Reading the LLC Gain Curve: How m and Q Shape Your Design

FHA replaces the square-wave bridge voltage and the rectifier with their fundamental components. The rectifier and load become an equivalent AC resistance R_ac, and the tank becomes a simple linear circuit. The result is a closed-form gain that depends on only three numbers: the normalized frequency F, the inductance ratio m and the quality factor Q.

Series resonance

f_r = 1 / (2π · √(Lr · Cr))

Lower resonance

f_r2 = 1 / (2π · √((Lr + Lm) · Cr))

Inductance ratio

m = (Lr + Lm) / Lr

Quality factor

Q = √(Lr / Cr) / R_ac, R_ac = 8 · n² · R_load / π²

FHA gain

K = (m−1)·F² / √((m·F² − 1)² + F²·(F² − 1)²·(m − 1)²·Q²), F = f_s / f_r

• Above f_r (F > 1): the gain is below 1 and falls as frequency rises. The primary switches keep zero-voltage switching (ZVS), but the secondary rectifier current is cut off sharply, so rectifier reverse recovery matters if you use silicon diodes.
• Between the gain peak and f_r: the gain is above 1. The primary still has ZVS and the rectifier current returns to zero naturally before each half-cycle ends (ZCS on the secondary). This is the region that lets an LLC boost at low input voltage.
• Below the gain peak: the tank input impedance becomes capacitive. ZVS is lost, the MOSFET body diodes can be hard-commutated, and the slope of the gain curve reverses, which turns the control loop into positive feedback. A design must never enter this region in steady state.

At exactly f_r the gain is 1 for every load (in FHA). That is why designers set the turns ratio so the nominal operating point sits at resonance: efficiency is best there and the operating frequency barely moves with load.

The table below shows the FHA peak gain and the normalized frequency where it occurs, computed from the formula above. It is the quickest way to see the trade-off.

m = 4, Q = 0.2 / 0.4 / 0.6

peak gain 3.39 / 1.79 / 1.32

m = 6, Q = 0.2 / 0.4 / 0.6

peak gain 2.52 / 1.39 / 1.11

m = 8, Q = 0.2 / 0.4 / 0.6

peak gain 2.10 / 1.21 / 1.05

• Lower m (smaller Lm relative to Lr) gives more peak gain and a narrower frequency range, but more magnetizing current circulates all the time, raising conduction loss at light load.
• Higher Q (heavier load or larger Lr/Cr impedance) flattens the curve. The maximum gain you can reach at full load is set by the Q of the full-load curve, so this is the curve that limits the minimum input voltage.
• A larger Lr relative to Cr raises Q for the same load. Choosing the characteristic impedance √(Lr/Cr) is therefore a choice about how much gain margin you keep at full load.

At no load (Q → 0) the FHA gain does not fall to zero as frequency rises. It approaches (m − 1)/m. For m = 6 that is about 0.83. If high input voltage and light load together require a gain lower than this, frequency control alone cannot regulate the output, no matter how high the frequency goes.

Real converters are often worse than FHA predicts here, because transformer winding capacitance and rectifier junction capacitance add their own resonances at high frequency. This is why many LLC controllers switch to burst mode at light load, and why a design with a very large m can struggle to regulate at no load.

ZVS on the primary requires enough current at the switching instant to charge and discharge the switch output capacitances during the dead time. Near resonance that current is essentially the peak magnetizing current, which is set by Lm, not by the load.

Peak magnetizing current (near f_r)

I_m,pk ≈ n · V_out / (4 · L_m · f_s)

Approximate ZVS condition (half bridge)

I_m,pk · t_dead ≥ 2 · C_oss,eq · V_in

The ZVS condition above is a simplified charge balance and treats the current as constant during the dead time. It is a sanity check, not a substitute for looking at the switch-node waveform. If it fails, you can lengthen the dead time, lower Lm (lower m), or choose switches with less output charge.

1. 1A half bridge with Lr = 38 µH, Cr = 66.6 nF and Lm = 190 µH gives f_r ≈ 100 kHz, f_r2 ≈ 41 kHz and m = 6.
2. 2With n = 4 and a 48 V / 500 W load (R_load = 4.6 Ω), R_ac ≈ 60 Ω and √(Lr/Cr) ≈ 23.9 Ω, so Q ≈ 0.40 at full load.
3. 3From the table, the full-load peak gain is about 1.39 at F ≈ 0.49, so the converter can reach roughly 39 % more output than at resonance before entering the capacitive region.
4. 4The peak magnetizing current at 100 kHz is about 4 · 48 / (4 · 190 µH · 100 kHz) ≈ 2.5 A, which is available for ZVS at every load.
5. 5At no load the minimum gain is about 5/6. The required gain is 2·n·V_out / V_in, so regulation by frequency alone is lost above roughly 460 V input (FHA estimate).

These are the default values of the LLC plant and PI Bode tool on this site, so you can reproduce the numbers and move the operating point yourself.

FHA is accurate close to resonance and becomes optimistic far below or far above it, at very light load, and wherever higher harmonics carry significant energy. Use it to choose m, Q and the turns ratio, then confirm the gain at the corners of your operating range with a time-domain simulation and, finally, with measurement.