Number of power supplies on one circuit

Status
Not open for further replies.
N

NotAnElectrician

Member
Location
Chicago
Occupation
Project Engineer
Hello!

We are using multiple Meanwell SE-450-12 power supplies on a project. From a system design standpoint, we're hoping to be able to power five of these power supplies from one 20A/110V circuit. The power supply data sheet says it has an inrush of 35A. Five of these would have a total inrush of 135A. Some of the equipment we are used to using says that the breaker can take 20X amp inrush, so up to 400A inrush on a 20A circuit. It's well documented and thus we can design with confidence.

On this current project, we're having a difficult time finding out what the power source is. We're trying to find out. We know it's a 20A/110V circuit, but we don't know the type of breaker, etc.

Assuming the breaker is a more typical 20A/110V breaker used for general facility lighting and such, would the inrush rating on this breaker be sufficient to run these 5 power supplies? I've been seeing numbers like 10X so up to 200A inrush. I would love to see a typical data sheet for a circuit breaker that shows something like inrush rating.

There are going to be 16 circuits, each potentially with 5 power supplies on each circuit. Would that have an impact on the main panel breaker?

Thanks for any direction and/or help you can provide.
 
You are generally limited to 16A of continuous load on a 20A circuit.

These supplies draw a nominal current of 10A, presumably because of poor power factor.

I don't see how you can run 5 of them on a single 20A circuit unless they are lightly loaded.

Jon
 
They are rated 450W. 450W x 5 = 2250W. Max continuous load on a 20A breaker would be 16A, so assuming 120V that would be 1920W (even less if it really is 110V, which is unlikely), so you cannot run 5 of these on a single circuit; 4 max. (assuming no other loads). Personally, I would split it 3 and 2, just to have some redundancy in whatever these are powering.

Inrush current is not likely a concern. The magnetic trip current on MOST circuit breakers will be factory set for 10x the rated amps, so in this case 200A. 35A inrush, even for all 5 (which you can't do of course) would be 175A, well within the safe zone.
 
They are rated for 450W _output_.

They are also are only rated at 83% efficiency.

They also do not meet harmonic current standards.

Which means that they draw considerably more input current then suggested by the 450A rating.

Jon
 
Good morning, Thank you everyone for your input!

I should have also added to this post that each power supply will only be loaded with 345.6 watts for a total of 1728 watts for the 20A circuit. In light of this information, does that change anyone's thoughts?

I was also interested in seeing the comment about the 10 amp nominal current. This is interesting. Can someone explain this further? How would that mathematically limit me on how many watts I could comfortably load each power supply with?

Thanks again!
 
In AC circuits, the power delivered depends not only on the current, but the timing of the current waveform and how it matches the voltage waveform. You could imagine an AC load where the AC current was particularly poorly timed, such that you have _lots_ of current flow but _zero_ true power flow. The relationship between true power delivered to the load and 'apparent power' (RMS volts * RMS amps) is called 'power factor'.

Your supply circuit is limited by current flow, not power flow. Your 20A circuit is rated to deliver 2400 VA of 'apparent power' (and is further limited to 16A or 1920 VA on a continuous basis), which sets a power limit of 2400W (1920W) if you have perfect power factor. But if you have a low power factor the delivered power can be much less.

The data sheet for these supplies says that they draw 10A at 120V input when delivering 450W. This 10A (1200VA) is caused by the efficiency of the supply and the power factor of the supply. I'd estimate that the supply is consuming 540W of real power to deliver 450W, and the rest is due to 'distortion power factor'. This is power factor caused by the load current not matching the sinusoidal supply voltage.

As a guess, I'd estimate that when delivering 350W the supply will consume 8A, but that is a number pulled out of my left ear. You would need detailed performance graphs to better quantify the number. But as I guess I figure it is reasonable to have 2 supplies delivering 350W each on a 20A 120V circuit.

-Jon
 
winnie said:
In AC circuits, the power delivered depends not only on the current, but the timing of the current waveform and how it matches the voltage waveform. You could imagine an AC load where the AC current was particularly poorly timed, such that you have _lots_ of current flow but _zero_ true power flow. The relationship between true power delivered to the load and 'apparent power' (RMS volts * RMS amps) is called 'power factor'.

Your supply circuit is limited by current flow, not power flow. Your 20A circuit is rated to deliver 2400 VA of 'apparent power' (and is further limited to 16A or 1920 VA on a continuous basis), which sets a power limit of 2400W (1920W) if you have perfect power factor. But if you have a low power factor the delivered power can be much less.

The data sheet for these supplies says that they draw 10A at 120V input when delivering 450W. This 10A (1200VA) is caused by the efficiency of the supply and the power factor of the supply. I'd estimate that the supply is consuming 540W of real power to deliver 450W, and the rest is due to 'distortion power factor'. This is power factor caused by the load current not matching the sinusoidal supply voltage.

As a guess, I'd estimate that when delivering 350W the supply will consume 8A, but that is a number pulled out of my left ear. You would need detailed performance graphs to better quantify the number. But as I guess I figure it is reasonable to have 2 supplies delivering 350W each on a 20A 120V circuit.

-Jon
Click to expand...
Thank you for this! Your previous comments make perfect sense now. I wasn't considering the inefficiency at all, but it's right there in the data sheet. Just didn't know what I was looking at. I really appreciate the help!
 
NotAnElectrician said:
…
I should have also added to this post that each power supply will only be loaded with 345.6 watts for a total of 1728 watts for the 20A circuit. In light of this information, does that change anyone's thoughts?
…
Click to expand...
Side issue: when sizing circuits and protective devices, the actual expected load is not relevant, everything must be sized for the max RATING of the devices consuming power.
 
Jraef said:
Side issue: when sizing circuits and protective devices, the actual expected load is not relevant, everything must be sized for the max RATING of the devices consuming power.
Click to expand...

But this gets fuzzy when the device is itself a power converter supplying other devices.

For a VFD code requires you to size to the input rating of the VFD, even if that VFD will only be used at lower than full rating.

But the same requirement doesn't apply to transformers; you can have a circuit sized for a 25 kVA load feeding a 50 kVA transformer which has 25 kVA of connected load.

-Jon
 
Status
Not open for further replies.
Top