearly the same power handling capability. That's another reason resistors are rated in watts and not by current -- the same power rating can apply to all available resistance values across the series.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 16, 2014, 11:21:42 pm
content: I was considering "per meter" to be implied, so just "watt". But you're right, the implied R increases with length intrinsically so A (/I) is simpler than W/m.By the way, does heat propagation speed matter in this regard?

author: Richard Head
author_rank: Frequent Contributor
timestamp: October 17, 2014, 07:09:15 am
content: Copper tube is fine for high power RF applications such as resonant inductors or induction heaters as the skin effect is significant, but for 50/60Hz applications solid copper is best as the skin depth at 60Hz is huge and generally insignificant.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 19, 2014, 01:49:01 am
content: So wires and resistors were covered. But why are other things rated for amps rather than watts? For example, switches.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: October 19, 2014, 02:57:31 am
content: Quote from: lpc32 on October 19, 2014, 01:49:01 amBecause the switch only "sees" the amps flowing through it. Let's say you have an AC switch rated to work at 120V and 15A, even if you decide to use it with a 12V transformer you're still only allowed to use the switch up to 15A, not 150A.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 19, 2014, 04:53:13 pm
content: Why's that, if heat is the problem?

author: rs20
author_rank: Super Contributor
timestamp: October 20, 2014, 06:46:30 am
content: Quote from: Alex Eisenhut on October 19, 2014, 02:57:31 amA switch "sees" the amps flowing through it when it's on, it "see" the volts across it when it's off, and it sees a funky complicated non-linear combination of the two when it's switching. You'll often see datasheets for switches that specify "This is a 15A switch" in the marketing on the front page, but then quietly say on subsequent pages that it cancarrythat, but notswitch it. Furthermore, the current that can be switched "safely" is a function of the open-circuit voltage (and whether the load is inductive, etc). Definitely read the fine print in the datasheets.

author: Richard Head
author_rank: Frequent Contributor
timestamp: October 20, 2014, 08:08:39 am
content: And don't forget to add that the switch probably cannot break DC above 24-48V as it'll just arc inside and destoy itself. Same with relays when it comes to breaking DC.

author: Neilm
author_rank: Super Contributor
timestamp: October 20, 2014, 08:53:25 pm
content: Sometimes you will see relay manufacturers express the switch rating as power - i.e. volts x current. I've seen that on high voltage (>2kV) relays quite often.

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 21, 2014, 12:04:10 am
content: Quote from: lpc32 on October 19, 2014, 04:53:13 pmHeat is rarely the primary problem with a switch. After all, a switch is ideally designed to be a perfect insulator when it's off, and a perfect conductor when it's on. Either way, an ideal switch dissipates zero power. Real switches aren't quite ideal, but they're usually close enough that they don't get significantly hot in normal steady-state use, at least not if the contacts are clean and untarnished.No, the most common problem with switches has to do with interrupting the current when switched from an "on" state to an "off" state, especially with an inductive load. There's a further possibility of arc if the switch is subject to a voltage beyond its limits, even while remaining in the "off" state. That's why switches have both a voltage limit and a current limit, and sometimes more detailed limits depending on the nature and inductance of the load, and maybe different ratings for AC vs. DC. Read the data sheet for details.But pretend all of that is wrong for a moment. Pretend the only problem with a switch is heat. Then switches would be rated by max current, pretty much as they are now -- it's just that voltage wouldn't enter into the ratings, and you wouldn't have separate ratings for AC vs. DC as you sometimes see.Why would they be rated by current? Because the heat produced while the switch is conducting is I^2*R, where R is the internal resistance of the switch itself. The voltage that would appear across the switch in its "open" state doesn't enter into the picture. The resistance of the switch itself is known by the manufacturer and is a property of the switch. The only parameter which is relevant to heating, and is both known by and under the control of the engineer choosing the switch, is the current.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 21, 2014, 09:29:46 pm
content: I see. So it's not heat. Why's the switching action particularly an issue with inductive loads?But if it were heat, why not use W like resistors? And with a known R, isn't V just as known and under control as A?Yeah, a string of questions.

author: rs20
author_rank: Super Contributor
timestamp: October 22, 2014, 01:15:44 pm
content: Quote from: lpc32 on October 21, 2014, 09:29:46 pmBecause the inductor "tries" to keep the current flowing, which causes voltage spiking --> arcing --> damage when the switch tries to open.Quote from: lpc32 on October 21, 2014, 09:29:46 pmBecause the circuit designer doesn't need to care what R and W is if he/she can just be given an I limit. And it looks like you're confusing open circuit voltage (switch open) with the I*R voltage across the switch (switch closed).

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 22, 2014, 03:25:41 pm
content: Quote from: lpc32 on October 21, 2014, 09:29:46 pmWell, with a known R, it would be, but R isn't typically something you know or care about when selecting a switch. You just want to select a switch with a closed-contact resistance that is very low compared to the load resistance.Suppose you're working in an automotive environment, selecting a switch for some lights. You might know that the lights draw 2.5 amps, and that they are powered by a voltage supply that varies between 12 and 14 volts. What's the R or W of the switch? Who knows?With the way switches are currently rated, you select a switch that is rated for at least 2.5 amps, and at least 14 volts (you might restrict your search to those rated for 5 amps and 24 volts, to get a factor of safety in there).Which of the following switches would work?0.10 ohms, 0.5W0.02 ohms, 0.5W0.25 ohms, 3WDoes it make it easier if I rate those switches as?2A3.4A5AFor extra credit, notice that the ratings of the first type aren't in the same order as the ratings of the second type. Figure out which switch rated the first way corresponds to which switch rated the second way.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 22, 2014, 09:30:32 pm
content: Right, thanks. I was considering a closed switch similar to a resistor, but indeed you don't care, or necessarily know, its resistance.And continuing the little off topic detour: so arcing is expected when closing a switch on capacitive loads, and not at all on resistive?

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 22, 2014, 11:02:25 pm
content: Quote from: lpc32 on October 22, 2014, 09:30:32 pmEvery conductor always has some capacitance and some inductance. So I'm not comfortable with your choice of the words "not at all" above. Maybe "not so much"?The biggest arcing problem is typically when opening a switch that's switching an inductive load. The inductor wants to keep the current flowing, and will generate a voltage spike in an attempt to maintain the current flow.Closing a switch that's switching a capacitive load, you have to consider the inrush current. You should select a switch that can handle that inrush current, and/or take steps to lower the inrush current.Another issue is that closing and opening switches are rarely the smooth operations that you'd like them to be. Because the contacts are dirty, oxidized, and not perfectly smooth, there's almost always some bounce. So the operation of closing a switch or opening it usually involves a whole lot of repeated closings and openings over a short time span. Each of them can involve some arcing.But unless you're going to design switches, you generally don't worry too much about the details of arcing. Just select a switch with the right ratings to handle the current, considering both the steady-state current and any initial inrush due to capacitors charging. Also, if you need to switch a significantly inductive load, you can use an appropriately rated snubber diode to provide a path for the inductor's current without forcing the inductor to put a huge voltage spike across the switch.When in doubt, see the data sheet.

author: lpc32
author_rank: Frequent Contributor
timestamp: October 26, 2014, 05:54:00 pm
content: Thanks.