# Low voltage across low resistance for heating elements

I am planning some low power heating devices in the 10 - 20 watt range and have been considering the usual high resistance alloys: nichrome, kanthal, and stainless steels (304, 316, 630). I have noticed that all of these materials are expensive. Is there some reason I could not just use steel or even aluminum?

I realize that high resistance metals are preferred because for a given voltage you can use either a thicker or shorter wire but, why must i put so much voltage across it? Could I just use cheap steel wire and low voltage? I was planning to use 12 volts before but now I am thinking of building a step down transformer with a 1 volt or even 0.5 volt secondary winding and putting that across a steel or even aluminum wire such that V^2/R is in the 10-20 range.

This got me thinking about who decided what voltage is high or low anyway and why higher voltages are generally preferred. Apparently from the conductor's pov only current matters. While P = VI in general in the context of heat losses in a conductor that is a misleading equation. As is V^2/R. Or is the current density per cross sectional area what really matters? Why is I^2/R the only form that matters here? I am curious about what is really going on in that conductor and what equations describe it quantitatively.

The pro high voltage arguments are usually in the form of high voltage means less current for a given V x I and that means less waste heat in a conductor because again apparently the EMF is only relevant insofar as it increases the current in a given conductor.

But wait. Waste heat is what I want. The more the better. So I am thinking maybe that secondary transformer winding should be at even lower voltage. 1 mV perhaps? Water analogies make me wonder if there is a point at which no current will flow at all at some low level of EMF. How about 1 picovolt across a superconductor? Would that make a good heater?

## Answers and Replies

Related Electrical Engineering News on Phys.org
jrmichler
Mentor
Copper makes a good heater in a soldering gun:

Weller has been making these for over 50 years. The photo above is from Amazon: https://www.amazon.com/dp/B000JEGEC0/?tag=pfamazon01-20. I had a Weller soldering gun at one time. It had a copper tip that failed, so I made a replacement tip from a piece of 12 gauge copper wire. The replacement worked as well as the factory tip.

Power is power, whether high volts and low amps, or low volts and high amps.

#### Attachments

• 111.7 KB Views: 428
osilmag
Gold Member
$$P=I^2 * R$$ is the equation used for power dissipation because it describes the current flowing through the heating element.

Tom.G
Science Advisor
V= I×R, pick any two and you can find the third.
P= V×I, pick any two and you can find the third.

As a result of the 'pick any two' above you can also get:
P= I2×R, pick any two and you can find the third.
P= V2/R, pick any two and you can find the third.

How you optimize the values depends on the individual situation.
You asked about using 1mV. lets see how that works out:
• Using P= V×I, that would work for 10 Watts if you had 10,000Amps available.
• To keep the wires from melting you would need 5/8 inch diameter Copper wires. Kind of hard to work with. And you might want them fatter to keep the temperature well below that of molten Copper.
• Using V= I×R and knowing V= 1mV and I= 10,000A, we use V/I= R. 1mV/10,000A= 0.1microΩ.
• This is a ridiculously low value, the 5/8 inch diameter wire would be 0.024 inches long, about 0.5mm; and that includes the wire that the transformer secondary is wound with.
Now lets try it with 10V:
• Using P= V×I, for 10 Watts we have 10V, and 1A
• To keep the wires from melting you would need 0.003 inch diameter Copper wires (about 40Gauge). Again you might want it fatter to keep the temperature down.
• Using V= I×R and knowing V= 10V and I= 1A, we use V/I= R. 10V/1A= 10Ω
• 40gauge wire has a resistance of 1.05 Ohms per foot. 10/1.05= 9.5ft. of 40gauge wire.
• You could use 22gauge or larger wire to connect your resistor to the transformer without much difference in performance. 22gauge wire resistance is less than 20mΩ per foot. Or you could even use 500ft of 20gauge wire entirely, and get the same result with a much lower wire temperature. (40gauge wire is also maddeningly hard to work with!)

So there you have some of the trade-offs. Hope it helps.

And keep your curiousity going, that's how we learn!

Cheers,
Tom

anorlunda and jim hardy
jim hardy
Science Advisor
Gold Member
2019 Award
Dearly Missed
Boat and airplane builders do exactly what you propose.
They get a big step down transformer , use it to heat a foot or two of wire stretched across a support, and cut foam with the hot wire.

You might experiment with a junked 20 amp battery charger and some baling wire from the hardware store.
I find battery chargers at my friendly metal recycle yard, usually all that's wrong with them is the ammeter is smashed or the battery connection clamps have got broken off . Sometimes the thermal overload has got rusty and is stuck open. Easy pickin's.

old jim

Lets make this more real world. Carbon steel has a resistivity of 17 x 10-8 ohm meters. So lets consider a 1 meter coil of 16 gauge steel. 16 awg has an area of 1.29 sqmm or 0.00000129 sqm. The resistance should be ##\mathrm R = \frac {\rho \cdot l} A = \frac {17 \times 10^{-8} \times 1} {0.00000129} = ~0.1318 ~ \Omega\\I = \frac V R = \frac {0.001} {0.13178} = 7.6 mA\\P = VI = 0.001 \times 0.0076 = 7.6 \mu W##
Well that would not do much. I need 10 - 20 watts. So do I raise the voltage to get more current? Nope. Lets use a thicker and maybe shorter wire such that R = ##\frac {V^2} P = \frac {0.001^2} {10} = 0.1 \mu \Omega##

So how thick does the steel wire need to be at 1 meter length? ## A = \frac {\rho \cdot l} R = \frac {17 \times 10^{-8} \times 1} {0.0000001} = 1.7 sqm ## Wait, what? Ok. So clearly resistance is a big problem. Even if we shorten the length ten times to 4" we still need a 0.17 sqm conductor like a 0.4m x 0.4m x 0.1m piece of solid steel. Haha no way. Well what about copper then? It has a resistivity of only ##1.7 \times 10^{-8}## or 10 times less than steel. I think that just means the cross section would be .017 sqm. This would mean my copper heating element would have to be a 130mm x 130mm x 100mm piece of solid copper. Doable I guess but not very practical. Would be an interesting experiment except that the copper wire connecting to it would create way too much resistance.

I think what I am learning here is to start with a practical but low resistance conductor like copper or aluminium and determine the lowest voltage I can use with that to get my 10 watts. To continue with extremes I will consider a 10mm x 100mm x 100mm copper plate as the heating element .

There will still be the problem of the resistance in the secondary transformer winding and the wire to connect the transformer to the heating element. Unless they are refrigerated superconductors they are going to add a huge amount of resistance killing the current and power even if the transformer is a part of the heater and only a few inches away, but I will ignore that voltage divider problem for now. To be even more extreme I will solder the secondary winding chilled superconductor connector wires to the centers of the two faces of the plate so that the element length is only 10mm. So how much voltage do I need to get 10 watts of heat from that element?

## R = \frac {\rho \times l} A = \frac {1.7 \times 10^{-8} \times 0.010} {0.01} = 1.7 \times 10^{-8} = 170 \mu \Omega\\
P =\frac {V^2} R ~~ V^2 = PR ~~ V = \sqrt {PR} = \sqrt {10 \times 1.7 \times 10^{-8}} = 0.412 mV ##

So due to practical material concerns half a volt is about as low as I could go if I ignore the resistance of the other conductors in the circuit. Lets consider just the copper wire connected to the transformer winding. Lets use 4" and 4" of AWG 6 Copper wire for the hot and neutral of the secondary winding for a total of 8" or 0.2 meters. ## R = \frac {1.7 \times 10^{-8} \times 0.2} {0.0000133} = 0.256 \Omega ##

So even assuming a superconducting transformer this will basically kill the current and power in the circuit. In fact the 170 ##\mu \Omega## for the copper plate can be ignored as it will have almost no effect on the overall current and power. So using the new value for R how much voltage would be needed for 10w? I think it does not matter because almost all of the power would be dissipated in the 6 gauge wire. Those wires might melt before getting enough power to the heating element itself.

So I guess to go with a low voltage low resistance heating element one really has to consider both the wiring to the transformer and the secondary transformer winding as they seem to be the limiting factors to how low you can go instead of the element itself. Its all series resistance voltage divider considerations. Clearly you would want the transformer as close to the element as possible. I was also thinking of those electrically heated gloves and socks. I guess how low you can go with those depends a lot on the internal resistance of the batteries used as well as on the conductor sizes.

Maybe the first thing to do in designing a low voltage heater is to make sure that the resistance of the heating element is much greater than that of the sum of the secondary transformer winding (or battery resistance) and the wires leading to it. Not sure how much greater but maybe 10-20 times as a minimum. I will redo my design with this in mind. Now I want to take apart a soldering gun and measure the resistances there.

Last edited:
anorlunda
Staff Emeritus
There's no need to over complicate. As others said, this is an exercise in trade-offs. Not all of them are electrical.

You want a reasonable voltage, current, strength, size, durability, temperature, stiffness, blah blah. Designers who have considered those things for different applications have selected the material that best fits, sometimes copper, sometimes nichrome. You know the application and the real electric plus non-electric requirements, not us.

I wrote half a volt for the theoretical minimum voltage across the copper plate. I should have said half a millivolt. I need to get back into transformer design before I can come up with a more accurate minimum, but i guess that example shows that one might be able to get 10 watts of heat with 1mV of voltage if the other resistances could somehow be nearly eliminated. Maybe by building the device on Pluto and using superconductors. Btw I think that is one practical application of this. Stuck on Pluto with only a 36v RTG? Yes you could use the heat of the RTG itself more efficiently but maybe the RTG is too big to move. So you want to design a low voltage heater.

Anorlunda, I was not particularly interested in non-electrical constraints. This is about theory. I am not looking for actual help designing a heater. Although I am actually building some. I wanted some comments on low voltage heating devices and on why they are so rarely used afaict. I posted this because the idea of using something like 1 volt to power a heating device just seems so odd and I am getting a better idea as to why.

Also the idea of something like a 1 volt, 1 kW heating element intrigued me. Power is proportional to the square of the current. So it seems like going with a lower resistance element may actually be more efficient if generating heat is what you want.

And the way people talk about these things online makes it seem like 1000 amps will have the same heating power regardless of the voltage used to create it and maybe that's true and if so it feels odd. I was sort of looking for something like an answer involving current density or something to do with the actual process of how electrons are pulled from their atoms with a low voltage vs a high voltage. I know there are smart people around here who sometimes have more in depth answers to questions that are usually only answered at a surface level.

I am assuming that current is mostly what creates heat and that can be increased by either raising the voltage or lowering the resistance, but what is actually going on in that wire and is it different with a low voltage and low resistance vs high voltage and high resistance? I am looking for a better intuitive understanding.

anorlunda
Staff Emeritus
And the way people talk about these things online makes it seem like 1000 amps will have the same heating power regardless of the voltage used to create it
That's certainly not true for the people responding in this thread.

Ibut what is actually going on in that wire and is it different with a low voltage and low resistance vs high voltage and high resistance? I am looking for a better intuitive understanding.
If you understand all those equations, then you should know that current is current. There is no difference as you describe. I don't understand why you still have questions. Other people may be thinking of those non-electrical things you want to ignore.

Are you asking about electrical conduction? If so, see
https://en.wikipedia.org/wiki/Drude_model
or
https://en.wikipedia.org/wiki/Free_electron_model
but I warn you that the topic is very advanced.

jim hardy
Science Advisor
Gold Member
2019 Award
Dearly Missed
And the way people talk about these things online makes it seem like 1000 amps will have the same heating power regardless of the voltage used to create it and maybe that's true and if so it feels odd.
Your fluency with algebra and Latex in calcuating resistances infers great familiarity with the very basics of electricity.
But your questions suggest the opposite.

I am assuming that current is mostly what creates heat
No it is the PRODUCT of current and voltage. Pwatts = Iamps X Volts

and that can be increased by either raising the voltage or lowering the resistance,
You are familiar with ohm's law V= Iamps X Rohms ?
So R = V / I
and I = V / R ?
Since P = VI
P also = V X V/R = V2 / R
and P = V X I = (IXR) X I = I2R
you'll see P= V2/R and P=I2R all over the place.

but what is actually going on in that wire and is it different with a low voltage and low resistance vs high voltage and high resistance? I am looking for a better intuitive understanding.
It's not different. It's just a math triad..

That's how i interpreted your questions . If i missed your point don't take offense.

There's another relevant concept - maximum power transfer theorem. You touched on it when you mentioned resistance of your transformer windings.

But that's a subject for another post, and there are plenty of PF'ers here more capable at explaining that than i. Probably yourself included.

summary - i think you already know your answer. After all, most of learning is discovering what we already knew but just weren't applying..

old jim

It might be useful for people to know that I do have an EE background. Studied it at uni for 3 years, but didnt finish for monetary reasons and of course never worked in the field. That was many years ago, but I am familiar with a lot of the first half of what is typically covered and that certainly includes Georg's Law. So my elderly brain is maybe just getting confused too easily.

My problem however is not with understanding the basic math of Ohms Law. I guess I am not succesfully communicating what I find puzzling or uh interesting about this subject. No problem. I certainly appreciate the answers so far and just exploring some real world examples has helped me a lot.

As I said for me this is now more about the resistances outside of the heating element itself being the largest factor in choosing a high resistance or low resistance heating element and yes the Maximum Power Transfer Theorem does seem relevant here. I did think of that but I am just designing a 10-20 watt heater. So I dont need the maximum possible power, and if the resistance of the element is the same as the external resistance then 10 watts will also be dissipated external to the heater which seems quite wasteful. I would like to keep Pext pretty low.

I assume similar reasoning is actually the main reason most heaters use nichrome instead of copper or aluminum or steel or copper clad steel. High resistance high voltage elements have the advantage of lower currents through the house wiring and therefore lower external power dissipation and of course if you just use mains voltage you dont need a transformer which I have noticed are extremely expensive now.

Thanks for the links to Drude and Sommerfield. I dont think I have read those before. Will look into them further. Particularly the more modern free electron model.

jim hardy
jim hardy
Science Advisor
Gold Member
2019 Award
Dearly Missed
I assume similar reasoning is actually the main reason most heaters use nichrome instead of copper or aluminum or steel or copper clad steel.
Stainless steels have better corrosion resistance than copper or aluminum, and plain carbon steel will rust.

I guess I am not succesfully communicating what I find puzzling or uh interesting about this subject. No problem.
Thanks, i sorta thought that might be the case... I hope you weren't put off by my response it wasn't intended derisive , i was honesty puzzled by your questions . Your reply is most gracious, thanks again.

I'd say go to your local fishing tackle shop and get an assortment of stainless steel leader wire
and get a 6 amp battery charger from Walmart or Tractor Supply (or at a yard sale)(or even a trash pile, i'm not too proud to dumpster-dive)
i much prefer the old fashioned heavy transformer type over the modern lightweight switched-mode type,

and experiment.

if you have a Variac so much the better .

my 2 cents and overpriced at that.

That's certainly not true for the people responding in this thread.
Are you saying it is not true? Hmm. I guess it isnt since power is also proportonal to resistance. So 1000 amps through a 10 ohm wire will dissipate more heat than it would through a 1 ohm wire.

The sort of experiment I would like to try would be more like building a small 1V Americium RTG and connecting the output to a large copper plate as a heating element. Or to build a transformer with a heavy gauge secondary winding with an open circuit voltage of only 1 mV and see how much current I can get from such low voltages. Large battery chargers may have interesting transformers inside though.

jim hardy
Science Advisor
Gold Member
2019 Award
Dearly Missed
I would like to try would be more like building a small 1V Americium RTG
i dont think i want to know anything about that DIY project.
Are you skilled in radiation protection methods and equipment?

Well Strontium 90 and Americium 241 and Caesium 137 etc may as well be unobtanium. So its mostly a moot point. It is not merely a regulatory issue either. Even if you can buy it legally the cost per gram is prohibitive. United Nuclear does sell Strontium 90, but in such small quantities as to be useless for generating enough heat for an RTG. I was just looking into if any naturally occuring elements have sufficiently high alpha or beta decay but so far it appears the answer is no. Anyway if it were doable as a hobby project someone a lot more an intelligent than I would have done it long ago.

As far as safety issues I would worry about respirators and metallized suits and glove boxes and lead foil and sheet once I found a cheap and plentiful source of active isotopes. It is just something I would love to do in theory. The voltage would be across flat peltier-like plates so it seemed like a good match for enormous copper barstock as low resistance conductors for the low resistance low voltage heating experiment I would like to do.

Just imagine. Once you built the RTG you would get free power from at least the Americium version for hundreds of years, but even a Strontium version would outlive me. Every home should have one. If I were king of the world I would not waste even one gram of Strontium 90 from nuclear waste. It would all be put into RTGs for the masses. Although maybe the process of extracting such useful isotopes would be prohibitively expensive.

Is there some reason I could not just use steel or even aluminum?...I was planning to use 12 volts before...
Then just get some scrapped old PC (or just the motherboard/CPU/heatsink...) Nothing will beat that price

It is just me or this topic really went a bit ... imaginary? What is the actual question, again?

NTL2009
Just an exploration of the parameter space for low voltage low resistance heating elements and why it is rarely done and why it might be a bad idea. Also exploring any difference between the heating effects of current at low resistance and high resistance and and any more fine grained explanations of the physics involved if you've got em. Not too ambitious for a single thread, right?

CWatters
Science Advisor
Homework Helper
Gold Member
The very low voltage/very high current approach is avoided because of losses in the wires between the source and heating element.

The very high voltage/very low current approach can cause issues with the fragility of the thin heating element needed to give it a high resistance.

Rive and NTL2009
Also, the mass/surface ratio. For effective heating you need that heat transferred, and with a big fat conductor it has less surface to similar weight (weight means price). So if you need that heat outside then even for big currents it is better to split the current => connect heating elements parallel.

Unless you want to use the heat directly, like in an arc furnace or such. Then high currents might be the way to go, but then you better think about 'cables' like these:

#### Attachments

• 107.2 KB Views: 308
jim hardy
anorlunda
Staff Emeritus
I think the OP questions have been adequately answered.

Thread closed.

If anyone sees value in continuing this thread, click my user name and "start a conversation".