Hi all

I bought a couple of 7.4 VDC batteries and I would like to run them with my Therm-ic brand insole footpads. The footpads are about 12 ohms of resistance each. The 7.3 volt batteries run the footpads too hot when hooked up directly. I am trying to figure out an easy way to lower the 7.3 volt batteries to put out somewhere between 5 and 6 volts. The insoles are a constant resistance at 12 ohms each. Is there an easy way I can accomplish this? Whatever it is, it needs to be very compact as it will hang on the back of a ski boot. A resistor, an LED? Ideas?
The batteries are Tenergy Li-ion 18650 at 7.4 volts, 2600 mAh and are rated at a max discharge current of 5.9 A and a continuous discharge current of less than 2.6 A

I dont mind if I loose a little efficiency as these batteries have plenty of power

 

Hi all

I bought a couple of 7.4 VDC batteries and I would like to run them with my Therm-ic brand insole footpads. The footpads are about 12 ohms of resistance each. The 7.3 volt batteries run the footpads too hot when hooked up directly. I am trying to figure out an easy way to lower the 7.3 volt batteries to put out somewhere between 5 and 6 volts. The insoles are a constant resistance at 12 ohms each. Is there an easy way I can accomplish this? Whatever it is, it needs to be very compact as it will hang on the back of a ski boot. A resistor, an LED? Ideas?
The batteries are Tenergy Li-ion 18650 at 7.4 volts, 2600 mAh and are rated at a max discharge current of 5.9 A and a continuous discharge current of less than 2.6 A

I dont mind if I loose a little efficiency as these batteries have plenty of power

This is not a reasonable DIY project for a beginner. You will need temp sensors and feedback. And multiple safety systems to make sure they do not run when you don't want them to run. You can run into an overheating problem and not realize - you feet are not sensitive to medium temps. You could easily turn your foot soles into slow-cooked meat killing capillaries and lead to amputation of your foot. A ski boot takes a long time to remove -especially if you are cruising down the hill when (if) you finally notice the issue.

Broken wires on sensors can cause an over temp, short circuits on the power can cause issues. You are essentially making a medical device - you should not be the experiment. If someone else wants to help you, that is on them. Good luck.

 

This is not a reasonable DIY project for a beginner. You will need temp sensors and feedback. And multiple safety systems to make sure they do not run when you don't want them to run. You can run into an overheating problem and not realize - you feet are not sensitive to medium temps. You could easily turn your foot soles into slow-cooked meat killing capillaries and lead to amputation of your foot. A ski boot takes a long time to remove -especially if you are cruising down the hill when (if) you finally notice the issue.

Broken wires on sensors can cause an over temp, short circuits on the power can cause issues. You are essentially making a medical device - you should not be the experiment.

Gopher- those are non issues. I have been wearing store bought boot heaters for years and the only problem with them is they dont create enough heat. I have quick plugs on the back of the boots that lets me unplug them in a second. Again, I am happy to use resistors, LEDS, diodes, whatever to lower the voltage. Just looking for some advice

 

Gopher- those are non issues. I have been wearing store bought boot heaters for years and the only problem with them is they dont create enough heat. I have quick plugs on the back of the boots that lets me unplug them in a second. Again, I am happy to use resistors, LEDS, diodes, whatever to lower the voltage. Just looking for some advice

Yes, the store-bought heaters are carefully calibrated to make them fail-safe. Therefore they do not deliver enough heat when needed but they are also unlikely to burn you when something goes wrong.

Your Goal makes bacon-foot much more likely. It is not a non-issue if you achieve your goal of more heat.

 

IMHO, as a former competitive skier and ski mountaineer, I would never consider boot heaters since wet feet are colder than dry feet. If your feet are getting cold, you need to replace your wet socks with dry socks. I'm with @GopherT, but there is no real reason for you to listen to us -- nobody else does.

 

7.3 volts (fully charged batteries) on a 12 ohm load produces 4.44 watts of heat. If that's too much, adding a diode in the circuit that can handle that kind of wattage will drop the voltage down to about 6.6 volts. That on a 12 ohm load will produce 3.6 watts.

If you dead short the diode you're still only going to get 4.44 watts. Your goal is to lower the amount of heat, not increase it, so I don't see why others are concerned with CFS (Cooked Foots Syndrome). An additional diode will drop the voltage another 0.7 volts; so your final voltage would be 5.9 volts resulting in 2.9 watts.

OR if you wanted to go with using a battery that has the lower voltage then plug it in and have lower wattage (heat). Your feet will be less likely to sweat and therefore less likely to be cold. Your goal is not to increase the heat, you said you want to lower the heat. I'm just not seeing where the others are concerned about overheating your feet comes from. If I'm wrong, someone correct me.

 

Try to connect heaters in series and use one battery. Maybe that will enough.

 

Yes, the store-bought heaters are carefully calibrated to make them fail-safe. Therefore they do not deliver enough heat when needed but they are also unlikely to burn you when something goes wrong.

Your Goal makes bacon-foot much more likely. It is not a non-issue if you achieve your goal of more heat.

Why do so many people that come to this forum insist on risking their health, life or property to save a buck? Or because they think they can do it better?

 

Voltage on Insole never exceed 3.7V, but always lower, depends on regulator position.
http://www.drwarm.com/product/838/cid/134.shtml
Specification
Description Dr.Warm Manual Control Insoles
Function Keep warn 4-8 hours; Recharged
Model M4
Material EVA bottom with cotton
Battery Voltage 3.7v
Battery Capacity 1500 mAh
Charging Voltage Peak 4.2+/-0.05v
Minimum Discharging Voltage 2.7v
Charging Mode CC/CV
Battery's Maximum Charging Current 1500mA(0°C~45°C)
Battery's Maximum Discharging Current 1500mA(-20°C~60°C)
Battery Internal Resitance < 10Ω
Szie 4 sizes: S/M/L/XL; 36-38/39-41/42-43/44-46
Logo Dr.Warm or custom logo
Certificate CE, RoHS, FCC
Instructions:
1) Charging: Connect the charger to the insoles with the USB cable and plug in. (Charging time: 3-4 hours)
2) Long press the on-off button on insole and then press I/II/III position to control the temperature.
I means low temp shows blue;
II means medium temp shows pink;
III means high temp show red.
3) Press the middle buttom to switch to the off mode and press again will switch to the work mode.
4) Out of battery or take off insoles, press the on-off buttom to close the insole.

 

Maybe I'm learning something here: What was the original battery that came with your ski boots? If it was 3.7 volts then suddenly I have to change my opinion. Not only are you running the heaters higher than they were designed to you're also risking CFS.

A good rule when posting here is to give us ALL the information. That way we can't make stupid assumptions that your boots are too hot and you want to cool them a little. My earlier advice is definitely in question now. I had no idea that they were designed to run on 3.7 volts.

Here are the numbers:

3.7^2 ÷ 12 = 1.14 watts.
7.3^2 ÷ 12 = 4.44 watts.

That's almost three times more heat. THAT'S A LOT!

And if you lower the voltage to 5.9 volts:

5.9^2 ÷ 12 = 2.9 watts. Still more than double the heat. Is that what you want? Double the heat? Can you stand double the heat? I think I'm having to agree more and more with others who've said it's a bad idea. I thought you wanted to lower the heat. I didn't know you wanted to lower the heat from way too much to just too much.

Really, we need all the details.

 

Are you using
Hotronic S4 Battery pack (voltage 4.8 nominal)
or
Hotronic S3 Battery pack (voltage 3.6 nominal)?

These are the batteries I'm finding on line.

3.6^2 ÷ 12 = 1.08 w
4.8^2 ÷ 12 = 1.92 w
6^2 ÷ 12 = 3 w
7.2^2 ÷ 12 = 4.32 w

Double the voltage is four times the wattage.

 

Please don't ruin bacon for me!