Converter LLC. A variety of voltage amplification formulas give different results. How to choose the right formula?

[Glebiys]

Hello,

After reviewing various LLC converter design books, I have identified several different voltage gain formulas.

1) Half-Bridge Gain Equation (YouTube)

Result: 0.90

2) TI LLC handbook (I did not understand what j is, so I did not take it into account in the calculations.)

Result: 0.87

3) Infineon

Result: 0.96

4) ST

Result: 0.49 (??? over low)

5)

Result: 0.96

Notes:
1) Lr = 40uH / Lm = 200uH / Cr = 7nF / Ln = 5 / Fn = 1.12 (normalized frequency) / Qe = 0.46 (quality factor) - used this data in all calculations.
2) In some formulas it was just "Ln" (Lm / Lr), in others it was "m" (Lm-Lr / Lr).

All but two formulas gave different gain values.

I am stuck.

I am calculating the LLC Full-Bridge converter, but these formulas were described in Half-Bridge topology designs. I did not find information about how the formulas depend on Full-Bridge / Half-Bridge.

I know that Half-Bridge is exactly half the power, and Full-B is everything.

Questions:
1) Why do these formulas give different results? Which one is more accurate? Perhaps there is an even better formula.
2) Does the voltage gain formula depend on the selected topology? (H-bridge / F-bridge). If so, what exactly is changing?


Thank you!

 

[Easy peasy]

Proper design of an LLC is not easy, 1/2 Bridge is good up to 1.5kW if you know what you are doing, and FB above that.

The max gain is really dependent on the worst case combination of Vin range, Vout range and max loading, going from a low Vin to a high Vout at max load requires more gain, higher Q, ( or adjust the turns ratio on the Tx ) - going to a low Vout requires a higher max freq and/or more use of the steep parts of the gain curves ( higher Q ) to get a low Vout at a reasonable max freq.

Going for a high Q design means there will be more circulating current at no load and at nominal full load.

Most engineers do all this in LT Spice and adjust, Lr, Lmag, Cr, turns ratio, Vin, Vout until they get a good compromise for their app

you also get a feel for all the interactions this way too

good luck.

 

[BradtheRad]

I know that Half-Bridge is exactly half the power, and Full-B is everything.

Did you intend to say Half-Bridge produces half the supply voltage? It typically appears at the join between a 2-capacitor stack as I've seen in push-pull arrangements using a half-bridge.
I could be wrong.

 

[Glebiys]

@Easy peasy , @BradtheRad ,

Thank you for the answer!


Yes, the power is about 2kW, so I chose Full-Bridge.

Didn't know about LT-Spice, it should help!


@BradtheRad , yes.



For example, a full-bridge gives a range of +Vin...- Vin (+400...-400VDC), and a half-bridge only half of +Vin...GND (+400...0VDC).

The half-bridge is silent for exactly half the duty cycle and half of the power is lost.

 

[mtwieg]

The first two equations are equivalent. The second one expresses it as the magnitude of a complex number (j is sqrt(-1) ).

Equations 3, 4, and 5 are also equivalent.

edit: and 1 and 5 are equivalent, so they all should be giving the exact same answer. Show your work!

 

[Easy peasy]

half bridge simply applies half the voltage and twice the current compared to the full bridge equivalent, it operates in exactly the same manner otherwise ...

 

[Glebiys]

After a while, I rechecked the formulas and it turned out yes, they are similar. In the first formula, I misidentified the last parenthesis of the divider. (I counted it as fn to the -1)

I have plotted gain plots in Octave on several formulas:

Spoiler: #1 - Youtube

Spoiler: #3 (Infineon)

Spoiler: #5 (Energies)

Octave code: https://pastebin.com/wbySVS2K

As a result, the formulas from Infineon and from YouTube are really equivalent and show the gain. I also checked the formula from Energies, but for some reason its gain is shifted to the right (I rewrote the formula several times).

The only thing in Infineon instead of Ln (Lm / Lr) uses m (Lr + Lm / Lr). (Ratio of total primary inductance to resonant inductance).

 

[Easy peasy]

the top 2 curves are virtually identical - a series of curves that reach up to 2 on the LH side are generally the most useful - note the Q for each curve and translate this to actual output load

 

[BradtheRad]

I rewrote the formula several times).

Did you try one of those fancy math routines that can rearrange a formula automatically as you move something to either side of the equals sign ? Example, Mathematica or Wolfram or similar?

 

[Glebiys]

@Easy peasy, Ok. Thank you!

Did you try one of those fancy math routines that can rearrange a formula automatically as you move something to either side of the equals sign ? Example, Mathematica or Wolfram or similar?

No, I've seen a lot of people solve in WolframAlpha, but haven't done it myself yet.

I'll take a look.