Electronic – Ferrite core StepUP Transformer Design

The DC resistance of the IRF540N could give a clue: -

If the gate drive is 5V, the volt drop will be 0.25V at 20A (4 devices in parallel). Given the numbers in your question I think you are using a 12V supply for the power to the primary centre-tap. That's going to be about a 2 or 3% loss of regulation just for the IRF540Ns.

You primary leakage inductance is a staggeringly high 21uH. At 50kHz this is an impedance of 6.6 ohms. This will be a significant problem to getting good regulation as this inductance can be regarded as not contributing to ampere-turns delivered to the secondary.

I expect you may mean 2.1 uH but even at this level it's going to be an issue (bigger than the MOSFETs).

Not mentioned in the question is the ability of the DC power supply to maintain 12V - this could be drooping to maybe 11.5 volts.

Also notable is the output capacitor - it looks like you may have over-stressed it off-load with the 800V DC output - it's a 450V capacitor - do you think it will be performing as well. Nevertheless I'd consider removing it for doing the lamp load tests because the output will be fairly flat due to the square wave drive and output.

Lets start with a few basics here

$$V = N \dfrac{d}{dt} \Phi = N \cdot A_e \dfrac{d}{dt}B$$

Here \$ B \$ is flux density and for a typical ferrite running at 20kHz I would personally be designing for a flux density swing of around \$200 \text{ mT}\$

Assuming the input voltage is is 27 volts peak to peak maximum, 50% duty cycle @ 20kHz \$ dt = 25 \mu s\$

We now know everything except \$A_e\$ and \$N\$ so we can rearrange the above equation to find \$ N \cdot A_e\$

$$N \cdot A_e = \dfrac{V \cdot dt}{dB} = \dfrac{27 \cdot 25 \times 10^{-6}}{0.2} = 3.38 \times 10^{-3}$$

Note that this product is using Area in SI units (\$m^2\$) and the data sheet probably gives \$A_e\$ in \$\text{mm}^2\$ so we want \$3380000\text{mm}^2\$

In theory any combination of \$N \cdot A_e\$ will work but we note the more turns we have the more area is needed for the windings and the size of the wire we need is set by the current it needs to carry. A typical figure for wire size is to assume a maximum current density of \$ J= 4.5 \cdot \dfrac{\text{amp}}{\text{mm}^2}\$

Given our current of 7.5 amps this requires a wire diameter of \$ \sqrt{\dfrac{4 \cdot I}{J \cdot \pi}}\approx 1.5\text{mm}\$

So we can have a core with a lot of turns and small \$A_e\$ or few turns and big \$A_e\$ the most economic design will be the smallest core into which the turns will fit and if we assume that the wire packs poorly the area required for a single turn is \$A_t = (1.5 \text{mm})^2 = 2.25 \text{mm}^2\$

We can thus work out the area required for the winding \$ A_w = N \cdot A_t \Rightarrow N = \dfrac{A_w}{A_t}\$

Substituting this into the equation above

$$N \cdot A_e = \dfrac{A_w \cdot A_e}{A_t}=\dfrac{V \cdot dt}{dB} \Rightarrow A_w \cdot A_e = \dfrac{A_t\cdot V \cdot dt}{dB} = 7605000 \text{mm}^4$$

Where \$A_e\$ is the core effective area and \$A_w\$ is the available winding area taking into account the bobbin.

Now the coil you have designed will have much more inductance than you want so you will need to introduce an air gap. I usually don't try to calculate this but ask my winding house to gap the core, in the center leg to give me the inductance I want.

The boost converter can be designed in a similar way but I'd ask further advice as at such high frequencies skin and proximity effects are significant.