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Induction heater work coil

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Ben Solon

Tue Aug 14 2012, 02:12AM

Registered Member #3900 Joined: Thu May 19 2011, 08:28PM
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Posts: 600

Other than adjusting coupling to the workpiece and changing the inductance, what are the trade offs between large and small amounts of turns. I know that more turns will induce a smaller voltage inside the piece and less turns will step up current x amount of times, creating more of a drop therefore increasing I^2R losses.

So what are more turns for anyways? Do people only do that when they want to drop their ~fres down to manageable levels?

IamSmooth

Tue Aug 14 2012, 08:39PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

More turns allows you to heat a bigger piece of metal. The coil should allow you to easily heat your workpiece, or to do so with small movements in and out of the field. The more turns, the less induced voltage, and less induced current in the workpiece. You will also have a lower Fres for the same tank capacitance. This results in deeper current penetration into the workpiece, which may or may not be desired depending on your application. All this means it will take longer to heat the metal for the same input power. To compensate you will need a higher voltage going to the work coil.

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Steve Conner

Tue Aug 14 2012, 09:26PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

What Iamsmooth says is true, but the effects he mentions are mostly explained by changes in coupling and inductance.

If you arrange it so that the overall coil dimensions and operating frequency don't change as you change the number of turns, then it all becomes a matter of amp-turns per unit length of coil. It doesn't really matter if the coil is made of 10 turns of thin conductor carrying 100A, or one massive turn carrying 1000A. If the frequency is the same, and the total length of the solenoid is the same, the heating effect should be the same too.

In practice, it is hard to get even current distribution in a really wide conductor, so longer coils are made by increasing the number of turns. But within reason, you can just choose the number of turns to suit whatever tank capacitors you have. Every capacitor has a particular frequency at which it gives its maximum kVAr.

Ben Solon

Tue Aug 14 2012, 10:05PM

Registered Member #3900 Joined: Thu May 19 2011, 08:28PM
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I read a paper on work coils that I can't seem to find now. It said that the spacing between turns must be less than or equal to the radius of the conductor to archive even heating. It also said plenty of other usefull things about coil dimensions and leakage, but that's what I got most out of it.

I always follow that rule, so in my book, increasing turns also means changing the dimensions. So what I really wanted to know was "what are the effects of more or less turns if all else stays the same."

So what I have from you Steve is that if you have a low current system, you can increase the turns to get equal AT to a higher current, fewer turn system? But what if I have a system that has a limit of 300A rms and 580v rms? If I want to heat al cu and fe, how do you think I should go about the work coil?

The reason is that the caps I have don't have a propped kva rating, just the "normal" A V and C. In case you have info that I don't, they're 2 3.75uF eurofarads in parallel. The ratings I've posted above.

IamSmooth

Wed Aug 15 2012, 03:39AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

ben123324 wrote ...

So what I really wanted to know was "what are the effects of more or less turns if all else stays the same."

I thought I just answered that above.

3-5 turns usually works.

Ben Solon

Wed Aug 15 2012, 04:04AM

Registered Member #3900 Joined: Thu May 19 2011, 08:28PM
Location:
Posts: 600

Yea, sorry if it sounded like I was still asking the same thing

I'm fine with that now.

I think I'm going to swap in a 4 turn coil give or take depending on Steves kva stuff. I don't know how to arrive to that, maybe A=C(dv/dt)? If the dv is from the positive peak max voltage that the cap can handle to the peak minimum(ie 1640), and the current is the peak current the cap can handle(ie 430), then I should adjust the time accordingly. But I don't know. Steve can you confirm this?

Steve Conner

Wed Aug 15 2012, 07:16AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

A capacitor suitable for induction heating should have RMS voltage and current ratings. The RMS current rating is usually much less than 0.7 times the peak current implied by C*dv/dt.

So, divide the voltage by the current to get an impedance. You also know the capacitance. So, you can plug these into the equation for impedance vs. frequency of a capacitor (Xc = 1/(2*pi*F*C)) and solve for F. That's the frequency at which your capacitor will perform best.

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