# Heating wire filament with low voltage / amps

Is it possible to heat either nichrome or copper wire (both 40 awg) with the following low-power delivery of 3.5 Volts, 7.4mW and 2.11mA?

This is very low power delivery and we're wondering if 3 inches of coiled wire (40 awg at .0031 inch diameter) could be heated to speed the evaporation of certian liquids?

Would a high resistant wire (nichrome) or a low resistant wire (copper) get hotter at low power? How hot would they get?

According to the wire resistance calculator found here http://www.cirris.com/testing/resistance/wire.html

The resistance of 3" of 40awg copper wire would be 0.262 ohms or just over 1 ohm per foot.

If you hook that to a 3.5V power supply, it will draw over 13 amps, not the 2.11mA you mention. You would need in the neighborhood of 1600 FEET of 40 awg copper wire before your current draw would approach 2mA

40 awg Nichrome wire is approximately 70ohm/foot according to this page http://hotwirefoamcutterinfo.com/_NiChromeData.html You would only need about 23 feet of this, to achieve 2mA of current from your 3.5V supply.

Your numbers are way out of the ballpark.

For a given dissipation (wattage), how hot the filament depends on how quickly heat is removed from it, which depends on the ambient temperature, surface area, emissivity, convection.

It might not matter how hot the filament gets. The wattage alone tells you how much energy it is dissipating into the liquid. The effect of heating a liquid can be estimated from its heat capacity and latent heat of evaporation.

• The current is constant up to 30 minutes. The liquid is actually a thin coating on the wire of volatile oils intended such to speed diffusion when the wire is heated. What we still don't know is if the low power will materially speed evaporation? – Greg May 7 '13 at 21:09

When you say your power supply is "3.5 Volts, 7.4mW and 2.11mA", what this means is that your power supply will provide 3.5V and a maximum current of 2.11mA (multiply both and you'll get its power). This means that you shouldn't connect it to something that'll draw more than 2.11mA at 3.5V: for example, a resistor of less than 1659 ohms. If you do, bad things will happen to the power supply (exactly which bad thing depends on your power supply).

So your wire must have a resistance of at least 1659 ohms. Lower than that and you could fry the power supply by demanding to much current. Much higher than that, though, and the current you'll draw will be much lower than 2.11mA, and so you'll get less power to heat your wire.

I can't really tell much more without details about your setup, but I can say that: if you want to achieve a high temperature in a wire with such a measly power supply, you are better off with thin wires of high resistance, because these will result in less metal (shorter and thinner wires) in order to get a 1659 ohm wire, and then you'll have less metal to heat with your 7.4mW of power.

• Thanks very much. Can you explain the calculation you used to derive 1659 ohms? – Greg May 8 '13 at 22:19
• Ohm's law: which resistance would let through 2.11mA at 3.5V? – FrancoVS May 9 '13 at 6:00

If you have a limit in the power supply, the same upper limit apply to the heat you can produce. So whatever you choose the wire resistance you can't do better than the small power of 7.4 mW that probably can't help to evaporation of visible amount of any liquid. With the data in your example, to sink 2.11 mA from your power supply at 3.5 v, you have to apply a load resistance of 1.666 kOhm. By the way, creating such a resistor out of nichrome or copper is absolutely a bad idea, since you would have either a too much long wire or a too thin one making things difficult to practical handling.

• Sorry, can you explain how to "apply a load resistance of 1.6666 kOhm" and further describe? – Greg May 7 '13 at 21:13

## protected by W5VO♦May 7 '13 at 21:41

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