I am trying to understand if it will be possible for an energy source with an output of 4.7mW at 47% efficiency to power a 35AWG Nichrome wire (think dime sized coil) to 700-800 degrees F.

If it is possible, how many seconds/minutes would it take to heat the coil?

Also would a capacitor be helpful with a 4.7mW output in order to store enough energy to heat the coil?

http://www.cecs.wright.edu/balloon/images/2/22/Nichrome_Wire_Heating_Element_Design_Basics.pdf I initially used the chart above to give me a basis of the amount of energy required to heat Nichrome wire to 700-800F but I am too unfamiliar with conversions to know if this is possible.

  • 1
    \$\begingroup\$ What are you trying to do? \$\endgroup\$ – starblue Jul 18 '13 at 7:55
  • \$\begingroup\$ @starblue's question is very deserving of an answer \$\endgroup\$ – John U Jul 18 '13 at 8:16
  • \$\begingroup\$ What is the ambient temperature you want to operate your device at? Is it room temperature? \$\endgroup\$ – Szymon Bęczkowski Jul 18 '13 at 11:01

Short answer: no.

50% longer answer: yes.

But seriously, how hot the wire will get for a given power input will depend on how well it's insulated. If the insulation is perfect, then there is theoretically no limit to how hot you can make it. Realistically, either at some temperature the heat radiated to the environment equals the power input, and it stops getting hotter, or the wire melts.

I'm not entirely sure what "4.7mW at 47% efficiency" means. 47% of what? Is 4.7mW the power output, or input of the power supply? All that's really relevant in the power output of your power supply, because the power input to the wire will equal exactly this, and 100% of this electrical power will be converted to (what else?) heat.

Anyhow, let's say 4.7mW was the power you put into the wire. How power translates to temperature rise for a particular component is usually expressed as an absolute thermal resistance in units \$K/W\$, or equivalently, \$^\circ C/W\$. It expresses how much temperature rises per unit power input.

So let's say you want a temperature of \$750^\circ F \approx 400^\circ C\$, and ambient temperature is \$20^\circ C\$, thus you need to raise the temperature \$ 400^\circ C - 20^\circ C = 380^\circ C\$, and you need to do it with \$4.7mW\$. The thermal resistance must then be:

$$ \frac{380^\circ C}{4.7mW} \approx 80800 ^\circ C / W $$

or more. That's not physically impossible, but it's some pretty serious insulation. For comparison, a typical thermal resistance to ambient for a typical TO-220 package, which is approximately the same size as you describe, is \$34 ^\circ C / W\$. This table of thermal resistances of common packages from Linear Technology may be helpful to put those numbers in perspective.

You also asked "how long", and again the answer is complex, and will depend on the environment and insulation. The ultimate speed of heating will be limited by heat capacity of your heating element and the rate of energy input (power). This will be offset by the rate at which heat is lost to the environment. But, having already established that you need some pretty incredible insulation to achieve the target temperature in your case, we can skip the rigorous analysis and intuitively know that the high temperature is reached by integrating the power over a long time to accumulate enough energy, and say simply the time required to reach the target temperature is "a long while".

  • \$\begingroup\$ Phil, your answers are always excellent and insightful. I'm going to have to bookmark the "page of recent Phil answers." \$\endgroup\$ – JYelton Jul 18 '13 at 7:48

Well, according to your datasheet, you need 0.72A through 35AWG Nichrome to heat it to 800F (or, as in the note below the table, 0.36A for coiled 35AWG Nichrome).

Also, it has a resistance (following page) of 20 ohms per foot. So... \$P=I^2R=I^2l\cdot20\frac{\Omega}{ft}\$, and hence 4.7mW will heat \$\frac{4.7mW}{0.36A\cdot 0.36A\cdot 20\frac{\Omega}{ft}}\approx0.00034 ft \approx 0.1mm\$ of that wire. So no, you are orders of magnitude away from being able to heat a dime sized coil to 800F.

  • \$\begingroup\$ Thank you for your help (as well as math) us2012. Is there any chance my tiny amount of energy with the aid of a capacitor could store and then output enough energy to reach the needed .036A? \$\endgroup\$ – notEnufEnergy Jul 18 '13 at 0:02
  • \$\begingroup\$ Energy is the primary issue here, but not the primary design concern. The current through your wire is mainly determined by the voltage applied. Theoretically, if your "4.7mW source" has high enough voltage, you could construct a circuit that would charge a large capacitor bank for a fairly long time and then discharge that to heat up the wire for a fairly short time. Not trivial though, and probably defeats the point of what you're trying to do. \$\endgroup\$ – us2012 Jul 18 '13 at 0:20

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