According to my understanding, this induction heater should be melted, but it's not. Why not?

Question

I opened up this induction heater to find that it only has maybe 12 AWG wire in its work coil, rated for 30 A RMS. I've been doing induction heater simulations for a project, and the two topologies I've looked at (series and parallel half-bridge inverters) step down voltage from the source and amplify current to something on the order of 100 A RMS. How is the induction hob not melting?

You need high current in the work coil to induce a large magnetic field to induce a large current in the workpiece, so it's not like using a different circuit would change this. It comes down to physics.

I am also baffled how there's an inductor inside that looks like it's rated for maybe 15 A, but I think it's just there for smoothing the supply voltage. Am I right? I doubt it has anything to do with the quasi-resonant inverter topology, but I'm looking into it now.

Answer 1

Where do you find that a 12 AWG wire is rated for just 30A?

That may be true for standard, PVC-insulated wire used for home electrical systems.

The ampacity of a wire doesn't depend solely on wire dimensions (cross-section) and construction, but also on the characteristics of the insulator and the environment in which it is deployed.

Unless you have a datasheet that says that a specific wire can carry at most 30A, your assumption may be wrong. For example, silicone-insulated copper wire can withstand higher temperatures for the same cross-sectional area, so their ampacity will be higher, all other factors being the same.

Just an example of your ampacity estimation being wrong in general, here is a datasheet of a high-temperature appliance wire, 12AWG (just the first I happened to find):

This is a relevant excerpt from further down in the page:

Note in the header the insulating material il not common PVC, but EPDM (Ethylene Propylene Diene Monomer).

As you can see, it is rated for almost 70A at 30°C. That's not your 100A, but it's getting close, and it is in free air, i.e. with just natural convection acting on it.

As Marla already pointed out in her answer, forced air-flow conditions could well allow higher ampacities.

EDIT

(To further clarify what ampacity really is, as the OP needed some explanation).

Think of a bare copper wire as a resistor: it disspates power as heat. That heat has to go somewhere and wherever it goes it could cause problems.

For example, it could heat up the wire insulation, other devices nearby, or even other parts of the wire itself if the wire is coiled up or is placed within a closed environment, like a cable conduit.

That's why common household extension cords usually come with a warning stating that their full power-carring rating (or their ampacity) is guaranteed only if the cable is not coiled up, but acutally extended and spread out.

All these factors must be factored in to determine the ampacity. To cite a more authoritative source, I found this technical report from the US Nuclear Regulatory Commission:

Ampacity Derating and Cable Functionality for Raceway Fire Barriers (Sandia National Laboratories)

Its scope is quite beyond what you asked, but it contains a nice and clear explanation of what is really meant by ampacity.

Excerpt (from page 4 – emphasis mine):

The term ampacity, as used in this report, is defined as the maximum current carrying capacity of a given cable conductor applied in a given installation configuration. A cable's ampacity is dependent on its routing and installation configuration. That is, the same cable will have a variety of individual ampacity values depending on how and where it is installed. For example, a cable may have ampacity values associated with open air, conduit, and cable tray applications, each of which will be unique. Furthermore, other factors beside the raceway type impact ampacity including environmental ambient temperature, loading conditions (number of cables in the raceway), and grouping of raceways. Hence, ampacity is not a single valued property of a given cable, but rather, is a context-driven value that must be determined (or conservatively bounded) for each application of interest.

Moreover, to address one of your concerns, actual damage to the conductor itself is fairly rare, since copper melting point is 1083°C.

It is very unlikely that such a temperature will be reached in normal operation, since probably there are other materials in the system that are much less resistant to temperature and that will fail earlier.

OTOH, if something catches fire, like the wire insulating material, for example, then you could have actual copper meltdown, but it's not caused by Joule heating itself.

Answer 2

The coil winding is cooled with forced air (fan). You can likely hear the fan working when the heater is on. How do I know? I have opened several of the small induction units to repair.

Your photo does not show the fan. The fan will be in the other half of the enclosure that you have disassembled.

Also, the insulation looks like it could handle higher temperatures.

Answer 3

As there are presently other answers answering the direct question, I will just address the remaining confusion:

The work might indeed be carrying hundreds of amperes; how much depends on the coupling factor and turns ratio. Simply enough, the work coil is an impedance matching element between the work and the power supply; it could use more turns of smaller cable to deliver the same power, at the same phase angle, merely at a different ratio of voltage to current.

Basically, it is the transformer action that allows 30A wire to do the job, and there are enough turns present to get a reasonable voltage rating, and thus the required power output.

The remaining consideration is the Q factor. Notice the coil is much closer to the work than its diameter, i.e. the distance is a few cm while the coil is >10cm across. Also notice the ferrite plates underneath the coil: these are pole pieces, which serve to direct magnetic field from underneath the coil, up towards the top. This increases coupling factor, reducing the loaded Q factor of the system, i.e., less reactive power must be handled for the same power output.

Induction heating is most feasible when the work is a modest resistivity material, or magnetic, such as steel or stainless. Titanium and graphite are also typically good load materials (at least when of adequate thickness). Aluminum and especially copper are less preferable, as they tend to reflect magnetic field as much as absorb it; the result is less inductance (operating frequency rises) but still a high Q factor (perhaps 10-20, versus 3-7 for steel work of the same dimensions and distance).

Compare to the unloaded Q (no work placed near the coil) which might be 100 or more. Which implies (but only loosely, for reasons that aren't worth going into right now) that, say:

• Suppose the coil delivers 1kW real power in normal (loaded) operation.
• Suppose normal (loaded) operation runs at a Q = 5. Then, there's 5kVA of reactive power circulating in the system.
• The coil dissipates its own fraction of that 5kVA, given by its unloaded Q, i.e., 5kVA / 100 = 50W.

Which sounds reasonable enough for a coil of that size, and considering it will be made of insulation that can handle a fairly high operating temperature. (Not to mention there's a frying pan immediately above it, running somewhere north of 100°C, raising the ambient temperature already. Harsh service!)

Answer 4

#12 certainly is OK for 30A in house wiring. (If).

12 AWG is rated 30A at 90 degrees C thermal.

For in-wall home wiring under NEC, you absolutely can do that, with two conditions. First, the terminals at both ends need to be rated 90C thermal - not a problem except for residential breaker panels which are 75C. Make a splice outside the panel. Second, the load needs to be one of the very few loads exempt from the restrictions in 240.4(D). We don't get to do this very often :)

The 240.4(D) limitation is an arbitrary one which applies to all small circuits using #18 through #10 wire, it goes away at #8. Then, no question - splice outside the service panel (to avoid the 75C limit there) and you can rock most modern wire types at the full 90C wire rating. This is a neat trick for upgrading a feeder without digging lol.

Much of the thermal load is the skillet

Even as it is, I would expect the coil to be able to convection cool. The impediment is that right on top of it, is a hot skillet.
That's the only reason it needs a cooling fan.