LC Parallel resonance induction heater

[StoyanElectroTech]

Consider parallel LC resonance induction circuit with solid iron workpiece, inserted in the inductor. Is it possible to obtain more heat power output from the heated solid iron workpiece in the inductor, compared to the electrical power consumed in an LC parallel resonance induction circuit?

Is it possible as a result, the electrical power input to LC circuit to be, for example, 100 watts, the heat power generated from the the heated solid iron workpiece to be higher, say 200 watts or more, depending on the efficiency of the system and the characteristics of the load?


I_L Is the current flowing through inductor.
I_In is the input current to the LC circuit.

Lets say (Parallel LC resonance), First I prepare the inductor and the workpiece , then i switch capacitors in parallel until I reach resonance state, so that the input current become very low (Very high ratio of I_L/I_in ).



Thank you.

 

[rjenkinsgb]

Very low input current requires a very high "Q" factor in the tuned circuit.

As soon as you put the object to be heated inside the coil, that is acting like a shorted turn and the Q drops to an extremely low value; the resonant frequency is almost irrelevant.

It's acting as a transformer with the item being heated as the secondary, like an induction hob.
Low input power just means low energy transfer to the load.

(And, the most fundamental principle of physics is that you can never get more energy out than in, overall. Everyone that has claimed to have done so was either mistaken or a fraudster).

 

[StoyanElectroTech]

As soon as you put the object to be heated inside the coil, that is acting like a shorted turn and the Q drops to an extremely low value; the resonant frequency is almost irrelevant.

First I prepare inductor coil (insert the item to be heated) , then I switch Capacitors in parallel until Parallel Resonance occurs (Very high ratio to I_L/I_Input . So High, that thermal energy from the inductor suprasses input electrical energy ).

 

[rjenkinsgb]

No input current means nothing is taking energy from the tuned circuit.

It's like tapping a tuning fork or plucking a guitar string - as long as nothing is taking energy from the resonant part, it will "ring" easily. As soon as you put any load on it (eg. touch it), it stops.

You don't get anything for nothing in physics.

 

[Tony Stewart]

You could never dissipate the peak loss in resistance inheat as the peak power stored in either the capacitors or the inductor. If you tried, you would find your Q is so low, it won't oscillate because it would be overdamped by the loss. This is true for both series and parallel tank circuits.

 

[StoyanElectroTech]

So it is not possible to achieve high "Q" in parallel resonance configuration if I first prepare the inductor with solid iron core (the inductor with the iron core will still have some L - inductance value), and then to switch capacitors in parallel until i reach parallel resonance state with high "Q" value?

 

[rjenkinsgb]

You can only achieve high Q if nothing is loading or drawing energy from the coil - so not in an induction heater output coil with any load (object being heated) in the coil.

 

[StoyanElectroTech]

I've got some pretty big values for Q factor here :

https://digitalcommons.mtu.edu/cgi/viewcontent.cgi?article=2621&context=etdr

Page 40.

Workpiece Material Q Factor
No Workpiece 64.16061
Aluminum 40.8304
Stainless Steel 22.11006
Steel 4.3

It is true that No Workpiece situation has the Highest Q, But I am still able to achieve high results for Q with other materials , higher frequency and additional capacitors (to compensate for the increased L-inductance value).

 

[rjenkinsgb]

've got some pretty big values for Q factor here :

Those are not high at all - a well made air cored RF coil can easily have a Q in the 500 - 800 range.

And, the power to the item being heated will always be less than the input power to the coil, no matter what you do re. tuning it.

 

[StoyanElectroTech]

This is also interesting https://zenodo.org/record/1334846/files/13504.pdf

 

[StoyanElectroTech]

https://www.researchgate.net/post/What-the-Q-factor-of-induction-heaters-are-calculated-based-on-electrical-measurements-you-obtain

 

[StoyanElectroTech]

Google Image Result for https://d3i71xaburhd42.cloudfront.net/ee58323c2ea9792a02baa24e980169a0e7ae0da7/2-Table1-1.png

 

[StoyanElectroTech]

http://revue.elth.pub.ro/upload/17849807_AKieronski_RRST_2_2017_pp_154-158.pdf

 

[Tony Stewart]

I can simulate any Q you want and show you the peak VAR and Watts with a variable resistor and the real power is a small percentage of the peak VAR (reactive VA.
Give some f , Z, R or Q and I'll prove it. Series or parallel the same.

 

[StoyanElectroTech]

Consider Parallel Resonance Induction Heater with Solid Metal Workpiece , Placed in the inductor. If I add capacitors in paralle is it possible to achieve Parallel resonance with high "Q" (Current amplification or high ratio between current in the inductor and input current)?

Lets say "Q" > 200
Thank you.

 

[Tony Stewart]

That depends on your specs for fo, current, voltage, stored energy and dissipated energy, DCR, ESR, load R , excitation f error and finally total stored energy and transfer energy efficiency or loaded Q.

A self resonant oscillator is essential with these high Qs and DCR increases with temp. so temperature coefficient if not copper. It could be water-cooled Cu tubing.

Certainly its a challenge , and conjugate matched source to load is ideal for MTTP or much lower Z(f) source impedance for max efficiency.

Pls draw a schematic.

I think you meant series resonance for stored current amplification. Q= Zo/Rs = √(Ls/Cs) / Rs
where ESR (C) + DCR (L) + RdsOn = Rs. But it depends on your RLC load is in parallel with C or L. You can have a series resonant source coupled to in parallel from one part to load in parallel.

Parallel hi-Q tanks amplify voltage.

 

[StoyanElectroTech]

Is it possible under any conditions ? Lets say i use very quality components
?

 

[Nigel Goodwin]

Is it possible under any conditions ? Lets say i use very quality components
?

You appear to be trying to achieve perpetual motion? (getting more energy out than you put in), so no, it doesn't matter how good your components are - it's not possible.

 

[StoyanElectroTech]

"Q" >= 200 ? Yes or No

 

[StoyanElectroTech]

That depends on your specs for fo, current, voltage, stored energy and dissipated energy, DCR, ESR, load R , excitation f error and finally total stored energy and transfer energy efficiency or loaded Q.

A self resonant oscillator is essential with these high Qs and DCR increases with temp. so temperature coefficient if not copper. It could be water-cooled Cu tubing.

Certainly its a challenge , and conjugate matched source to load is ideal for MTTP or much lower Z(f) source impedance for max efficiency.

I think you meant series resonance for stored current amplification. Q= Zo/Rs = √(Ls/Cs) / Rs
where ESR (C) + DCR (L) + RdsOn = Rs

Parallel hi-Q tanks amplify voltage.

You have current amplifier only in Parallel Resonance . Under ideal conditions I_Inductor = I_Capacitor >> I_Input (Q times)