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I want to operate a 1-inch length of nichrome 80 wire, 26 gauge, at 200-600 degrees Farenheit. This type of wire has 2.657 Ohms of resistance per foot. So, for 1", there would be 0.22 Ohms of resistance in the wire. According to my amperage chart I will need between about 1 amp and 2 amps to operate in my desired temperature range.

The problem is that according to my calculations I would have to run at 0.5 Volts if using 2 amps and 0.25 Volts if using 1 amp. This seems like a very small voltage to me.

For example, my power supply, a standard benchtop supply has 0-3 amps and 0-50 volts. The Voltage meter on my power supply has gradations in 2 volts. In other words between 0 and 10 there are 5 tick marks. So, to run at 0.25 Volts, for example, I would have to have the gauge at 1/8th of tick mark--a tiny amount on the gauge. It seems like if I just nudged the Voltage knob a tad too much I could blow out the wire.

I am doing something wrong here? Do I need some kind of special, ultra-low voltage power supply, or are my calculations wrong in some way?

What if I put a resistor in series with the wire? That would increase the voltage needed, but I would still need to operate at a very exact voltage, right? For example, if I added a 5-Ohm resistor, then operating at 1 Amp would seem to require 5-Volts, a more normal voltage, but once again am I risking a burnout if I nudge the know slightly too far?

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    \$\begingroup\$ A constant current supply will allow you to set current as desired |Tryintg to use any system at a tiny fraction of full output with mechanical controls will always be hard and usually not sensible. If variable vltage HD been what you want then making something to provide it would be easy and sensible. \$\endgroup\$ – Russell McMahon May 13 '15 at 4:52
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    \$\begingroup\$ Your supplied data is inconsistent and neither of the possible interpretations or other stated facts seem right. || If 1 foot = 0.2657 Ohms as you say then the 1 " figure is wrong. As 2.657/12 ~+ 0.22 Ohms I'll assume the 0.22 Ohms/inch is correct. Use 1 A to start I^2 R = power = 1 x 0.22 = 220 mW. In a 1 inch bare wire in air I'd be surprised if you got it more than warm. For 4A you get 880 mW which is still suspect. | Adding a series resistor will help but a variable current supply helps more. \$\endgroup\$ – Russell McMahon May 13 '15 at 4:58
  • \$\begingroup\$ Note that whatever that one inch of wire is attached to will draw the heat off the wire (as well as getting hot also). So, you'll have a hot spot in the center of the wire with cooler regions at each end. \$\endgroup\$ – gbarry May 13 '15 at 5:49
  • \$\begingroup\$ Did you mean 0.022 Ohms per inch, and 0.025 volts? \$\endgroup\$ – Reversed Engineer May 13 '15 at 6:12
  • \$\begingroup\$ There was a typo in the original post reading 0.2657 Ohms per foot, but it is actually 2.657 Ohms per foot. The post has been corrected. \$\endgroup\$ – Tyler Durden May 13 '15 at 10:21
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Get a power supply that has a current limit, and drive the wire in current mode. The voltage will end up at whatever it needs to be for the current you want. All you need to do is set the voltage high enough so that the power supply hits the current limit first. I would not recommend putting a resistor in series, it would dissipate a huge amount of heat. You probably want to make sure you're using very thick wires as well so that they don't dissipate too much power.

Another possible solution to consider is getting a longer piece of nichrome and folding it back on itself a few times or twisting it into a narrow coil. The idea is to increase the length of the wire (and the the overall resistance) so that you don't have to use such a large current.

Also, you're off by a factor of 10 somewhere, 0.2657 ohms per foot would be 0.022 ohms for the wire. Which is quite low.

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