First of all you need to realize that a 12V SLA battery is not, ever 12V, if it's charging then it sits around 14.3V (but that's dependent on chemistry and temperature) when it's discharging then it can be as low as 10V.
The most robust solution is to make sure that all the equipment can tolerate 10-15 V, because that will allow you to get rid of any output regulator which will waste some energy while on battery.
Almost all electronics that says 12V will easily tolerate 10-15V, much of it will be happy with 8-24V as well, the main exception are computers that feed the 12V input directly to harddisks, those devices really like a regulated 12V.
A good charger will be needed that regulates the float voltage according to temperature and also limits the charging current.
One solution would be:
- A beefy AC/DC off-the-shelf mains supply, a 19V power brick for a laptop could do.
- A Buck DC/DC converter handling charging, but a simple LM317 could also be used.
- A Buck DC/DC converter handling the on-line output regulation.
- A switch (2 FETs) which switches over to battery if the output of the online-converter drops out of regulation.
Buck converters tend to be simpler and more efficient than buck/boost converters, so that's a good reason to prefer those in a design, but if you really want a regulated 12V output then there is no way to avoid a buck/boost converter and the plan becomes:
- A beefy AC/DC off-the-shelf mains supply, a 19V power brick for a laptop would do nicely.
- A Buck DC/DC converter handling charging, but a simple LM317 could also be used, especially if the battery is stored at a fixed temperature, (look up "SLA charger lm317).
- A switch (an opamp controllring two 2 FETs) which switches the input of the output regulator from the primary input DC over to battery if it drops below the battery voltage or simply two diodes.
- A Buck/boost DC/DC converter handling the output regulation.
A good buck/boost converter topology is SEPIC, because you only need one FET and a single coil, so it's cheaper than two converters back-to-back: http://dren.dk/carpower.html the linked design will output the same voltage no matter what the input voltage is (8-24 V)
... or, if you are lucky, you can just buy one:
http://www.mini-box.com/micro-UPS-load-sharing
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.
Best Answer
Presumably you can find a power supply capable of delivering 12V at 3A to the two devices? Plenty of power supplies like this on ebay etc..
Use a relay powered from AC - when the AC drops out the relay contact feeds your load from the battery. When the AC returns, the relay contact swaps back to feeding power to the load from the 12V power supply.
A slight refinement is to have a schottky diode connected between battery and load (in effect across the relay contacts) - the benefit of this is to always guarantee that as the contact switches there will be about 11.5 volts to power your load. Also recommended is a large ish hold up capacitor across your load (maybe 1000 uF).