How to have a load pull more current from a low voltage high current battery

I am designing a pair of heated pants and have run into an issue with having to use batteries (or any other portable power source) to power them.

What I have now is about 50ft of Teflon wire with 5Ω resistance. The wire is 30AWG PTFE W-16878 Stranded wire rated for 600 volts.

For the power source, I have two of these 1.2 V 3.1AH batteries As they were the best option for a high current battery that can also handle higher temperatures.

The issue is with the two batteries in series they only provide 2.73V so when I connect the wire to the batteries it only pulls 0.546 amps which is not even enough to produce a noticeable change in temperature. Ideally the power draw should probably be around 30 - 60 watts but is subject to change based on how much heat the wiring actually ends up producing.

My first thought was to use an op amp and create a current to voltage converter, but there is the issue of if the op amp can handle the current I put in, and if it can continue operating when the circuit gets heated as the pants are heated. To be safe, any circuit made for the pants will use the same teflon wiring so I can avoid any wires burning up.

However, there is likely another easier solution that I am missing which is why I am posting here. I have tried researching battery powered motors to see if they might use circuitry that may help solve this issue, but had no luck finding anything useful.

Also please keep in mind that any solutions need to be reasonably portable as these pants will be used all winter.

Edit: The specifications for the wire is MIL-W-16878/4 Type E Stranded Wire with a temperature rating of: -55°C to +200°C. I can't seem to find a specific current rating.

Also as stated here: "M16878/4 is used in high temperature electronic applications M16878/4 has excellent thermal aging, solder damage, flame, and moisture resistance." Which is why I choose this specific type of wire.

• Assuming that the wire is rated for such a current, you may want to consider using a shorter piece of wire, or placing multiple pieces in parallel. The op-amp idea is unlikely to work out well in any way as op-amps are generally designed to precisely operate on small signals but unsuitable for driving power on their own. Nov 5 '19 at 16:06
• @ζ--: the part about multiple short sections in parallel in a good answer; you should make it one. That's pretty much how car rear-window defrosters are done. The OP should figure out the power per unit length of wire they want, then work out what that means in terms of the length of wire in each section, then work out how to arrange that in the heated pants. Nov 5 '19 at 16:14
• @TimWescott I'll actually start by drawing out how to connect the wires in parallel while keeping the wire properly spread out within the pants. But yeah it definitely seems like the best solution assuming that I can make the shorter sections of wire work within the pants. Nov 5 '19 at 16:43

To start, well-insulated heated pants might not need 30-60 watts of power. I'd be concerned about overheating at that point, although I don't have a strong thermodynamics or bio/metabolism background to analyze this.

Assuming that the wire and battery are able to handle a larger current, you will want to either consider a shorter piece of wire, or multiple pieces in parallel. This will cause more current to be drawn from the battery, at roughly the same voltage (with a modest decrease due to load/output impedance). I'll continue with this assumption for the remainder of the answer, to keep math simple.

Given a battery voltage of 2.5 V (rounded for convenience), if you use 50 feet (5 ohms) of wire directly, you'll draw 0.5 amps for a total power of 1.25 watts (0.025 watt per foot).

If you use 25 feet of wire (2.5 ohms) you'll draw 1 amp, for total power of 2.5 watts (0.1 watt per foot). Now combine the two halves (25 feet each) in parallel, for a total of 2 amps and 5 watts.

Notice that if you split the wire into $$\n\$$ sections combined in parallel, you'll get $$\n^2\$$ times the current and thus $$\n^2\$$ times the power--thus, if you split the wire into six sections in parallel, you'll get roughly 45 watts, assuming that the battery and wire can handle the load: the current will be 18 amperes in total (3 amps per section). Each wire will need to handle 3 amps plus a safety factor, and will need to dissipate 7.5 watts over its length (almost a watt per foot). Make sure that the wires are suitably bonded to the load to effectively deliver heat, and that hotspots are avoided.

Since these are heated pants, safety is paramount and hence you'll need to ensure that no burns occur.

• I didn't even consider putting the wiring in parallel. As for the heating side of the pants alone, I have to use standard warm pants because I can not tolerate the cold or being too hot. With insulated ski pants the second I walk inside I start overheating. I also need to keep my legs above 65 F as any lower and my skin burns from Raynaud's. The Batteries output around 10 amp when checked directly with the multimeter so it can handle a lot. The wire is also made for this very specific application so I'm not worried about it. I'll see if i can make the wires in parallel within the pants Nov 5 '19 at 16:37
• @Brittany: If you measured 10 Amps by connecting the multimeter (set to measure current) directly across the battery, you were measuring the short-circuit current of the battery - for normal operation, you shouldn't draw anywhere near the short-circuit currrent. At 10 Amps, your 3.1 Ah battery will last less than 15 minutes. Nov 5 '19 at 16:57
• Follow up question: When you say ensure that the wires are suitably bonded to the load do you just mean make sure they're properly soldered on? Or is there anything I can do specifically to get a strong connection? I plan to at the very least use a switch to be able to turn the pants off easily as the battery holder holds the batteries in too well. and possibly some sort of dimmer to adjust the heat if needed. Nov 5 '19 at 16:59
• @PeterBennett I mentioned the 10 amps as the batteries can output that amount of current without heating up or catching fire. I'll likely end up having to buy another set of batteries to provide enough current. Nov 5 '19 at 17:07
• @Brittany With regard to bonding, I mean that the wire needs to effectively and evenly heat the pants. If the heat isn't spread properly you'll get burns where the wires lie and cold spots between the wires. Nov 5 '19 at 17:19

for your wire to dissipate at least 30W of power:

1. you need to cut the wire up into smaller lengths.

2. all the cut pieces put together in parallel must have a resistance of 0.25 Ohms.

3. from there apply your 2.73V battery source to the ends of the wire and viola 30Watts of power will be dissipated with regards to V=IR, thus V = 2.73V, R = 0.25 Ohms, I = 10.989A and Power(P) = I^2R or IV = 10.989 x 2.73 = 29.99Watts.

• kindly make sure your wire can withstand the high current flowing. Nov 5 '19 at 17:04
• How long with the batteries last at 11 Amps? Nov 5 '19 at 19:21
• from the formula C=xT where C is battery capacity in milli Amp Hour, x is current drawn in amps and T is time in Hour. T = 3.1mAH/11A = 0.28 hours = approximately 17 minutes. Nov 6 '19 at 20:19
• To get much more time on for the heating you need to put more cells in parallel to increase the battery capacity. Nov 6 '19 at 20:23