Overheating Batteries

Discussion in 'The Projects Forum' started by sfountain9000, Jul 1, 2013.

  1. sfountain9000

    Thread Starter New Member

    Jul 1, 2013
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    I am looking to make a basic heating element 1" x 1" square that utilizes at max 4 AA batteries. Then problem I am having is that I have very little electrical knowledge. I am a Civil Engineering student and circuits is not part of our degree plan. When I connect the series the element heats up properly but the batteries become really really hot. I am wondering how I go about reducing the power flow back to the batteries so that the element remains warm but the batteries stay cool. I am curious as to if a resistor between the coil and the (-) will work.
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    How long do you hope to keep the element warm?

    Let's do the math.

    A capacity of a typical AA battery is 2500mAh.
    Four batteries in series would give a voltage of 4 x 1.5V = 6V.
    If the resistance of your heater element is 2.4Ω, it would draw 2.5A.
    The power dissipated by the element is 6V x 2.5A = 15W and the batteries would last for about an hour.

    If you reduce the current to 1/5 of 2.5A, i.e. 0.5A,
    the power dissipated is 3W and the batteries would last for about 5 hours.

    The reason the batteries are getting hot is because they are supplying too much current. The way to reduce the current is by increasing the resistance of the heater element, i.e. put a number of heater elements in series.

    Adding an external resistor is simply wasting energy as heat dissipated by the resistor.
    The solution is to make the additional resistor part of the heater element.

    Another method is to reduce the number of batteries. Instead of four batteries, try three or two batteries, assuming that you originally have the batteries connected in series.
     
  3. sfountain9000

    Thread Starter New Member

    Jul 1, 2013
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    I would like the element to switch off once it has been on for a certain amount of time or once it reaches a certain degree. It just needs to heat up long enough to vaporize a liquid and then it can cool down. Possibly 5 mins on 10 mins off. I understand for this I would need an internal timer.

    The question I have is if I slow down the current will that make the heating element not heat up as much?
     
  4. Shagas

    Active Member

    May 13, 2013
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    What heating element are you using? If you found an element that has a higher resistance (or use two of those elements in series) then you battery problem would be solved because not so much current would be pulled by the element and they wouldnt heat up. Batteries heating up is not good because it wastes alot of energy inside them and it also deteriorates them.

    You can make it switch off after a certain time using a 555 timer chip in monostable configuration .
    Or you could put a heat-dependant resistor circuit which switches it off once a certain temperature is reached .
    Sorry i'm going to sleep soon and too tired to draw up a circuit that you can use for this .
    You can try googling '555 Timer one-shot' connected to a mosfet as a switch or something or just wait for someone else to provide .
    I'll drop in tomorrow if no one helps you out
     
  5. sfountain9000

    Thread Starter New Member

    Jul 1, 2013
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    Shagas,
    I cannot find any place to by a heating element on the scale that I need it. We were making a coil out of thin wire. When I place more than 1 coil the elements do not heat up. Any help is much appreciated. Thank you.
     
  6. LDC3

    Active Member

    Apr 27, 2013
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    Is the wire insulated or non-insulated. Most likely it is non-insulated and your coil is not a coil, but a short and long path for the current flow. The short path would heat up quite a bit, but the long path probably won't.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    Right. For any given resistor, including your heating coil the power it dissipates is P = I^2*R or watts = amps squared times resistance in ohms. So a small change in current causes a large change in power.

    It's a little more complicated by the fact that a resistive wire usually changes (increases) resistance as it heats up. That's why light bulbs don't blow up like fuses - the current is self-limited by the increased resistance of the hot filament.
     
  8. LDC3

    Active Member

    Apr 27, 2013
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    If you change to 4 D-cell batteries, then they won't heat up as much since the current is distributed over a larger cross section.
     
  9. Shagas

    Active Member

    May 13, 2013
    802
    74
    As mentioned above :
    Check that your wire is laquered (has a thin insulating layer of varnish on it )
    If you are not sure then try scraping the wire with a sharp knife , if white stuff doesnt come off it then you got naked wire and you aren't making a coil but a short circuit and that is why your coil isn't heating up.


    Edit :
    If you don't have a shop nearby then you can find a crappy little electric motor and open it up and you will find plenty of insulated wire inside .
     
  10. wayneh

    Expert

    Sep 9, 2010
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    Insulated wire for a heating coil? That sounds like trouble. A heating coil is wound - normally with bare wire - so that individual loops do not overlap.
     
  11. wayneh

    Expert

    Sep 9, 2010
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    Many rechargeable D cells are just AA cells in a larger package, so I don't think that would help. A "true" D cell would help a lot.
     
  12. sfountain9000

    Thread Starter New Member

    Jul 1, 2013
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    I am not using insulated wire for the coil. I was using bare wire. I will try the wire with the lacquer on it and see what happens.
     
  13. MrChips

    Moderator

    Oct 2, 2009
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    I hope your non-insulated wire do not have overlapping turns or adjacent turns that are not touching.

    As I said earlier, you can either make your element with more wire or reduce the number of batteries.
     
  14. Shagas

    Active Member

    May 13, 2013
    802
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    True but if we are talking about relatively low temperature then I'd say that it wouldnt matter so much.

    ----------
    If you manage to coil up the wire without the wire windings touching each other you might want to try Nichrome or Constantan wire ( they have a much higher resistance than copper wire and it will be easier to get them to heat up ) . They are easily available in school labs and school science shops if you have one of those nearby .

    ----------
    Even if you do it with laquered wire you will probably get it to work but your batteries will be heating up because the coil of copper wire will have very low resistance and alot of current will be pulled.
    Try finding the thinnest wire you can .

    By the way ... why are you using a coil of wire at all?
    If you have 1x1 inch then why not use 4 resistors , drown them in conductive paste and put the ceramic plate on top of it. Much simpler than using a coil in my opinion
     
    Last edited: Jul 2, 2013
  15. wayneh

    Expert

    Sep 9, 2010
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    Or just get a 1" square Peltier. ;)
     
  16. sfountain9000

    Thread Starter New Member

    Jul 1, 2013
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    A peltier might work great! Also, using the 4 resistors with the ceramic plate, what size resistors should I use?
     
  17. Shagas

    Active Member

    May 13, 2013
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    Well a peltier would be optimal but you said something earlier about not having access to 'exotic' components :)
    If you use 1/4 watt resistors which are your standard resistors and 6volts then for starters you can Four 150-170 ohm resistors connected in parallel and see how that goes . It should give you 0,7- 1 watt of heat .
     
  18. blueroomelectronics

    AAC Fanatic!

    Jul 22, 2007
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    More info on what you're building would help. What's this device for? How much heat are you after? What sort of battery life are you expecting? Do you plan to regulate the heat?
     
  19. wayneh

    Expert

    Sep 9, 2010
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    Haha, do you think so? :p
    A peltier without control circuits probably wouldn't work so well either. A 1" square peltier has a resistance of ~1Ω, so would look pretty much like a dead short to the battery pack. And without a sink it would probably overheat unless you keep the current below 2A or so.
     
  20. blueroomelectronics

    AAC Fanatic!

    Jul 22, 2007
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    Gotta say I've always hated the mystery project posts. A Peltier will chew through 4 AA's like nothing else. Better yet use the batteries as the heating element, just short em all out! :)
     
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