Induction heater design - Current transformer windings ratio

I am currently building an induction heater and would like to verify my approach to this as I'm uncertain if it is correct. Below is the topology of the resonant tank circuit I have constructed. The tank circuit illustrated is coupled to a H bridge driver via a current transformer, with a resonant frequency of 12.841KHz. simulate this circuit – Schematic created using CircuitLab

Each capacitor is rated at a peak current loading of 25A at 10KHz, achieving 50A in their current configuration. Therefore, considering peak current flow and factoring it against the ESR of the circuit (0.013 Ohms), the peak voltage of the secondary circuit must not exceed 0.65V assuming no impedance due to reactance is present. However, this would only be capable of producing 32.5W (0.65V x 50A) which is far below what I am aiming to achieve.

The primary side of the circuit supplies 48V (960VA) to the primary coil of the CT. The transformer ratio calculated gives: Vp/Vs = 48V/0.65V = 73.8 turns. This seems excessive and likely to result in a very high inductance/impedance on the primary side of the circuit, inhibiting current flow / overall power.

My questions are:

1) Is my approach to designing this circuit correct so far? If not, what mistakes am I making?

2) Does my power calculation reflect the total power of the circuit or just the power lost?

• Just an aside, having done something like this before. Use polypropylene capacitors. Low ESD and losses. My first caps overheated and failed. And just an "gut feeling" - smaller caps and larger inductance, using thicker wire/tubing if necessary – Dirk Bruere Jun 14 '17 at 14:22
• Why must the secondary's peak voltage not exceed 0.65v? – Ben G. Jun 14 '17 at 15:03
• Because 50A x 0.013 = 0.65V. If the voltage exceeds this the capacitors are likely to be damaged. – Junkers Jun 14 '17 at 22:48
• "assuming no impedance due to reactance is present" <- and why would you assume that? This circuit is entirely built out of reactive components, after all.. – immibis Mar 26 '18 at 4:50
• Because the circuit is at resonance, hence the capacitive and inductive reactances are of opposite and equal magnitude and therefore should cancel out. – Junkers Mar 27 '18 at 5:56

It is not clear where you got the "ESR of the circuit (0.013 Ohms)". Your circuit shows a parallel tank circuit. Perhaps you intended to drive the load as a series circuit?

Maybe you meant the load resistance (equivalent resistance of the coil). A value of .013 ohms would be reasonable for a coil with 4 uHy. This gives a coil Q of about 23 (XL/R).

A coil Q of 23 is quite high for induction heating.

A coil Q of 23 might be an empty coil or a lightly loaded coil with non-ferrous load.

Using an input voltage of 18 volts peak keeps your capacitor current (red color) below 50 amps peak. Green color is current in R1.

The power into the resistor is only about 16 watts. But that is to be expected with such a lightly loaded induction coil.

To achieve higher power you should try to more tightly couple your heated load to the coil. This will increase R and decrease the coil Q (XL/R).

You are likely to need capacitors that can handle higher currents if you expect to achieve higher power to your heated load.

• The ESR stated is based on measurements I made with a LCR meter. Each capacitor measured 0.01 Ohms and the induction coil measured 0.008 Ohms. These measurements must be skewed as when I measure the Q of the system (placing the probes across the tank circuit) it measures 14 unloaded. With the workpiece in the coil it drops to around 4. Since posting this I have run this circuit using a H bridge inverter. I've found that depending on the number of turns of the primary transfromer, Vpk-pk voltage can reach 80V while the work coil isn't loaded. I've yet to understand why this is. – Junkers Jun 27 '17 at 9:55

I know very little about induction heaters but I think I hear a small misconception in your question.

Therefore, considering peak current flow and factoring it against the ESR of the circuit (0.013 ohms), the peak voltage of the secondary circuit must not exceed 0.65 V assuming no impedance due to reactance is present.

ESR is equivalent series resistance. When you include this in your circuit simulation you will see that it causes a slight lag on the "internal capacitor" due to its effective R-C time constant. Let's have a look at a simulation: simulate this circuit – Schematic created using CircuitLab

Figure 1. ESR model of one capacitor. Figure 2. Note the slight lag between the NODE1 and NODE2 voltages. The voltage across the ESR is the difference between those two lines.

For the 10 V peak signal we can measure in the simulation that there is about 1.57 V difference between the two at the zero-cross point. This is more than you would like but you should see the idea that the peak capacitor voltage can be greater than 0.65 V and more like 3.5 V.

Your problem now is that you are only running about 12 A peak through the capacitors as shown by the current curve.

• Where do you get the 3.5V value from? – Junkers Jun 27 '17 at 9:39
• If 10 V peak gives you 1.57 V across the ESR then dividing both roughly by three would get you inside your 0.65 V limit. Note that I've calculated using peak whereas your ESR will be rated for RMS. – Transistor Jun 27 '17 at 10:03