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I'm trying to design a battery back-up power supply for my home network devices. I need these:

  • 48 V / 40 A
  • 24 V / 10 A
  • 12 V / 2.5 A
  • 5 V / 5 A

I got a 48 V / 60 A industrial power supply with battery backup (four 12 V car batteries in series) and need to divide the voltages.

After digging through Google and Electrical Engineering Stack Exchange, I want to use bunch of diodes (like 1N4007, but of type able to withstand the current) and simply take out needed voltages somewhere between these like this:

diodes http://i59.tinypic.com/nn8w1d.png

The second option are high current transistors like BC547B driven by zener diodes, but I'm not really sure if that would be better than the first option. Is my idea OK, or I should do it another way?

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    \$\begingroup\$ Looks like a recipe for a fire to me... What do you think the current through those diodes would be with nothing connected to the outputs? \$\endgroup\$ – Majenko Oct 25 '14 at 18:21
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    \$\begingroup\$ also note that car batteries when fully charged are not a nice 12V, they are like 13.8V maximum. \$\endgroup\$ – KyranF Oct 25 '14 at 19:09
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    \$\begingroup\$ so your total voltage could be as high as 55.2V. Your devices will not be able to handle the extremely low tolerance and unknown voltage drop at unknown loads of the diodes, plus the fact the batteries will not be a fancy and steady 48V total. You must therefore rethink everything, and learn about voltage regulation. Also, you want 48V at 40A, that is some serious shit! \$\endgroup\$ – KyranF Oct 25 '14 at 19:15
  • \$\begingroup\$ I don't have to care about batteries here since they are connected to said industrial power supply, which gives stabilised 48 V on its output. \$\endgroup\$ – Myghael Hallgrímursson Oct 25 '14 at 20:46
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This is a no-go. The power which you have to dissipate in the diodes will be enormous.

diodes between 12V and 5V dissipate 7V @5A = 35W
diodes between 24V and 12V dissipate 12V @7.5A = 90W
diodes between 48V and 24V dissipate 24V @17.5A = 420W
Total amount of dissipated power (in diodes) = 545W

The diodes able to withstand such power need to be extremely big, the absorbed power will all be transfered into heat so you would need giant cooling fans or a very big airflow.

This kinds of power really need to be converted by a switch mode conversion which will give you an efficiency in the order of 85%+.
Instead of dissipating 545W in the electronics you now only have to dissipate 15% of the output power, so 295Wx0.15=44.25W
Still a lot but way more feasible.

So you better go looking for three switched mode PSU for 48V>24V, 48V>12V and 48V>5V

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  • \$\begingroup\$ Just wondering if using 48V>24V, 24V>12V and 12V>5V would be more/same/less efficient than 48V>24V, 48V>12V and 48V>5V? \$\endgroup\$ – Biduleohm Oct 25 '14 at 19:02
  • \$\begingroup\$ Very little difference in efficiency, but your 24V regulator would need to handle its own current, plus the 12V regulator's current and the 5V regulator's current, and the 12V would need to handle its own and the 5V current. By having them separate you are keeping those current requirements separate. \$\endgroup\$ – Majenko Oct 25 '14 at 19:04
  • \$\begingroup\$ Ok, thanks. You're also more immune to failures (if the 24V fails the 12 and 5V doesn't if the PSU are separate) \$\endgroup\$ – Biduleohm Oct 25 '14 at 19:11
  • \$\begingroup\$ I thought that enormous heat production is done when lowering voltage with resistors, that's why I wanted to use diodes. By the way, what about high-power transistors, like these used in CPU power supply on PC motherboard? \$\endgroup\$ – Myghael Hallgrímursson Oct 25 '14 at 20:41
  • \$\begingroup\$ If you convert 48V>24V>12V>5V The power for the 5V will be converted 3 times, every time with 15% loss (and 15% heat) and the 12V will loose 2 times 15%. In total you loose an additional 15W in heat which makes around 1/3 of your loses. Both in heat and efficiency I would not call this very little. \$\endgroup\$ – Requist Oct 25 '14 at 20:50

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