# Half Bridge Impedance Matching

I’m currently building a half bridge induction heater and can’t understand impedance matching for the life of me.

Here’s an example of a basic half bridge induction heater. How and why was 45uH calculated for the matching inductor? Note that I do know how to calculate and simplify impedances, I’m just confused on how the matching inductor value was calculated.

• What frequency does it switch at? Have you tried to simulate it? Jan 13 at 8:02
• I did simulate it, running at resonance frequency, in this case 211.9 kHz. It works perfectly. However I’m not using the schematic’s resonant tank values so I need to know how the matching inductor value was chosen. Jan 13 at 8:40
• It was chosen to ensure that the load on the half-bridge was always inductive. Jan 13 at 9:59
• The value depends on work coil inductance and Q factor. Jan 13 at 13:43
• @Andyaka is probably right -- but the impedance seen by the inverter can still go capacitive, at frequencies near work coil resonance and ignoring the blocking cap, when the Q factor is high; in effect the circuit reduces to a series-resonant equivalent. As Q isn't given here, it isn't strictly clear if this can happen. Jan 13 at 13:53

simulate this circuit – Schematic created using CircuitLab

Note we omit the coupling capacitor for convenience. Choose a value such that reactance is much less than (say <10%) the minimum inverter load impedance, at minimum operating frequency. (At such low ripple fraction, it can even be polyester type, though higher current handling can be found in polypropylene.)

This is an implicit L-match network, the equations of which can be referenced from RF tools, or solved directly:

$${L_M}^2 = \frac{1}{\frac{\omega_r}{Q R_I L_W} - \frac{1}{Q^2 {L_W}^2}}$$

Where $$\L_M\$$ is the matching inductance, $$\L_W\$$ is the work coil inductance, $$\\omega_r\$$ is the resonant frequency, $$\Q\$$ is the work coil Q factor (other components assumed ideal, or their losses rolled into an equivalent value here), and $$\R_I\$$ is the desired inverter load resistance.

We can rearrange this to an alternative form,

$$L_M = L_W Q \sqrt{\frac{R_I}{R - R_I}}$$

where $$\R\$$ is the resistance of the tank circuit, $$\R = Q \omega_r L_W\$$. Of course either sign is an acceptable solution to the equation, but we choose positive for convenience.

Note that we can split the circuit into the cascade of two resonant circuits, series matching to parallel:

simulate this circuit

At resonance, the parallel tank reactances disappear, and we have an R-loaded series resonant L-match circuit, LM-CM-R. We can virtually split $$\C_T\$$ into corresponding series and parallel resonant fractions $$\C_T = C_M + C_W\$$, and this condition occurs when $$\C_W = C_T \frac{L_M}{L_W + L_M}\$$ and $$\C_M = C_T \frac{L_W}{L_W + L_M}\$$.

We can express the resonant frequency in this way, or by noting that the circuit has an inductor divider, the Thevenin equivalent of which has them acting in parallel. Or even more directly, since the inverter is assumed zero impedance (a good approximation for a high-efficiency switching circuit), by superposition, they are simply in parallel. So we have:

$$\omega_r = \sqrt{\frac{L_M + L_W}{C L_M L_W}}$$

Note that LM is a term in ωr, so we have a circular reference if we try to solve based on given values; this goes away if we assume a fixed operating frequency and allow CT for example to shift, and adjust our calculations until close to the intended value.

We can simulate some values to prove it out:

simulate this circuit

Without loss of generality, $$\C_T\$$ and $$\L_W\$$ have been fixed ($$\Z_0 = 1\,\Omega\$$), and four values of $$\R\$$, $$\L_M\$$ have been chosen. The peak input current in each was measured. They give RI of:

1. LM = 47µH, R = 10Ω (Q = 10): RI = 20.33Ω; expected LM: 8.52Ω
2. LM = 47µH, R = 50Ω (Q = 50): RI = 9.747Ω; expected LM: 9.36Ω
3. LM = 10µH, R = 50Ω (Q = 50): RI = 0.5963Ω; expected LM: 0.593Ω
4. LM = 10µH, R = 10Ω (Q = 10): RI = 2.489Ω; expected LM: 2.308Ω

The low-Q solutions don't work well, more or less because no real solution exists (notice the second relation blows up when RI ≈ R); this is a byproduct of the assumption used to derive these relations (inverter current is real at resonance). In general there are three conditions: inverter current always inductive, inverter current just reaches zero reactance (double root), and inverter current reaches zero at two points (going capacitive inbetween). This at least requires a quadratic to express, but an exact solution of LM requires solving a quartic or quintic polynomial, which obviously isn't very convenient to write out and is best solved numerically. Alternately, you can repeat the calculation and adjust values in a simulator until impedance and frequency are as expected; the system should be stable with quickly converging results.

Note that typical work coil Q factors range from 3-10 (close fit between coil and work, ferrous material below Curie temperature), to 10-30 (tight fit, highly conductive materials; modest fit, most materials), to 50+ (loose fit). I find it easy enough to estimate Q based on work coil and material, and an exact figure can be established later.

Note also, the Q generally varies during operation, as work temperature rises, and magnetism disappears (if applicable). The load on the inverter therefore varies as well.

Typically, the inverter is designed for excess capacity: for an inverter factor of 2, it can deliver nominal power at up to twice the nominal load current (i.e. Pmax / Vout), or down to half the nominal supply (if it's a variable-input type control).

There is some gain, in terms of inverter factor, from making use of frequency-shift control, but notice this has a significant downside: the control loop has a double pole, which goes as the difference between driven and resonant frequencies, with the same time constant as the tank. This makes compensation quite challenging, especially if you have to operate into high-Q loads.

The original source to that schematic, by the way, appears to be Richie Burnett's work. The page is a classic in the internet induction heating community:
High Frequency Induction Heating | richieburnett.co.uk

Sadly it appears to have been widely plaigarized now. I see a number of sites the schematic alone appears in:

A more responsible search would repeat steps for each figure, and passages of text. Just from this result, I imagine quite many other instances can be found.

• I read your induction heater series as well as Richie's before asking this question but was still confused as to how the matching inductance was chosen. This certainly sheds some light and will play around with these calculations. Also, I assumed the second 𝐿𝑀 in the second paragraph was meant to be 𝐿W. Please correct me if I'm wrong. Thanks! Jan 13 at 21:11
• @ElectronicsNoob Hah, good catch. Back in the day, I think none of us had the know-how to calculate, and it was a matter of just picking values until it worked. I suggest preferring series resonant anyway, but with simulation or a more detailed solution (the polynomial I hinted at, but stopped short of providing for risk of errors), LCRC can be solved as well. Jan 13 at 22:11
• I did indeed plan to start out with series resonant but after reading your and the other guy's notes I came to the realization that the series resonant half-bridge forces the transistors to take on the full resonant current... Or am I wrong? I'm still learning. Jan 13 at 22:52
• Most likely a matching transformer is used. Which also accounts for range, i.e. a number of tap settings, avoiding needing excessive inverter capacity, or having very limited Lw and Q range. Typical examples use a single-turn secondary and various primary tappings. (For LCRC, usually a tapped or selectable Lm is used.) Jan 13 at 23:34

The "Matching" inductor is chosen for the "appropriate" transformation of the voltage-switched input.
See the reported currents.

It modifies only the "low" resonant frequency.