Trying to heat a wire to around 94F for a few seconds using Pi3 and Nichrome

First off I am trying to figure out if this is possible. Trying to use it to work with thermochromatic pigment that changes at 92F The pigment would be on a fabric with the heated wire on the other side. The fabric is quite thin, similar to a morphsuit. I would like to power it or at least control it with my Pi3 because I want to flip it on just long enough for it to heat up to that ~94F for a moment. My main issue I think I am going to run into is the max current across GPIO pins which seems to be around 16mA per pin or 50mA across all 3.3V pins in the rail. I want to power with a portable batter pack If there is a good way to control it from the Pi and use the power straight from one of the outputs on that battery pack that would work well too.

I will post back here if I find a solution on my own. Thanks for taking the time to read and or respond.

EDIT I guess I forgot to clearly post what my main question for this community is. Does anyone know if that low current output would even be enough to get something like 32 AWG Nichrome to around 100 degrees. Looking at the problem I think a good way to go about it would be to use the Pi to control some other circuit powered either straight from the battery pack or using a AA or 9V battery pack. However I have no idea what I need to purchase and how I would make such a circuit. I have an idea of what I want to do but I do not have the know how to accomplish it. Thank you again for taking the time to help me.

• Yes, it's possible - if the wire is thin enough and short enough. What size wire do you need, how long is 'a moment', and how will the wire heat up the pigment? – Bruce Abbott Oct 25 '16 at 19:07
• @BruceAbbott I am only trying to get the temperature long enough for the change to start which is almost instantaneous. So having it to that temperature for around a second should be plenty. The pigment is on the other side of a very thin cloth that I would glue/tape it to. I just purchased some 32 AWG Nichrome wire amazon.com/gp/product/B00MRASISC/… I can make that whatever length I need it to be to work. – user3642085 Oct 25 '16 at 19:38
• Bear in mind that you are actually heating the cloth, which heats the pigment. So thermal conductivity of that cloth is important here. The first step should be to experiment with the wire, cloth and pigment physically to see how it all responds. Then worry about controlling it when you know how much current, for how long, is needed to get the response you want. This will tell you whether your idea is practical. – Ian Bland Oct 25 '16 at 20:23
• @IanBland I fully agree with you. I edited my post to make what information I am looking for more clear. Thank you for your input. – user3642085 Oct 25 '16 at 21:05
• jacobs-online.biz/nichrome/NichromeCalc.html says:- to get 32g Nichrome wire up to 104ºC requires 0.49A. – Bruce Abbott Oct 25 '16 at 23:01

You must define Power and load resistance for a given surface area.

• TBD: Pd [mW], Area[mm2]
• unknowns : Rjc, Rca for thermal resistance of junction, j of heat source, case,c or the human interfaces and ambient, a, temperatures.

Summary Proposal

• ARM Cortex GPIO ports use ~25Ω drivers for RdsOn using datasheet $R=\frac{V_{OL}}{I_{OL}}$

• Using 1/8W carbon film , CF18, resistors , I calculated thermal resistance as 0.68 °C/mW

• this is the case temperature rise above ambient
• thus for 10°C rise , only Pd = 15mW is required
• NiChr wire would be useful to trace pattern from track
• but must be pulsed current from a MOSFET !!

How to generate 15mW power in a resistor , I leave up to you.

• But as an initial test, I would 74ALVCxx gates @3.3V into 700Ω using CF18

• Proof: $V^2/W = R= 3.3^2/15mW = 726Ω$ then subtract 25Ω driver and choose 700 Ω
• then record time to rise 10'C
• to speed up time, use 25Ω load which causes Voh=Vcc/2 and pulse at 14% duty cycle for same Pd in resistor
• then compare response times.

Record measurements faithfully in logbook. 1: https://i.stack.imgur.com/26J0a.png

• The area [$\text{mm}^2$] is important here. If large enough, the nichrome wire's resistance will play an important role and may dictate which type of battery can be used. If the cloth is worn, the OP will want to insulate it from the skin to both prevent burns, and prevent the skin from sinking too much of the thermal energy - requiring more power and increasing the likelihood of burns. – rdtsc Oct 25 '16 at 23:18
• Using a carefully selected and driven PTC thermistor directly as the heated component would allow for self regulation of temperature. Comment about thermal sensitivity and mass of skin are germane. – KalleMP Oct 25 '16 at 23:21
• I agree but PTC's but span of 10'C is not enough to make a dent. – Tony Stewart Sunnyskyguy EE75 Oct 25 '16 at 23:28
• @TonyStewart.EEsince'75 Thank you very much for the information you provided here. So far I am expecting I will need to simply use my Pi to control a separate circuit and its own power source. Most my supplies arrive tomorrow and I can get started testing. – user3642085 Oct 26 '16 at 20:27

I think that you will need way more current than the GPIO will be able to provide. In this case you should use an external device that can be controlled by the IO. Simplest is to just use a N-Channel MOSFET on the south end of the heater wire with a resistor in series to limit the current. The IO pin would just attach to the gate of the MOSFET, and when driven high will make the MOSFET essentially a "short circuit".

You will have to experiment as to how much current is required and for what duration to accomplish what you are looking for.

Cheers

• Reed relay may be convenient small form factor that could provide simultaneous isolation of control circuitry from exposed heating wire with independent power source. – KalleMP Oct 25 '16 at 23:22
• @KalleMP A kit I purchased has one of these in it, ghielectronics.com/downloads/man/20084141716341001RelayX1.pdf would that perhaps work? – user3642085 Oct 26 '16 at 20:22
• You will still need some extra muscle to drive the relay coil, the coil chart indicates the needed current and voltage for the relay version you have. The contacts will be able to drive sizeable resistive load of 7A at DC (28V max). – KalleMP Oct 29 '16 at 17:18