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I am using a 26AWG wire for an application. As per specifications a 26 AWG wire can carry, 2Amps max. Supposing I connect a load of 3Amps. For how long will tis be sustained by the said wire ?

The current is applied in pulses in this application, of 25 mSec max.

I mean is there any way I can calculate the time taken before the wire burns out ?

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  • \$\begingroup\$ Is the wire insulated by some plastic or is it naked? \$\endgroup\$ Commented Jul 16, 2017 at 4:09
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    \$\begingroup\$ Short pulses in large distance might be no problem, but pulses of 25 ms with a gap of only 5 ms will be. Calculation of the time to burn out is very difficult and requires differential equations. \$\endgroup\$
    – Uwe
    Commented Jul 16, 2017 at 10:35
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    \$\begingroup\$ The current applies in 25 ms pulses... but how frequently your pulses are coming? \$\endgroup\$ Commented Jul 16, 2017 at 15:38
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    \$\begingroup\$ @Board-Man just don't read those long blankets of overcleaver answers. If you are doing an experiment in a lab- just go for it. If you are working on an industrial system- then you just can't afford a risk. \$\endgroup\$
    – user76844
    Commented Jul 16, 2017 at 17:43
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    \$\begingroup\$ A wire is basically a resistor. The peak current is less important than the average current, which you haven't specified. \$\endgroup\$
    – Dave Tweed
    Commented Jul 16, 2017 at 17:47

4 Answers 4

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It is nice to see all theoretical considerations and suggestions for experiments to determine thermal constant of insulated 26 gauge wire. Of course the precise result will depend on ambient condition, whether it is in still air or under some airflow, etc. However, all this work already has been done in Electrical Engineering, and results are well documented. For practical considerations I would suggest the following Wikipedia page.

From this page, the 26 AWG wire will burn up at about 20 A in 10 seconds, and hold up to 218 A for 32 ms.

So, to answer the direct OP questions, there is nothing to worry about 3 A for a surge of 25 ms long.

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  • \$\begingroup\$ How about an estimate of length effect, where short wires can dump heat out the ends? \$\endgroup\$ Commented Jul 16, 2017 at 21:06
  • \$\begingroup\$ @analogsystemsrf, if the wire is short and has good heat sinks, the current carrying capability of wire segment will be even better, so less worries. \$\endgroup\$ Commented Jul 16, 2017 at 21:47
  • \$\begingroup\$ @AliChen. The surge is for 25msecs. But it is periodic at every 200msecs or so. How will this impact the performance ? Also, this will impact the temperature as well, right ? How can I model that ? Likewise the length of the wire ? \$\endgroup\$
    – Board-Man
    Commented Jul 17, 2017 at 4:36
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    \$\begingroup\$ @Board-Man What to model? The thermal time constant of your wire is in a range of seconds, see nepsi.com/resources/calculators/… , which is much longer than 0.2s period of your process. So the wire will average your current. 3 A over 25/200 duty cycle is just 375 mA on average, so the temperature rise will be only about 1/5th of 10C, or about 2C. \$\endgroup\$ Commented Jul 17, 2017 at 5:19
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The maximum current you can pass through a wire continuously depends on the maximum temperature that it's permitted to reach.

That equation usually involves so many variables that it's easier to measure it than to try to calculate it. Usually, we simply go from guidelines with a good margin of error.

In the steady state, the wire gets hot enough to be able to dissipate all the energy that's arriving. It loses heat by convection to the air around it, conduction to its supports, and by radiation.

A short wire loses most of its heat to the supports, the centre section of a long wire has to convect most of its heat away. Radiation makes up little of the heat loss, except at very high temperatures such as a bulb filament.

The maximum temperature of a plastic coated wire is usually limited by the softening point of the plastic, usually 70C for PVC. A bare wire can run hotter, but not so hot that it melts its solder attachment, or oxidises in the air. Remember that the temperature rise is above ambient, so the ambient temperature has to be taken into account as well.

No one guideline can handle all those complexities, and your max temperature rise might be different from those of any particular guideline.

In practice, at room temperature, and if you're not trying to get stuff UL rated so you can sell it, you may well not see any ill effects with 3A where the general guidelines say 2A.

But remember, a continuous 3A gives 2.25x the dissipation of 2A (power goes as \$I^2\$), so more than twice the temperature rise that the guideline is assuming. You may not burn the insulation, but it may soften to the point where it loses mechanical strength, and something can push through the insulation and cause a short circuit.

As the main limitation for low current wires like this is thermal, if you stick to 2A RMS, then you will get the same heating effect as 2A DC. If you pass pulses of current through the wire, you can compute the RMS or heating effect as the duty cycle * current^2. So a 3A pulse with a 4/9ths = 45% duty cycle is about 2A RMS. Similarly 4A pulses at 25%, 10A pulses at 2% duty cycle, all have the same heating effect as 2A DC.

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  • \$\begingroup\$ Thank you. I will be passing the 3Amps in the form of surges. These surges will last only for about a max of 25mSecs. So, will this be ok ? Will the plastic/insulator be as rigid ? Canyou please point me to any application note with the formulaes please ? \$\endgroup\$
    – Board-Man
    Commented Jul 16, 2017 at 7:53
  • \$\begingroup\$ @Board-Man answered in the question \$\endgroup\$
    – Neil_UK
    Commented Jul 16, 2017 at 9:57
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If there is no plastic insulation, then you can use some Junction Temperature Equations \$T_j = T_a + R_{tj}*current*current\$, and you would only need to find out \$R_{tj}\$. Keep in mind that different length's of wire give different \$R_{tj}\$. So measure the one you will be using.

\$T_j\$ = The temperature you heat up to
\$T_A\$ = Room temperature
\$R_{tj}\$ = Thermal resistance of the wire.. which sounds very weird..

In other words, pass through 2 amps, measure the temperature after 10 minutes so it has settled down, The \$R_{tj}\$ would then be equal to \$\frac{measured temp - T_A}{2*2}\$. The measurement should be on a naked copper wire.

It won't be perfect (because the resistance of the wire changes with the temperature), also you will use some plastic insulation which will thermally insulate the wire like jacket so it will get hotter (imagine a dusty heatsink). But it will give you a good approximation.

And... I'm no expert, but in my opinion, heating something with constant current from 10 degrees to 20 degrees goes faster than 110 to 120, because at 110 degrees the temperature bleeds out. So I would make use of a capacitor-discharging approximation. The same way you charge a capacitor, it goes slower and slower until it reaches its final voltage.

So I'll use two approximations, which are just my gut feeling, it will probably put you in the ball park at least. Right now I'm using witch craft.

The voltage across a capacitor with a resistor at its terminals bleeds out according to this equation:

\$V(t) = V_0e^{-\frac{t}{RC}}\$

apply witchcraft and identify the temperatures

\$T(t) = T_je^{-\frac{t}{K}}\$

\$ln(\frac{T(t)}{Tj}) = -\frac{t}{K}\$

\$K = \frac{-t}{ln(\frac{T(t)}{Tj})}\$

\$T_j\$ = The temperature you heat up to
\$T(t)\$ = Temperature at time t since you stopped heating it up
\$t\$ = time t since you stopped heating it up

So heat it up to 50 °C, wait until it reaches 40 °C, t would be the time it took for it to go from 50 to 40 °C in seconds.

\$K = \frac{-t}{ln(\frac{40}{50})}\$

Once we've acquired the K value and the \$R_{tj}\$ value, we're done with the measuring. And the capacitor equation is for discharging, let's change it to charging and get the time it takes until it reaches some dangerous temperature. I'd say 100 °C is dangerous. Paper self ignites at 220°C or is it 210.. whatever. Also your insulation starts melting around 100-200°C.

So, let's go back to this equation:

\$T(t) = T_je^{-\frac{t}{K}}\$

If I'm charging it up, then it's like this:

\$T(t) = T_j(1-e^{-\frac{t}{K}})\$

Now we want to know the t until it reaches 100 °C.

\$100 = T_j(1-e^{-\frac{t}{K}})\$

\$t = -K*ln(1-\frac{100}{T_j}) \$

So say that you measured K to be 50 and calculated \$T_j\$ to 500°C, then it would take \$-50*ln(1-\frac{100}{500})\$ = 11.15 seconds

Keep in mind, I used two approximations that are very approximative, they will put you within the ballpark, they are nowhere near perfect or exact.

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If you have a very short wire, with lots of copper foil at each end, or big terminal blocks with large screws and pieces of chrome-plated steel to remove the heat, then this next analysis is useful.

AWG 26 has 0.129 square milliMeters crosssection. Lets examine that for its heat-removal behavior.

Standard copper foil ---- 1ounce per square foot ---- moves 1 watt at 70 degree temperature difference, per square of foil. That foil can be any size square.

Lets convert that round wire into a flat piece of foil. Standard foil is 35 microns thick. We have wire area of 0.129 mm^2 or 0.129 * 1000micron * 1000 micron, or 129,000 square microns. Dividing 129,000 by 35, we find 4,000 microns is the flattened foil equivalent (ok 3,680 microns, or 3.68 milliMeters). Just call it 4mm.

We now have a useful number: 4mm of wire will have 70 degreeC of temperature drop, if one watt of heat is flowing.

Resistance of #26 is 0.133 ohms per meter. At 2 amps, the heat per meter is (using I^2 * R) 2 * 2 * 0.133 = 4 * 0.133 = 0.53 watts.

One meter of wire contains 250 regions of size 4mm. The heat can flow left and right to the two ends of the wire, with 0.27 watts of heat flowing left & right.

At the extreme ends, right at the heatsinks, each 4mm adds 70 degreeC * 0.27 = ~~ 20 degreeC temperature rise. Next 4mm adds almost 20degreeC, tapering off to zero added heat right at middle of the 1 meter of wire.

With 125 sections of 4mm, the temperature rise will be 125 /2 * 20 degreeC, or 60 * 20 = 1,200 degreeC rise.

Again, this assume the heat can only flow ALONG THE WIRE.

Good news! If your wire is only 2" long (50 mm, or 12 sections of 4mm, or 6 + 6 sections left and right, with that linear taper buying 50% less heating, equivalent to only 3 sections, the rise will be 20 degreeC * 3 = 60 degreeC.

So................if your wire is only 2" long, and well heatsinked at the ends, you are safe.

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    \$\begingroup\$ -1. This is an eclectic mix of wrong ideas. 1. Copper foil ability to dissipate heat strongly depends on its orientation relative to gravity field (if it is in still air), and on color of coating, etc., and will be different under forced air flow; 2. You can't equate the round wire with flat pieces of foil - internal layers are not exposed to the same heat transfer conditions; 3. The calculated 1200 C rise is shear nonsense - if the 26 AWG wire can carry 2 A, it means the standard 10 C rise, by definition. Heat doesn't flow only along the wire. I would seriously suggest to delete this answer. \$\endgroup\$ Commented Jul 16, 2017 at 19:38
  • \$\begingroup\$ As I clearly stated in my answer, I am modeling only heat flows along the wire, to heatsinks at the two ends. Respond to that, OK? None of the other answers put a safe LENGTH value out to the OP. \$\endgroup\$ Commented Jul 16, 2017 at 21:04
  • \$\begingroup\$ The result of your model (1200 degreeC rise at 2 A over 26 AWG copper) is in clear disagreement with reality. Your model assumes absolutely unrealistic conditions of total thermal isolation of wire from surroundings. Even the energy sink via thermal radiation in total vacuum will stabilize the wire temperature at much lower level. \$\endgroup\$ Commented Jul 16, 2017 at 21:57
  • \$\begingroup\$ And read the answer----the sentence immediately following states "assumes heat flows only along the wire". Then I analyze a 2" piece. No one else provided a length effect. \$\endgroup\$ Commented Jul 17, 2017 at 3:46

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