LLC resonant converter — first-harmonic design

A worked design for a half-bridge LLC stage, carried out the way the textbooks do it: replace the square wave by its first harmonic, replace the rectifier and its load by the equivalent resistance they present at that frequency, and what is left is a linear network that can be written down. Steigerwald's 1988 comparison of resonant topologies is where the method comes from, and every converter application note since repeats it. It is here because it is the shape a real engineering worksheet has: a page of stated inputs, a chain of definitions that never mutates anything, a complex impedance, three gain curves that have to be compared against each other, and a verdict at the end that says whether the design meets the specification.

The specification.

V_{bus}

400 V

V_{out}

48 V

P_{out}

1000 W

f_{sw_min}

60 kHz

f_{sw_max}

200 kHz

The bridge drives a square wave between 0 and V_bus. Only its first harmonic is kept: peak 4/π of the half-swing, and the network sees the rms of that.

V_{sq_pk}

\frac{4}{π} \cdot \frac{V_{bus}}{2}

\frac{4}{π} \cdot \frac{400 V}{2}

254.648 V

V_{sq_rms}

\frac{V_{sq_pk}}{\sqrt{2}}

\frac{254.648 V}{\sqrt{2}}

180.063 V

The rectifier and the output load together look resistive to that harmonic. The 8/π² is the same 4/π twice over — once for the voltage the bridge applies and once for the current the rectifier draws.

R_{load}

\frac{{V_{out}}^{2}}{P_{out}}

\frac{{(48 V)}^{2}}{1000 W}

2.304 Ω

I_{out}

\frac{P_{out}}{V_{out}}

\frac{1000 W}{48 V}

20.8333 A

n

\frac{V_{bus}}{V_{out}}

\frac{400 V}{48 V}

8.33333

R_{ac}

\frac{8}{π^{2}} \cdot n^{2} \cdot R_{load}

\frac{8}{π^{2}} \cdot {8.33333}^{2} \cdot 2.304 Ω

129.691 Ω

The tank is chosen from three numbers: where it resonates, how heavily it is damped at full load, and how much magnetising inductance sits across it.

f0

100 kHz

Q_{nom}

0.45

Ln

4

Everything else follows. Z0 is the characteristic impedance the chosen Q asks for at this load; the resonant frequency then fixes the two components.

Z0

Q_{nom} \cdot R_{ac}

0.45 \cdot 129.691 Ω

58.361 Ω

T0

\frac{1}{2 \cdot π \cdot f0}

\frac{1}{2 \cdot π \cdot 100 kHz}

1.59155 \times 10^{−6} s

Lr

Z0 \cdot T0

58.361 Ω \cdot 1.59155 \times 10^{−6} s

92.8844 µH

Cr

\frac{T0}{Z0}

\frac{1.59155 \times 10^{−6} s}{58.361 Ω}

27.2708 nF

Lm

Ln \cdot Lr

4 \cdot 92.8844 µH

371.538 µH

Worth checking that the parts that came out reproduce the numbers they were derived from, since a slip in the algebra above would not show up anywhere else on the page.

{f0}_{check}

\frac{1}{2 \cdot π \cdot \sqrt{Lr \cdot Cr}}

\frac{1}{2 \cdot π \cdot \sqrt{92.8844 µH \cdot 27.2708 nF}}

100 kHz

Q_{check}

\frac{\sqrt{\frac{Lr}{Cr}}}{R_{ac}}

\frac{\sqrt{\frac{92.8844 µH}{27.2708 nF}}}{129.691 Ω}

0.45

The second resonance, with the magnetising inductance in series. The converter cannot be operated below it, so it is the floor on the sweep.

f_{min_res}

\frac{1}{2 \cdot π \cdot \sqrt{Cr \cdot (Lr + Lm)}}

\frac{1}{2 \cdot π \cdot \sqrt{27.2708 nF \cdot (92.8844 µH + 371.538 µH)}}

44.7214 kHz

The tank impedance seen by the bridge: Lr and Cr in series with Lm in parallel with the reflected load. Complex, because the phase is the whole point — the bridge has to see an inductive load to switch softly.

fn X_Lr defined

fn X_Lm defined

fn X_Cr defined

fn Z_par defined

fn Z_in defined

Taken apart the three ways a worksheet asks for.

Z_{at_f0}

Z_in (f0)

Z_in (100 kHz)

=(99.1036 + 55.0575i) ohm

R_{in}

Re (Z_{at_f0})

Re ((99.1036 + 55.0575i) Ω)

99.1036 Ω

X_{in}

Im (Z_{at_f0})

Im ((99.1036 + 55.0575i) Ω)

55.0575 Ω

mag

| Z_{at_f0} |

| (99.1036 + 55.0575i) Ω |

113.37 ohm

phase

\frac{\arg (Z_{at_f0}) \cdot 180}{π}

\frac{\arg ((99.1036 + 55.0575i) Ω) \cdot 180}{π}

29.0546

Positive is inductive, which is the half of the plane the bridge must stay in.

fn phase_deg defined

{soft}_{switching}

phase_deg (f_{sw_min}) > 0

phase_deg (60 kHz) > 0

1

The gain, normalised to the resonant frequency. This is the expression the whole first-harmonic method exists to produce.

fn gain defined

A family of curves is written by naming its members, so the load each one stands for has a name on the page rather than a number in a call.

fn light defined

fn nominal defined

fn heavy defined

plot (light, nominal, heavy, f_{sw_min}, f_{sw_max})

Unity at resonance, whatever the load — the property that makes f0 the design point. Below it the converter boosts, above it it bucks.

nominal (f0)

nominal (100 kHz)

1

light (f0)

light (100 kHz)

1

heavy (f0)

heavy (100 kHz)

1

Where the nominal curve peaks, and how much gain is there. The slope is exact rather than a difference quotient, so the peak is found by asking where the slope crosses zero.

fn slope defined

f_{peak}

roots (slope, f_{min_res}, f0)

roots (slope, 44.7214 kHz, 100 kHz)

56.0867 kHz

M_{peak}

nominal (f_{peak})

nominal (56.0867 kHz)

1.62792

The gain the specification actually demands, at both ends of the input range.

V_{bus_min}

340 V

V_{bus_max}

420 V

M_{needed_max}

\frac{n \cdot V_{out}}{V_{bus_min}}

\frac{8.33333 \cdot 48 V}{340 V}

1.17647

M_{needed_min}

\frac{n \cdot V_{out}}{V_{bus_max}}

\frac{8.33333 \cdot 48 V}{420 V}

0.952381

margin

M_{peak} - M_{needed_max}

1.62792 - 1.17647

0.451451

verdict

{\begin{cases} if margin > 0 then "gain margin available" else "redesign: peak gain too low" & if margin > 0 \\ if margin > 0 then "gain margin available" else "redesign: peak gain too low" & otherwise \end{cases}

{\begin{cases} if 0.451451 > 0 then "gain margin available" else "redesign: peak gain too low" & if 0.451451 > 0 \\ if 0.451451 > 0 then "gain margin available" else "redesign: peak gain too low" & otherwise \end{cases}

="gain margin available"

The impedance curve, drawn as a magnitude in ohms so the axis is a number.

fn Z_mag defined

fn Z_ohm defined

plot (Z_{ohm}, f_{sw_min}, f_{sw_max})

The lowest impedance across the operating range bounds the tank current, and min over a sampled vector is how a worksheet asks that without a loop.

sweep

range (f_{sw_min}, f_{sw_max}, 10 kHz)

range (60 kHz, 200 kHz, 10 kHz)

=[60000 Hz, 70000 Hz, 80000 Hz, 90000 Hz, 100000 Hz, 110000 Hz, 120000 Hz, 130000 Hz, 140000 Hz, 150000 Hz, 160000 Hz, 170000 Hz, 180000 Hz, 190000 Hz, 200000 Hz]

Z_{sweep}

map (Z_{ohm}, sweep)

map (Z_{ohm}, [60000 Hz, 70000 Hz, 80000 Hz, 90000 Hz, 100000 Hz, 110000 Hz, 120000 Hz, 130000 Hz, 140000 Hz, 150000 Hz, 160000 Hz, 170000 Hz, 180000 Hz, 190000 Hz, 200000 Hz])

=[69.8676, 82.2021, 94.0515, 104.415, 113.37, 121.196, 128.16, 134.482, 140.327, 145.824, 151.067, 156.128, 161.059, 165.902, 170.687]

Z_{low}

\min (Z_{sweep})

\min ([69.8676, 82.2021, 94.0515, 104.415, 113.37, 121.196, 128.16, 134.482, 140.327, 145.824, 151.067, 156.128, 161.059, 165.902, 170.687])

69.8676

I_{tank_max}

\frac{V_{sq_rms}}{Z_{low} ohm}

\frac{180.063 V}{69.8676 ohm}

2.57721 A

Currents and stresses, which is what the design is finally judged on.

I_{pri}

\frac{P_{out}}{V_{sq_rms}}

\frac{1000 W}{180.063 V}

5.5536 A

V_{Cr_pk}

I_{pri} \cdot \sqrt{2} \cdot X_Cr (f0)

5.5536 A \cdot \sqrt{2} \cdot X_Cr (100 kHz)

458.366 V

E_{Lr}

\frac{1}{2} \cdot Lr \cdot {(I_{pri} \cdot \sqrt{2})}^{2}

\frac{1}{2} \cdot 92.8844 µH \cdot {(5.5536 A \cdot \sqrt{2})}^{2}

2.86479 mJ

Rated against the specification.

V_{Cr_rating}

630 V

{capacitor}_{ok}

V_{Cr_pk} < V_{Cr_rating}

458.366 V < 630 V

1