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Redundant Power Supply - Electrical Load Behaviour

citpaj
Occasional Contributor

Redundant Power Supply - Electrical Load Behaviour

I'd like to understand what happens to the electrical load of a surviving power supply in a redundant configuration, when the electrical feed to one of the two power supplies fails.  Normally, I assume that both power supplies share the load of running the server.  If there's an electrical feed failure to one side of the rack, do the surviving power supplies (connected to a second feed) draw extra power to handle the additional load of running the systems on they're own?  In this case, that would increase the load on the remaining power feed circuit, and possibly cause a breaker to trip. If my assumptions are correct, where can I find information on the load increase of the remaining power supplies, so that I may calculate the addiitional overall load on the surviving feed circuit.?

 

We're in the process of planning a new rack with redundant power distribution circuits and need to specify the power distribution strips to be able to handle any extra load imposed by a power failure on one feed circuit.

 

Thanks.

 

PJ.

1 REPLY
Matti_Kurkela
Honored Contributor

Re: Redundant Power Supply - Electrical Load Behaviour

The simple assumption would be that the total power consumption of the server would be divided across the remaining PSUs - so if the server has 2 PSUs which normally draw 400 Watts apiece, that is a total draw of 800 Watts. If one of the two PSUs were to fail, the full 800 Watts would be drawn by the remaining PSU.

 

In practice, you might want to leave yourself a little headroom, since the efficiency of the PSUs is very likely optimized for the expected "normal" load level. At full power output, the PSU may be somewhat less efficient, so in the above-mentioned case the total draw in the single-PSU operation might be something like 805..810 W instead of the simple expectation of 800 W.

 

Basically, if you need to specify 2 circuits, you specify the first circuit assuming that the second one does not exist, then replicate the first circuit.

MK