10-02-2008 07:19 AM - edited 03-06-2019 01:43 AM
We have a cisco 7604 with 2 x 2700W DC power modules (main + redundant). Does the redundant unit present a load? So for example we have the main unit connected to 60 Amp fused 48v DC supply. If I jumper this supply to the redundant unit will I need to increase the fuse size and possibly the cable size?
10-03-2008 08:21 AM
Hello Tony,
If I've understood correctly you would like to connect both power supplies to the same external DC supply.
First of all from a redundancy point of view this is not a good idea.
depending on power management strategy only one power supply of both provide power to chassis and linecards.
you can check this with
show power
how much current is needed ?
2700 W = 48V * I
I = 56,25 A
the second supply even without a load will draw some current to energize itself.
I don't recommend to do this on permanent basis, it could be acceptable to do it for some time during electrical works:
you don't achieve redundancy in this way
I would just power off the standby PS.
Hope to help
Giuseppe
10-03-2008 08:41 AM
Thanks Guiseppe,
I have since found the Cisco doc regarding redundant power, and some small print stating the A Battery is connected to PSU1, and B Battery connected to PSU2 (redundant).
Is also says that either PSU should be connected to 2 pairs (-48v and 0v) supplied off the same battery. The question I now have is:-
Is it necessary to connect 2 separately fused pairs per PSU or is it OK to connect 1 pair (fused) and then jumper that to the other pair on the back of the PSU?
Would the jumper option require a larger fuse on the (single) Battery supply (i.e. 2 x 60A = 120A)?
Your help is much appreciated,
Tony
10-03-2008 10:31 AM
Hello Tony,
separated cables pair and fuse
PSA ============ batteryA
PSB ============ batteryB
otherwise you are connecting them in series and trying to supply
96V on the two in series and you miss redundancy.
Hope to help
Giuseppe
10-06-2008 02:59 AM
Thanks Guiseppe,
Considering 1 Cisco Power Unit, the options we are looking at are:-
PSUA1 and PSUA2 are the two pairs of terminals on the SAME Power Unit.
PSUA +ve1 ================ BattA +ve == 0v
PSUA -ve1 =====FUSEX====== BattA -ve
PSUA +ve2 ===JUMPER======= PSUA +ve1
PSUA -ve2 ===JUMPER======= PSUA -ve2
So the 2 terminals are in parallel.
+ve ==========[+ve1]===========[+ve2]
| | |
| | |
BattA (R=PSUA1) (R=PSUA2)
| | |
| | |
-ve ==FUSEX====[-ve1]===========[-ve2]
The Alternative is:-
+ve ==========[+ve1]
| |
| |
BattA (R=PSUA1)
| |
| |
-ve ==FUSEY====[-ve1]
And...
+ve ==========[+ve2]
| |
| |
BattA (R=PSUA2)
| |
| |
-ve ==FUSEY====[-ve2]
Assuming a single feed is fused at 60Amps.
Does FUSE X = 60 Amps or Higher?
Does FUSE Y = 60 Amps?
The problem is that just adding a second power pair will be difficult at the site, so the top is the preferred option.
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