Hello Im currently in the process of building a efficient small DC powered heater and would like suggestions. It needs to be the least amount of watts vs. maximum heat output. The closest thing I found so far are the 10 and 8 ohm ceramic resisters wrapped with aluminum. When they are hooked in series and parallel and shorted out with a dc power source they get hot. The problem is shorting out uses lots of power. especially when hooked to a battery. Ive managed to get the watt usage down to around 200 watts with an Iota 500 watt ac/dc inverter but the other problem is when you blow air on them they cool off. Thats probably the biggest factor. Thats probably why you would want a vacuum sealed element and perhaps a better thermal conductor. If you have a DC motor that uses around 300 watts and hook the resisters in parallel to reduce current to the motor works to but the resisters get hot but the motor doesn't have much power. I got the frame built with a duel fan PCB type brushless motors from an old Xbox360 which is highly efficient at 12 volts like 5 or 10 watts with high air output. The frame came out really nice, made with good quality metal. I'm probably going to use the 18650 lithium ion batteries to run it. So I basically need to how to build a circuit that can cancel out the short to the battery and any advice on types of efficient DC heating elements. I would like to get the wattage down under 100 watts with a high amount of heat output. I think its ridicules that your average ceramic heater uses around 1500 watts when a 900 degree soldiering iron uses 40 watts


closed as unclear what you're asking by PeterJ, Ricardo, JRE, Dwayne Reid, Daniel Grillo Aug 5 '15 at 14:44

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    \$\begingroup\$ Very poor question. Very long and still fails to tell us the most basic thing : What do you need to heat, to what temperature, and for how long? Note that "maximum heat output" = "most amount of watts" so the question appears to demonstrate no understanding of physics and MAY be impossible to answer. \$\endgroup\$ – Brian Drummond Jan 18 '15 at 13:03
  • \$\begingroup\$ WHAT ARE YOU TRYING TO ACHIEVE? NOT HOW BUT WHAT?Read what Brian says. Do it. NBNBNB ALL resistive heaters are essentially 100% efficient at turning electrical power into heat (maybe 99% if they glow in the dark and make some light BUT even that is essentially "heat"). To get more TEMPERATURE you want to trap the thermal energy with insulation within the space to be heated. Essentially temperature = energy_in / (amount_of_stuff_heated) / (thermal mass constant of stuff). More "stuff" = lower temperature. Higher heat content per mass = lower temperature (water is about lowest). \$\endgroup\$ – Russell McMahon Jan 18 '15 at 13:25
  • \$\begingroup\$ Tell us what you WANT and we'll give you what you NEED. | Tell us how you intend to do some unknown thing and we have no clue what you want. \$\endgroup\$ – Russell McMahon Jan 18 '15 at 13:26
  • \$\begingroup\$ I think this guy is messing up heat and temperature. \$\endgroup\$ – Vladimir Cravero Jan 18 '15 at 13:31
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    \$\begingroup\$ "I think its ridicules that your average ceramic heater uses around 1500 watts when a 900 degree soldiering iron uses 40 watts" In a nutshell, this is where you are going wrong. The heater turns 1500 Watts of electricity into 1500 Watts of heat - heating up a whole room. The iron turns 40 Watts of electricity into 40 Watts of heat, heating up a soldering iron tip to 900deg. The difference between them is not efficiency (which in both is 100%), but the size (more accurately thermal capacity) of what they are heating up. You can't heat the room up with the soldering iron . . . \$\endgroup\$ – peterG Jan 18 '15 at 19:35

Let's say you have a normal room in a normal house with some basic insulation. Let's assume that the temperature difference between the inside of the room and the outside of the room at the current time is diffusing 500 watts worth of heat to the outside of the room through the walls, floor, door, windows, etc. (There are far more complicated calculations that take into account the thermal conductivity of these surfaces and the actual temperature differential but let's just ignore that for now and assume we're losing 500W of power.)

So to heat the room we need to have a heater that generates more heat over time than the room is losing over time.

Most resistive heaters will be very close to 100% efficiency in converting electricity to heat. Most efficiency losses in electrical systems are caused by waste heat being produced by components in the circuit, and reducing that waste heat will increase the efficiency of the circuit. Heaters don't really have the same problem, producing waste heat is precisely what they aim to do.

Going past 100% efficiency is physically impossible. That would mean your heater is producing energy out of nothing. What you need to do instead is better insulate the space you are attempting to heat to reduce the heat lost to the outside space and thereby reduce the power required to heat the room.

The other part of your question is about current limiting. If your circuit is nothing but a resistor then you can simply use Ohm's law and tune your input voltage to get your desired power output. If you cannot change your input voltage then you'll need to consider other options such as current limiter circuitry, which you can find hundreds of questions about with a quick search.

  • \$\begingroup\$ I admire the way you managed to understand what he was asking, and still posted after the OP showed so little effort. Side note: waste heat in the thermostat is not lost as it still contributes to the heating. Efficiency is therefore actually 100%. Also, instead of tuning the input voltage you can also wire it to mains through a relay, and change the resistance instead. \$\endgroup\$ – Mister Mystère Jan 18 '15 at 19:40
  • \$\begingroup\$ I'm very bored. And yeah any heat lost in the circuit due to parts that emit waste heat is not lost energy since you want to make heat anyway, I kind of mentioned that in the answer. But there would be weird places you'd lose energy that isn't through heat. With an AC heater you'd lose energy to vibrations and EMF losses and all sorts of nonsense. I'm not sure where it might go in a DC system but I'm almost certain there would be something that would pull the tiniest amount of energy from somewhere. I don't think you can reach 100% efficiency. Could be wrong though. \$\endgroup\$ – Nimphious Jan 18 '15 at 19:55
  • \$\begingroup\$ I'm with you on that one, I should have said "virtually" - I don't believe either 100% can ever be reached. That would be an interesting question to post! I'm tempted to post it, but it's your idea so after you. \$\endgroup\$ – Mister Mystère Jan 18 '15 at 21:41
  • \$\begingroup\$ You guys are clearly not seeing what the discussion is about. If you read the original post in regards to the topic it would make sense maybe obviously not you or arguing because your board and have nothing better to do which is a waste of my time. you are defining efficiency differently to fit your own words that does not pertain to subject. Im referring to efficiency as watt usage. im done with this conversation this has been a complete waste of time to entertain you so that you make fun of my topic i'll seek for the answers somewhere else thanks for wasting my time. ALL WRONG OFF TOPIC bye \$\endgroup\$ – Triphex Jan 18 '15 at 22:33
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    \$\begingroup\$ This is a super old question, but I thought it should be mentioned that there do exist heaters that are technically over 100% efficient. They transfer heat from the outside environment, as well as providing heat from the inefficiency of this heat transfer. An example of this is a peltier heating system, which can have an "efficiency" exceeding 110% . \$\endgroup\$ – BeB00 Feb 17 '18 at 3:30

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