If it is for heating I guess AC is as good as DC. I would construct a toroidal transformer with just 1 secondary winding (depending on input voltage). To achieve the high current place several secondary windings in parallel, and make sure their wire length is exactly the same.

**edit**

You can make to output voltage/current variable by feeding the transformer's input from a variac:

**edit 2** (re your digital regulation)

I've been thinking about this for a while and I think the best idea is not having to switch the high current in the first place. Any other components than the metal strips themselves and the connections to them will cause losses of hundreds of Watts at least.

Maybe we still can use our transformer, and do the switching on the primary side, then we won't have to worry about sub-milliohm transition resistances. I would use a DC voltage on the transformer's primary, chopped by a MOSFET. The duty cycle will determine the secondary's current.

**edit 3** (merge with other answer upon KV's suggestion)

First thing to make a note of is the **vacuum**. It means that all cooling will have to go through **conduction** through the wall of your vacuum chamber, since your temperatures won't be high enough to lose much heat through radiation, and of course there's no convection in a vacuum. This is also an issue for the heat dissipated in the load (the metal foil).

Going from 12V DC is a tall order. The standard way to go from a higher voltage and lower current to a lower voltage at higher current is of course an **SMPS**. Even at a low-ish 66% efficiency the 12V supply would only need to deliver 6.25A (for 75W). Piece of cake, it seems. However, the coil current is in the range of the output current, with peaks going higher. There are power coils which can handle 100A, but these have such low inductance that they need very **fast switching**, which causes very **high switching losses** in the MOSFETs. And then there's also the power lost as radiation, which may be *a lot*. Normal Schottly diodes are also out, so you'll need **synchronous rectification** using MOSFETs.

Talking about synchronous rectification: this is also an option for an AC power supply. You'll have a few voltage drops, however low, so you'll have to start with a voltage a bit higher than the 0.1V. The efficiency won't be high either, though even an extra 100mV drop will cause only 50W loss, so I think this is acceptable.
A classic diode rectifier is out due to the high power losses, and that's where the **synchronous rectification** comes in. You'll get a rectified sine, which is the closest you'll get to a proper DC source. (Don't even *think* about capacitors to smooth 500A currents!)

To **measure the current** you can use a couple of these sense resistors from Isabellenhütte.

(Despite several parasitic voltage drops neither of those is suitable for current measurement since we have no control over the resistances involved.) The 0.1m\$\Omega\$ current sense resistor is specified for 200A, so you'll need a number of them in parallel.
The power in the resistors is low, they're specified at 5W maximum, but count on a multiple of that for parasitic resistances. Best would be to weld as much as possible, and mount on a metal wall of your vacuum chamber.

If you use three 0.1m\$\Omega\$ resistors in parallel theoretically you'll have 17mV at 500A. That's not much, but in practice the value may be higher, like 25 or 30mV, due to parasitic resistances. At 100A that will be 5 to 6mV. An **instrumentation amplifier** will help you bring this to a level which is easier for the PWM chopper to work with.

The rest is in the feedback regulator, which is actually a **class D amplifier**, after the measured current is averaged by a low-pass filter.

Don't use a too high chopping frequency; it will only increase the switching dissipation in the MOSFETs, and besides heat is slow, so you won't need sub-millisecond switching.

Plumbing: You'll need a battery of parallel MOSFETs, which I would solder as much as possible on copper bars, to reduce parasitic resistances as much as possible.

The turns ratio will give you the impedance ratio and the voltage ratio. The choice of turns depends on the core **type and size** and the expected voltage that you need to drive through the transformer so that it does not saturate.

Typically one tried to have an equal size coil for primary and secondary for an isolation transformer. There will be some clever calculation to work out the optimal power transfer for a given coil size as more turns will eventually increase the DC resistance too high.

Your turns ratio will be quite high so the primary will be very fine wire. Winding so much on a toroid may be uneconomical in production volumes, pot cores and E-I stacks are still common.

Here is a link to two reference manuals Wolpert's and Whitlocks that can offer you some further guidelines. You will need to find out the parameters of the core you plan to use and see if you can cope with the required turns in the toroid.

## Best Answer

The problem is that the cores being used have no gap, so the inductor saturates. A topology with Flyback type operation stores energy in the core when the switch is on, moving the core up the BH curve. But, for the simple example of discontinuous conduction (DCM), when the switch turns off and current falls to zero, B does not return to zero but to an elevated residual point. So, the usable \$\Delta B\$ is very small and the core is walked into saturation. Continuous conduction (CCM) is even worse in this regard.

Adding a gap moves the residual point nearer to zero, giving a usable \$\Delta B\$. In the case of a gap, the inductance will be determined by the gap, not so much the core \$\mu\$. Consider the inductance of a gapped core inductor; with core cross section \$A_c\$ and gap length \$l_g\$ and winding turns \$n\$:

\$L_g\$ = \$\frac{n^2 A_c \mu _o}{l_g}\$

also relating maximum winding current (\$I_{\text{max}}\$) to gap flux (\$B_{\text{max}}\$):

\$n I_{\text{max}}\$ = \$\frac{B_{\max } l_g}{\mu _o}\$

By starting with a value for \$L_g\$, \$B_{\text{max}}\$, \$A_c\$, and \$I_{\text{max}}\$, it's possible to get an idea what \$l_g\$ and \$n\$ for the inductor would need to be. Let \$L_g\$=100\$\mu H\$, \$B_{\text{max}}\$=0.2T, \$A_c\$=20\$mm^2\$

\$l_g\$ = \$\frac{I_{\max }^2 L_g \mu _o}{A_c B_{\max }^2}\$ = \$\frac{1 Amp^2 100\mu H \mu _o}{20 mm^2 0.2T^2}\$ ~ \$0.16 mm\$

and

\$n\$ = \$\frac{i_{\max } L_g}{A_c B_{\max }}\$ = \$\frac{1Amp 100\mu H}{20 mm^2 0.2T}\$ = \$25 turns\$

This analysis is pretty simplified, leaving a lot out, but gives an idea of what to expect. Designing these types of inductors gets very involved. You might look at "Inductor and Flyback Transformer Design" as a reference.