Question on NuWave induction cooktop hardware

[wolfrogers]

I'm on the journey of creating a far more versatile/useable version of the induction heating tool used by mechanics to heat stuck nuts/bolts. The tools available to buy are not only clunky, big, not small space friendly, and freaking expensive. I'm using a single burner induction cooktop from NuWave. I already burned one unit out, and am hoping to avoid this again, as they aren't easy to come by for free. In my first attempt, I pulled the coil from it's pancake setup, wrapped part of it into a smaller cylinder coil, and tried it that way. Only took about four seconds to burn the controls out. I'm hoping to get some input as to why this happened. The coil I wrapped wasn't real pretty...the wraps were overlapping...2 or 3 layers in spots. There were also some of the strands that broke when I pulled it from the pancake formation. Would either of these have been the cause of the burnout, or perhaps something else?

 

[Paperdoc]

I sounds like you stripped the coil wires from a cooktop stove element and then wrapped them around a metal (maybe even not metal) cylinder to form a coil. Then you connected it to a power source, probably 120 VAC. Is that about right?

If that's what you did, you basically built a simple electrical induction coil and gave it a power source with a very high ability to deliver current. For an induction coil powered from an AC (changing voltage, not DC) source, the amount of current flowing though it for any given voltage supply is determined by the INDUCTANCE of the coil you built. For a coil with no metallic core inside, the inductance is determined by the number of turns (more turns = higher inductance) and the diameter of the cylindrical coil. For a coil with a magnetically susceptible (say, soft iron) core inside, the magnetic properties of the core increase the inductance substantially. In either case, higher inductnace is a little like higher resistance in a DC circuit - it reduces the current that will flow though the coil wires. So a very simple coil with few turns and no metal core will have a low inductance and allow a VERY high current to flow though it, almost as if the wires were not even coiled up at all. That's what happened to you. The low inductance of the coil you built allowed such a high current to flow through the wire that it overheated it and burned it out. In the ORIGINAL form as installed in the cooktop the inductance of that wire in its coil form was much higher and limited the max current that could flow, preventing this catastrophe. That's all part of the engineering design of the original heating coil.

For a coil designed to heat up automobile parts, the rate of heating (how long to heat up the part) depends on the strength of the alternating magnetic field generated by the induction coil, and that depends in large part on the current flowing though the coil. In turn that max current dictates how the coil will be constructed - what wire gauge, what coil size, how many turns, what core material, etc. And all of this is limited by what electrical power source might reasonably be expected to be used to plug this thing in. All of that engineering expertise, plus the manufacturing exppertise and marketing costs, etc. are what you pay for when you buy a commercial product. If you want to do it yourself instead, then you FIRST need to learn all that stuff so you can do a competent job of designing and building it.

 

[IronWing]

How would the nut or bolt being placed into the center of the induction coil affect the current flow? Would a control circuit of some sort be needed to adjust the current as the tool was placed over, removed from a bolt?

 

[mindless1]

I'd have to see the schematic of the cooktop to know for certain but offhand I suspect that it is designed with very many coils over a broad area to create even cooktop heat, but this means a higher working voltage and more coil turns, so if you just use a few turns you are going to put more current through them than the controller can handle.

Frankly an induction stovetop is not a good way to start the project. Too complex and set up for a high voltage that should stay enclosed. My vision for this would be something like the following:

Pick an 18V cordless donor tool, compatible with a brand /batteries I already have. Gut it except keep the trigger as an on/off switch. It needs to be a tool with sufficient space inside to house the circuit board, and might be nice if it had room to stuff an LED driver board in it and mount an LED pointing at the work.

Pick a far more basic inverter board with matching coil included. For example the link below is one or search through many on ebay as "induction heating module coil", seeking one that at 18V or more (12V cordless probably can't provide a satisfactory wattage output), has a current spec within what typical cordless tool batteries can do for short duration, say 20 amps. You might easily find this example (or one even better suited) cheaper on another ebay listing or elsewhere:

https://www.ebay.com/itm/New-DC-12-...ard-Module-Coil-Fan-DIY-Kit-S0M3/303156284610

All these are set up to mount the induction coil to the PCB but you don't have to do that. You could mount the coil on an extension of flexible metal pipe (or tubing) and run a couple of low(ish) gauge, high strand count, silicone insulated (highly flexible and heat resistant) copper wire to the induction coil, allowing it to get into tight spaces. Run a couple more wires if you want the LED remote mounted near the coil.

If the cordless tool trigger switch is low current rated and just biases a transistor for motor power, leave the transistor/circuit in and take the output from that to the induction board. This will only necessarily work with brushed tools, brushless would depend on the design and reverse engineering it, and you don't really *need* variable trigger control, just an on off switch to run at full power and then stop when it's hot enough.

If the budget stretches higher than the above example and you're willing to have a mains AC power cord to it, there are other examples on ebay of those, but do take care to properly insulate it if dealing with lethal (above say 36V would be a conservative limit) voltages. You could keep that safer as a homebrew project by having an external brick AC/DC PSU tethered to the tool, something like a 24VDC, 20A PSU can be had for $18 delivered on ebay. https://www.ebay.com/itm/AC-110V-22...h-Power-Supply-Adapter-LED-Strip/352657328504

 

[Paperdoc]

I suggest the proposal above has a serious power problem. In order to heat a nut or bolt or similar metal part rapidly by induction, I expect you need a pretty high heating RATE - that is, input of watts of heat, which is a unit of Joules (energy) per second. A consumer induction cooktop is able to provide heating rates very much the same as a standard stove element, and those run 2000 to 3500 W. That is why such elements are fed power at 240 VAC and up to 15 amps. If we assume that a useful automobile part induction heater needs even half that power, there is NO way to get that from a 20 VDC battery, and I REALLY doubt you will find an inverter able to convert 20 VDC to 120 AC or 240 VAC at a power rating over 1500 W!

For another comparison, think of what a standard automobile battery (a lot bigger than a common electrical tool battery) at 12 VDC can do for its toughest job - starting the engine when really cold. Such batteries come with a rating for ability to deal with that strain called Cold Cranking Amps, and a typical value is 600. That is defined as the amps current the battery can deliver at 0F without having the voltage drop below 7.2 VDC, and only for a period of 30 seconds. So suppose the battery does this over 30 seconds at an average output voltage of 9 VDC, and a current of 600 A. That's 5400 W for HALF a minute, and then the battery is SEVERELY drained! Now, if you draw only one third of that amperage (that's a rate of 1800W), the battery might last three times as long - 1½ minutes - before it's drained. Do you think the part you were trying to heat up at that rate (only HALF what a stovetop element can do) will be really hot in 1½ minutes? And don't forget, now you have no second chance - the battery is badly drained.

That is why these tools are made to plug into a heavy-duty 120 VAC or 240 VAC power source for the electrical system of the building, and not from a small portable battery.

 

[mindless1]

^ Fair argument but not all bolts need rapidly heated to red hot, merely heated up. You can get at least 500W out of paralleled series cells in Li-Ion tool batteries. It is enough to do useful work.

Yes a typical bolt would be really hot in 90 seconds from 500W. An induction stovetop (I assume portable unit for wolfrogers to be experimenting with it) is about 1600W but over a much larger area than inside the loops of an automotive induction heater. Suppose the cooktop has a 7" diameter element, then that's 38 cubic inches. Now suppose a 1.5" coil, that's only 1.8 cu in. Granted you can't directly equate a one-dimensional area like this but it does show that the magnetic density can be higher with 500W than stovetop has.

As far as battery life, suppose a 4Ah battery, at 500W is almost 28A. This is achievable with modern Li-Ion packs. That's over 8 minutes just diving rate by Ah but let's assume the high drain rate cuts that in half, then it's still over 4 minutes. Maybe it's only suited for a couple bolts per charge but for DIY automotive work, how many in a row do you really need to do and how hard is it to own more than one battery?