I designed an inductance heater. Its inductor is calculated 50 uH as below:


And I draw 9.4 A AC current at 33 Khz of it.

AC current fft

But, I want to know how to calculate the transferred energy to the working piece. Take this schematic into account:


Its power graph as below:

power graph

As you see some of the power is Reactance but there is always magnetic flux even when the power on the inductor is negative. Do I need to take the absolute value of the power as below?

absolute vale of the power

Or it would be wrong and there is another calculation for the energy transfer ratio?

  • \$\begingroup\$ You can compute energy in an inductance with the formula 1/2*L*i^2 \$\endgroup\$ – MathieuL May 23 '16 at 22:34
  • \$\begingroup\$ The power consume by the inductance is only the resistive part. You can't multiply directly the voltage by the current. The result of that is the apparent power in VA. If you want to get the real power you need to know the power factor \$\endgroup\$ – MathieuL May 23 '16 at 22:38
  • \$\begingroup\$ The power is transfered to the work piece by inducing a current in the workpiece. An inductive heater is sort of like a transformer with a shorted secondary (and weak coupling between primary and secondary). In order to establish how much power is transferred, you probably need to measure the impedance with and without the work piece. Any increase in the resistive component of the impedance represents the effective resistance of the load. That is what I think. Not being an expert. You cannot calculate power transfer without considering the effect of the workpiece. \$\endgroup\$ – mkeith May 24 '16 at 2:05

Energy transfer in an inductive heater happens only due to the coupling of the primary inductance with the secondary load, which appears as a weakly coupled inductive load with a near-dead-short across the coil.

You can see an attempt to reverse-engineer a commercially available induction cooker here, which includes a SPICE simulation.

The inductive heating element is modelled like so:


simulate this circuit – Schematic created using CircuitLab

Note that K1 here (the coupling between L1, the primary coil and L2, the load model) is 0.99 - i.e. very strongly coupled. I am not sure why this model was chosen - there is no evidence of it having been measured. To then find the power being transferred, you can simply measure the RMS power in the heating load.

As a side-note - to calculate the real power being dissipated in the inductor in your circuit, you need to calculate the average of the instantaneous power dissipation over a whole number of cycles (as it seems you have almost done in the green power graph - make sure you average over an integer number of peak-to-peak or trough-to-trough cycles, not the 11.5 cycles you have in that diagram). You will find it is a low number (all your real power is dissipated in that sharp peak) - and this makes sense, because there is no load in your schematic.


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