# Proper Voltage for small Nichrome heating element.



## AllanJ (Nov 24, 2007)

As the wire heats up its resistance increases. Since you took an arbitrary length of wire, you have to do some electrical engineering to figure out the proper voltage to feed it so it gives off a reasonable amount of heat and does not burn out like an incandescent lamp filament.

You can look for and analyze information on the gauge of the Nichrome wire versus its resistance versus temperature. 

At all times the current flowing through any portion of a circuit equals the voltage applied divided by the resistance. (Or voltage equals current times resistance.) This is "Ohm's Law".

Or you could just use trial and error increasing the voltage until the element turns orange (not yellow)and being satisfied with that level of heating.

Ideally you want a length of Nichrome wire such that 120 volts (no transformer or rheostat needed) gives the desired heating. 

Do not use "potentiometers" such as are used for volume controls in audio equipment. They only handle low voltages and milliamperes of current, and will burn out when used with what you are doing here. Also, power resistors sold in electronics stores are not what you want. What they do is reduce voltage by turning electrical power into heat and the whole purpose of this exercise is to use the Nichrome element and not other components to generate the heat. Furthermore the typical power resistors sold in consumer electronics stores such as Radio Shack will either burn out or not pass sufficient current to heat up the Nichrome wire.


----------



## Yoyizit (Jul 11, 2008)

A 10A, 120v toaster has a resistance of 12 ohms. If you want the same heating per inch of wire I guess you need to apply (0.8/12)120 = 8v, assuming your wire has the same cross-sectional area as toaster wire.

Try a 12v car battery and scale your power up or down from there. A continuous 10A draw rules out using smaller batteries.

If needed, you can make a voltage dropping resistor with #26 AWG; 10' = 0.4 ohms.
Iron [coat-hanger wire] has 6x the resistance of copper but the gauge is a bit heavy for this application.

Once you know your required power per inch of wire, the rest of the design should follow easily.


----------



## Dacr0n (Jan 16, 2009)

Thanks for the reply!


I had a hunch that it needed to be lower voltage.

So lets say I go to the salvation army and find a small 12V AC to AC power supply..

I would imagine that that can only pump out like 500 mAh.

That would be insufficient correct? I would probably need a transformer that gives me like 8-12 VAC and can handle at least 10 amps?


So lets say that I did get a nice powerful 12VAC transformer...

I know that AllanJ said not to use a POT, at least the volume type.

Would it be okay to use the household light dimmer type once I get the voltage to 12VAC?

Thanks again for all the help.


----------



## Dacr0n (Jan 16, 2009)

Here is some info on the wire.

Approximate WATTS, AMPS calculation at 120 volts :
10 Feet generates 1700 WATTS 15 AMPS 
12 Feet generates 1500 WATTS 12 AMPS
16 Feet generates 1000 WATTS 9 AMPS
20 FEET generates 867 WATTS 7 AMPS



So if I use .... like an 8 inch piece.... @ 12VAC...???

I don't know which forumla to apply



Watts is how much power a given length of wire will use right or output?


----------



## Yoyizit (Jul 11, 2008)

So 1700/15=113V (~120v), and 7.5 ohms.

113x8"/[(12)(10)] = 7.5v at 15A = 113w and 0.5 ohm.

You may want to parallel and properly phase a bunch of power xformers from Hosfelt.com or allelectronics or Mouser or Digikey or Jameco. 

Probably you can gang the shafts of three or four lamp dimmers to control this beast. You'll need each dimmer is series with a skinny wire to ensure current sharing.
Another way to control it is to use a 555 timer and power MOSFET (on a heatsink); there are electronics' forums who already have circuits that will do this.

If you want limited control you could put a skinny wire in series with the heater element and tap off on the wire for the desired voltage. 

There is no shock hazard with this low voltage.

If you only want 100w of heat use an incand. lamp.


----------



## Dacr0n (Jan 16, 2009)

Thanks yoyizit....


So ideally I would want a transformer that dropped 120VAC to like 8.0VAC and it needs to supply 15 amps? is that right?

90% of all transformers I find on the net are 3 amps or less....


Would a 3 amp transformer work?

like a 12VAC 3 amp transformer be okay?

also... is there a name for the specific type of potentiometer that would be well suited for this application... ideally i'd like to use just one.. instead of a bunch of them.


----------



## Yoyizit (Jul 11, 2008)

Dacr0n said:


> Thanks yoyizit....
> 
> 
> So ideally I would want a transformer that dropped 120VAC to like 8.0VAC and it needs to supply 15 amps? is that right?
> ...


You have to parallel 5 ea. of (preferably) identical 3A xformers, properly phased. I'd go for 6.3vac and use a shorter wire. You know, this is not much heat.

There is something here that I haven't mentioned. For reasonable power transfer from the wall outlet to the load, impedances have to respected. Your wire having so many inches per ohm is already constraining what this design will look like.
Once you decide on the watts you want, and knowing that xformers come in only a few standard output voltages, you won't have much wiggle room.

It's like E = IR. Once you decide on any two values, you're stuck with the third value.

One constraint is that your heater wire is 860 milliohms/foot. Another is that its life will probably be shortened at currents above 15A.

We are going to rapidly get into permutations and combinations of paralleling heater element lengths, xformer outputs, and dropping resistor values. You may want to fire up your spreadsheet.

E.g.: 
If you want to use 8", this is (8/12)(0.86) = 0.57 ohm. With a 15A max you can put a max voltage into it of 0.57x15 = 8.6 volts and get 8.6x15 = 129W.
If you want 100w, at 15A this is 100/15=6.7v, which is 6.7/15=0.44 ohms which is (0.44/0.86) x 12 = 6.2".

12v at 3A is 36w, max. and 12/3= 4 ohms. 4/0.86 = 4.7' of wire.


----------



## Yoyizit (Jul 11, 2008)

*Disregard my previous post*

I used to know this stuff! :huh:

Try L = V(1.2)/I

For a V = 12v and an I = 3A, the length L is 4.8".
For a V = 120v and an I = 15A, L is 9.6".
For a V = 120v and an I = 7A, the length is 20.6".


----------

