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How can I get 8V 30A power supply from a 12V 30A DC supply? I would be glad if you could illustrate on your point since i don't know much about this yet.

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  • \$\begingroup\$ Hint: use a switching regulator as power loss will be enormous when using a linear regulator at 30 A. 30 A regulators are not easy to find (in switching or linear form). 30 A regulators are really not for beginners. Start with 8 V 3 A using an LM2596 module. \$\endgroup\$ Dec 14, 2017 at 16:34

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How can I get 8V 30A power supply from a 12V 30A DC supply?

In principle, a linear regulator would allow you to do this. Typically they only consume a few micro or milliamps themselves and pass the vast majority of input current to their load.

In practice, 30 A through a linear regulator is a horrible idea due to the power inefficiency. 12 V x 30 A is 360 W while 8 V x 30 A is 240 W. So if you use a linear regulator that takes 360 W from the supply, and delivers 240 W to the load, where does the extra 120 W go? The linear regulator turns it into heat. Which means you need a truly massive heat sink, and a lot of careful design to avoid the regulator burning itself up.

This can be overcome by using a switching regulator. A switching regulator would be able to deliver 30 A to an 8 V load, while only drawing about 20 A (maybe 22-24 A in reality) from the 12 V input.

But designing a 240 W switching regulator is not a trivial job, and not a good choice for a first design project. If your goal is to learn, start with something smaller like mentioned in the comments, and try this design later. If your goal is just to provide power to some device you want to use, you'd probably be best off just buying an off-the-shelf 8 V 30 A mains-powered supply.

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Check out this appnote by National Semiconductor. This is how to do it with maximum electric efficiency. For 12V to 8V conversion, you would probably need inductors with less inductance and modify the feedback to throttle the PWM at a lower output voltage. And, you would need to beef up the primary-side PCB traces and capacitors, as the original PSU's input is rated at 33V+ = less current. Generally you're looking for an SMPS the size of a PC PSU. If you could get at 90% of efficiency, that's about 12W of dissipated heat. Fairly easy to dissipate with a "book-size" passive heatsink (smaller if you like adventure).

If you tried doing this in a linear fashion, you would have 4V * 30 A = 120 W of wasted heat. That's pretty difficult to dissipate by a passive heatsink. Think of the fan-equipped heatsinks on high-end PC CPU's and graphical cards. And yes there are discrete transistors that would be able to handle this current (maybe several of them in parallel if a single transie of this size is expensive - but that would require some matching).

If you wanted something "off the shelf", this reminds me of some MeanWell SD-350 or RSD-300 switchers, except that none of them features a 12V input. The output could possibly be tweaked down = that's less of a problem. Now if you asked for 230V AC down to 12-ish Volts / 30A, that would probably be easier to buy.

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    \$\begingroup\$ The user doesn't know about switch mode regulators and you're throwing them a 50 part switch mode regulator with the wrong voltage and without any explanation at all. Then recommend products that you've used, but doesn't have a 12 volt input. \$\endgroup\$
    – pipe
    Dec 14, 2017 at 17:47
  • \$\begingroup\$ @pipe: well I haven't found anything closer than what I've mentioned. In practical off-the-shelf terms, 8 V output is a little unusual particular voltage level, and so is 12V input at 30 A. My response gives a rough idea what level of complexity the OP is asking about, and what practical issues he/she will have to solve. That information also has value :-) Perhaps the summary is: off the shelf, you would need an adjustable lab supply, and it will have a mains input to achieve this output power level. Thanks for your verdict though - appreciated. \$\endgroup\$
    – frr
    Dec 15, 2017 at 14:10

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