### Author Topic: Help figure how induction heater really works  (Read 1452 times)

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#### SnipTheCat

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##### Help figure how induction heater really works
« on: October 12, 2016, 09:30:23 pm »
Hi,

I have been playing with induction heaters based on the circuit attached. And they working pretty good. However I have some issues to figure out the parameters and how it works.

In my circuit I use IRFP4710 Mosfets on one circuit and K58E06 on others, they all see to do the job. Inductors are 150µH all other parts are as in the schematic with the only variables left being the C tank and the work inductor:

1) C = 0.44µF, Work inductor = 1µH
I get 240Khz, 51Vpp / 17.8VRMS from one coil to the other
Power supply gives 1.3A/10V

2) C=0.66µF, Work inductor = 1µH
I Get 197Khz, 53Vpp/18.7VRMS
Supply gives 1.54A / 10V

3) C=1.44uF, Work inductor = 1µH
I Get 133Khz, 51Vpp/18VRMS
Supplyu gives 3.5A/10V

4) 3µF tank, 0.9µH coil
97Khz, 40Vpp, 13.8VRMS
It takes ~5A/10V

These measurements are with no load, once I put a load, lots of things change depending on the what it is (a screw driver, a graphite crucible), it can go up to 10-12A @12V, up to 20-25A @ 20V

What I don't understand is that when I measure the inductor (working coil) with a LCR meter, I get, let's say 1µF, if I put the load inside, the inductance is very slightly changed but nothing to justify hugh power changed in real utilisation.

For example my 1µH Coil show 1.2µH with the screw driver inside and 0.9µH with the crucicle inside. in practice, the screwdriver will make the circuit 4 pump 12A@20V while the graphit crucicle will mek it pump 22A@20V

I have a hard time to understand how to calculate things "in action". What are the ideal parameters?

I have tried to find technical information online but nothing was really clear.

With one of the circuit I have pretty good result melting aluminium in a small crucicle @ ~800°C . Working coil is watercooled (copper tube), and I reach ~20A @ 20V so about 400W. I noticed more caps on the bank means more amps at the same voltage and load but also lower frequency and higher "idle" current, beside trials and errors I'm not sure in which direction to go (I'd like slightly more power at 20V)
« Last Edit: October 12, 2016, 09:40:36 pm by SnipTheCat »

#### T3sl4co1l

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##### Re: Help figure how induction heater really works
« Reply #1 on: October 12, 2016, 10:48:28 pm »
At a given (constant, sine wave, AC) frequency, there are two parameters to every (two-terminal) component: resistance and reactance.

An ideal inductance or capacitance has no resistance.  Real components always have some of both.

What's more, the amount of each changes with frequency.

A component we call a "capacitor", is one which behaves like a capacitance over a useful frequency range.  And that frequency range might be very different for different kinds!  A small ceramic capacitor works very well from ~kHz to 1GHz or more; a supercapacitor might only be usable at very low frequencies, ~Hz to uHz!

So first of all, be careful about using components, because they are only useful (for stated purpose) over a range.

(Your MKP capacitors are probably just fine, here.  If they get awfully hot -- too much resistance -- try using pulse-rated or snubber type capacitors instead.)

Inductors are interesting, because they couple to other inductors nearby.  And if those inductors happen to have resistance connected to them, then the first inductor also looks like it has more resistance.  This is simply transformer action: a load on the secondary, is reflected as load on the primary.

So, what's happening is, the resistance of the workpiece is coupled to the work coil, and since the transistors are driving it alternately, they alternately supply power into that resistance.  The current through the coil is alternating in polarity, but because the transistors also alternate, you get positive and positive current draw from the DC power source.  Which means we're drawing real power, and dissipating real power in a resistive load.

It's a nice property, of this type of circuit, that it has a constant-voltage characteristic.  That is, the load on the power supply varies with the load on the coil.  It doesn't draw much power when the coil is unloaded, it just sits there oscillating.  When you add a load, almost all the extra power that's drawn from the supply, goes right into the load!  (Not all, because of losses in the transistors, coil and capacitors.)

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!

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#### JS

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##### Re: Help figure how induction heater really works
« Reply #2 on: October 12, 2016, 10:49:29 pm »
Parallel resistance (better model of the inductor with losses)... here is some more details.

When you add a load (some conductive material inside the coil) is like you change the core of the inductor. The inductance will only change as the permeability of the core changes, so, a screw driver is iron, will help something, if you put some laminations just to try it will increase even more.

The thing you are missing is the losses in the core. The important factor of the load isn't to be ferromagnetic or have high permeability, but to have good conductivity. Eddy currents through the load makes it heat and to take account for that load you need to use the parallel model of an inductor, with a resistance in parallel which will represent the load as they were the losses in the core of an inductor. You could also add the resistance of your windings in series which is responsable for the heat in the winding (which you are cooling with water) The inductance is not an important factor at all here, you could also thing it as a transformer, with the load as a resistance in the secondary or a single shorted turn, which is exactly as eddy currents are modelized. Both models will work, the first one is mathematically simpler, maybe the transformer is easier to understand why.

JS
If I don't know how it works, I prefer not to turn it on.

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#### SnipTheCat

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##### Re: Help figure how induction heater really works
« Reply #3 on: October 13, 2016, 08:39:53 am »

I think I get a better idea, my mistake was to think of the work coil + load as a single inductor when in fact I should see it as a transformer with a load on the secondary.

the MKPH capacitors are not overheating, they are specially designed for induction heating, however I do air cool them anyway (as well as the mosfets + radiators). I have tried polyester film caps rated for 630V, they work but they are very hard to keep cool enough.

And here's a little picture of the monster (yes I know it's quite horrible), working coil is not connected as I tested with other coils
« Last Edit: October 13, 2016, 08:41:40 am by SnipTheCat »

Smf