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Induction heater from Instructables, or what determines the current draw?

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Artlav

Sun Mar 23 2014, 07:14PM

Registered Member #8120 Joined: Thu Nov 15 2012, 06:06PM
Location: Moscow, Russia
Posts: 94

Hi.

I wanted to build a high-power induction heater, in range of a few KW, and there is a nice and simple design on Instructables, presumably going to 30KW:

So i started experimenting along the lines of that design.
But the more i go on, the less sense it makes.

Now, i only had experience making induction heaters based on ZVS driver before, a 300W one.
So this design have several elements that i'm not familiar with, so i might have simply be making something wrong.

I made a work coil LC tank of ~6uH and 1.5uF, aiming for a frequency of 50KHz.
This is coupled with a 20:1 transformer to an inverter from the above design, sans IGBTs (since it's low power experimenting, i'm using the MOSFETS that are supposed to drive IGBTs as the actual inverter bridge).

In the design above, he have a work tank at 65KHz and 2uF capacitor, suggesting a coil of 3uH.
Supposedly, it should draw a few hundred W at 30V (5-10A?), and 40A at 200V (8KW) unloaded.

In my case, i'm drawing 1.1A at 30V.

Now, here is where the questions start.
What determines the current draw?

If i understand it right, then it's the surge impedance of the LC tank, sqrt(L/C).

In my case it's about 1.5 Ohm, translated to 30 Ohm by the transformer, resulting in about 1A at 30V draw.
It adds up.

In his case, it's about 1.25Ohm, translated to 25 Ohm by the transformer, resulting in about 1.2A at 30V or 8A at 200V draw.
Which does not add up with his claims.
He need either 3 turns on the transformer, or completely different LC tank for these numbers to add up.

What is going on here?
Have i missed something?
Am i using the wrong math entirely?

I did look at

, which have a nicely explained similar sort of a design, only with a more sophisticated driver, and i can't seem to find how to determine the current draw.
It just says "Let's assume that the load across the secondary is 1 ohm", not why it's 1 ohm.

Second question, the thing draws more current unloaded than loaded.
Why is that?

Even more interesting is that his design is supposed to draw 40A at 200V UNLOADED - where do these 8KW of power go then?
I don't see any part in there that can take that much heat without melting or exploding.

Triple mystery is that mine draws about 30W unloaded at 30V apparently into nowhere.
I let it run for a few minutes, and poke around.
Nothing. Is. Even. Warm.
Where does the power go?

BigBad

Mon Mar 24 2014, 12:48AM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600

I had a quick look at the circuit, but not long enough to prove it, but isn't this just an inductive energy thing?

i.e. you're just cycling the same power in and out of the coil

Coils are only lossy to the extent that the resistive losses lose you power; just because there's 30A going in/out of a coil/circuit, doesn't mean it's consuming much power, it just means it's flowing in and out; in other words the voltage and the current are at nearly 90 degrees to each other; you can't just multiply 40A by 200 V if they're orthogonal.

Uspring

Mon Mar 24 2014, 10:48AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Dunno much about induction heaters, but:

If i understand it right, then it's the surge impedance of the LC tank, sqrt(L/C).

The impedance of an unloaded series tank at resonance is 0 or more precisely the copper resistance of the inductance.

In my case it's about 1.5 Ohm, translated to 30 Ohm by the transformer, resulting in about 1A at 30V draw.

A down transformer multiplies impedances by the square of its transformation ratio, i.e. 20^2 = 400. This is due to the fact that voltages are divided by 20 and currents multiplied by 20 on the secondary side. Together that results in resistances multiplied by 400 as seen on the primary side.

Artlav

Mon Mar 24 2014, 12:37PM

Registered Member #8120 Joined: Thu Nov 15 2012, 06:06PM
Location: Moscow, Russia
Posts: 94

BigBad wrote ...
Coils are only lossy to the extent that the resistive losses lose you power; just because there's 30A going in/out of a coil/circuit, doesn't mean it's consuming much power, it just means it's flowing in and out; in other words the voltage and the current are at nearly 90 degrees to each other; you can't just multiply 40A by 200 V if they're orthogonal.

But if it's 1A at 30V out of a DC power supply, before any filtering caps, then it should be actual power, right?
Where does it go then?

Uspring wrote ...
The impedance of an unloaded series tank at resonance is 0 or more precisely the copper resistance of the inductance.

Ah.
So i'm limited by the total resistance of the tank circuit?

Uspring wrote ...
A down transformer multiplies impedances by the square of its transformation ratio, i.e. 20^2 = 400.

Great.
Then the math does not add up in either case, and i'm clueless.

Ash Small

Mon Mar 24 2014, 01:03PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

I think part of the answer here, in the 'unloaded' state, is that power (Watts) = Volts x Amps.

When the current is greatest, voltage accross the coil is zero, and when the voltage is greatest, current flow is zero (ie current flow is changing direction).

Anything multiplied by zero is zero, so power consumed is zero (or just copper losses)

There are times when there is voltage and current at the same time, but as it is first in one direction, and then the other, then overall current flow (sum of current flow in both directions during time, t) equals zero.

Any transformer will have losses, though, even when unloaded.

I'm not sure how helpful this is, but I think it's pretty much a correct description.

EDIT: There is also the 'ringing up' that occurs as the voltage rises when driven at resonance (where power is being transferred into the LC tank circuit), and there are losses if it's not being driven at resonance.

IamSmooth

Mon Mar 24 2014, 04:11PM

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

The load quenches the rlc resonant circuit
Inductive reactants increases significantly with a load

However the quenching decreases significantly when you reach the Curie point

Artlav

Mon Mar 24 2014, 06:23PM

Registered Member #8120 Joined: Thu Nov 15 2012, 06:06PM
Location: Moscow, Russia
Posts: 94

All right.
Since i'm now completely lost i'll ask for directions. :)

Loaded with a big bolt, the thing consumes 0.8A at 30VDC.
What should i change to make it accept more input current/deliver more power to the bolt?

I'm reading 45V peak-to-peak across the tank capacitor, and 25A peak-to-peak along the pipe (64 turn current transformer loaded with 2 ohms give 0.75V).
On the scope they are almost perfectly 90 degrees out of phase, as described by Ash Small, so all is fine here.
It just does not want to "ring up" any higher.

And a complementary question - am i right to extrapolate this to a draw of about 8.2A at 310VDC, that is 2.5KW of power at full input?

IamSmooth

Mon Mar 24 2014, 09:26PM

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

You never read my tutorial?

Increase the input voltage
Decrease turns on the work coil

Dr. Dark Current

Tue Mar 25 2014, 12:33AM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

This is a series resonant heater driven from a fixed frequency source. Did anyone say brutal topology fail?

Artlav

Tue Mar 25 2014, 05:26AM

Registered Member #8120 Joined: Thu Nov 15 2012, 06:06PM
Location: Moscow, Russia
Posts: 94

IamSmooth wrote ...
You never read my tutorial?

I did, but it's a bit too much to take in one pass.
Increasing the input voltage is exactly what i'm trying not to do, and changing work coil turns is not exactly an easy operation.

Reducing turns on the coupling transformer helps, however.
For that case you warned to "make sure you do not saturate the core" - what would be the symptoms of that?

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