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Are the fuses required for a redundant power system, which takes power from a battery?

If yes, how to calculate the characteristics?

I am trying to isolate the faulty PSU, not the load. Load must have uninterrupted power.

Thanks.

enter image description here

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  • \$\begingroup\$ Do you know what does a fuse does? It interrupts the flow of current when too much current flows and then protects against fatal damage like fire. If you can guarantee that the DCDC converters will never fail and short the battery then indeed you don't need the fuses. Fuses are like the safety belt in your car, you don't need them when you never crash your car. But when your car crashes, they can save your life. \$\endgroup\$ – Bimpelrekkie Feb 10 '18 at 22:05
  • \$\begingroup\$ what do you want to happen if there is a fault? \$\endgroup\$ – Jasen Feb 10 '18 at 22:05
  • \$\begingroup\$ I want the load to continue to receive power, also to disconnect the faulty PSU. \$\endgroup\$ – Roman Simonyan Feb 10 '18 at 22:07
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If you are using the schematic as depicted in your image, then if one PSU fails then its fuse will blow and the other PSU will take over. This is a good design.

Fuses come with ampere ratings, so if your load uses 1 A then you might want to use a 3 A fuse, this gives you a 2 A margin. And once something horribly goes wrong, say someone stick a wrench between positive and ground making tens of ampere to flow, then the fuse will blow. Or if something else happens, maybe some transistor breaks.

The more you know about your load, the better fuse you can use. Say your load has some decoupling capacitors, this means that every time you turn on your device, there will be an enormous surge current which might blow the fuse. For this you might want to use a slow fuse instead of a fast fuse.


We still don't know anything about what your load is. But let's say it is a computer with a 400 W PSU. And that your mains voltage is 240 V.

Then the RMS current will be, during the worst case scenario, \$\frac{400\text{ W}}{240\text{ V}}\approx 1.6 \text{ A}\$.

In this case I would use a 5 A fuse. Or even a 3 A fuse. My gut feeling tells me that I should seek a fuse that is rated for at least 100%, or around there of what your load will normally use, but not ridiculously high. Like you would be an idiot to use a 1000 A fuse in this scenario. \$240 \text{ V} × 1000 \text{ A} = 240 \text{ kW}\$, one of your redundant PSU's will most likely not consume a fourth of a MW during a failure, so the fuse will never blow.

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  • \$\begingroup\$ if the top PSU fails with the input shorted having two fuses would help \$\endgroup\$ – Jasen Feb 10 '18 at 22:07
  • \$\begingroup\$ @Jasen how would it help? \$\endgroup\$ – Harry Svensson Feb 10 '18 at 22:09
  • \$\begingroup\$ I am trying to supply uninterrupted power for the load. \$\endgroup\$ – Roman Simonyan Feb 10 '18 at 22:12
  • \$\begingroup\$ @Jasen Gah, realized my derp. \$\endgroup\$ – Harry Svensson Feb 10 '18 at 22:16

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