I need to build 0.1V@100-500A power supply - and I wonder what's the best way to approach this task? I guess here are not much specific IC's suitable for such DCDC... Should be current-stabilized, 10% ripple is ok, no requirements on switching frequency...

What's the rough estimation for the size/quantity of components we always need in DCDC (inductors/caps/number of channels) for such PSU?

This is for heating metal stripes, which have very low resistance. 20mm^2 copper connections is not a problem.

Some thoughts:

Even my AWG4 wire would give me 0.8mOhm per meter... If I place DCDC right next to consumer, I can have like 0.2m connections and 0.16mOhm resistance... But If I want to have switching regulator, I'll need lots of FETs & inductors... Fortunately, I already have bunch of 5mOhm N-FETs, so I can run 20-30 of them in parallel, and bunch of 1uH inductors out of some 1.5mm^2 wire which I also can run in parallel (all this from a batch of damaged PC motherboards)... The question is what's the most realistic way of driving all these things - I am thinking about having microcontroller driving all these FETs (using simple discrete driving circuits) and feedback through onboard 1msps ADC.. I wonder what should be switching frequency given that inductors are only 1uH?

Bounty comments: Still want to hear thoughts on DCDC way, without transformers. Compact DCDC might fit right inside vacuum chamber, which is impossible for transformer. And yes, the task is to heat tungsten foil of different geometry in vacuum (up to some 1000C).

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    \$\begingroup\$ @100 to 500 Amps? I use power supplies like this, do you realize how hard a job it is to supply to .1V at 500A, 1mOhm will give you a .5V drop. That is more then your output by 5 times. There are industrial power supplies designed for this and they are very very expensive and often need water cooling. \$\endgroup\$
    – Kortuk
    Jul 26 '11 at 16:59
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    \$\begingroup\$ Lemme Guess - Are you building a welder? \$\endgroup\$ Jul 26 '11 at 23:34
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    \$\begingroup\$ @Fake Name No, welders require more than 50W of power :-) This is to heat tiny strips of metal to controlled temperature. \$\endgroup\$ Jul 27 '11 at 1:09
  • \$\begingroup\$ You might consider using a current-sense transformer and regulating current instead of voltage. As Kortuk pointed out, a current-sense resistor isn't going to work. Or does it have to be exactly 0.1 V? \$\endgroup\$ Jul 27 '11 at 12:01
  • \$\begingroup\$ I can't think of a compact DC/DC doing 500A, and by no means more compact than a toroidal transformer. \$\endgroup\$ Aug 4 '11 at 14:33

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.

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.

Current sense resistor

(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.

  • \$\begingroup\$ And if I want to regulate the current, I'll have to adjust input voltage through one more DCDC? Sensing current is another interesting problem for such range... \$\endgroup\$ Jul 27 '11 at 1:27
  • \$\begingroup\$ @BarsMonster - maybe you can sense current inductively or with a hall effect sensor as used in a current probe - bandwidths in the kilohertz are possible. This does sound like a job where current regulation (to a value set by operator experience) may be desired. Or even use a temperature sensor to adjust the current up and down within an allowed range. \$\endgroup\$ Jul 27 '11 at 6:01
  • \$\begingroup\$ I need digital regulation - metal strips are quite small, and slow human would not be able to reduce current before it evaporates :-) Same for temperature sensing - such small things will have very fast temperature gradients.. \$\endgroup\$ Jul 27 '11 at 7:48
  • \$\begingroup\$ @BarsMonster, we do this at my job and we use a transformer and a water cooled rectifier. We know exactly what the transformer is giving as a gain and we use a current controlled source. This is not a simple task. We have water cooled lines to the filament but they do not provide a problem because we are using a current control mechanism. \$\endgroup\$
    – Kortuk
    Aug 4 '11 at 14:17

Starting with a welder may be a very good idea. That's an immense amount of current by most standards. A welder has power to spare (as you note) and a rewind of a welder transformer may be as easy as anything you can do. Um. Scratch that, probably. Welder will be "volt per turns" and you need at least 5 turns per Volt (0.2V output to allow 50% loss along the way. Maybe a bit more than that.

Take modern semiconductor based welder. Count turns on secondary OR place one turn on secondary (liable to be "not too hard to find a way to do this) and see what volts per turn is. If you have < ~= 0.5 V/turn you may have a starter. If so it is possible much easier than most alternatives.


I see you're starting with 12V DC, so you can't just use a transformer. Steven is right in that the output can be AC if it's just for heating. A decent size toroidal transformer should do it. The primary is driven by a H bridge from the 12V, and the secondary is used directly as the high current AC output.

Don't expect super efficiency. The characteristics of the transformer will be key. It will need to be designed for the high current and low voltage output.

  • \$\begingroup\$ Well.... Making transformer using some AWG4 wire (and probably even more if we would need not just 1 loop)? That's gonna be huge... \$\endgroup\$ Jul 27 '11 at 1:26

Is it possible to reconsider your design requirements, and use a different (bi-)metal for the heating element? Nichrome wire is a couple ohms per meter, and is available in many gauges. I use a piece of 16 gauge with a transformer that gives me 30A AC at 2v or so, so thats 60W, same ballpark as you are talking about. Perhaps you could use a different method to accomplish the same task. Heat the thing either directly, or indirectly to heat your other object.

  • \$\begingroup\$ Nichrome is ok only for low-ish temperatures, in my case temperatures will be 1000C and higher. At 1000C nichrome will oxidate/evaporate depending on conditions. \$\endgroup\$ Jul 27 '11 at 7:46
  • \$\begingroup\$ so your element is what? Tungsten? Sounds like you are in for some interesting work just connecting your big conductors to the element without melting the connection point. Copper melting point is just under 1100C. \$\endgroup\$ Jul 27 '11 at 16:23
  • \$\begingroup\$ Yes, tungsten. It's going to be interesting indeed :-) \$\endgroup\$ Jul 27 '11 at 22:22

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