Protection of ELV circuits

[faraway73]

Hi Everyone

I am new to the forums and would like some advice for circuit design. I have an AC/DC (240VAC/24VDC) power supply unit supplying one DC circuit. My question is, should I have a circuit breaker on the DC side to protect the DC circuit? And if a circuit breaker is required on the DC side, in what way is it different to a laptop charger, which does not have protection on the DC side? I am interested in best practice from an Australian standards point of view.

 

[VEBill]

"...a laptop charger, which does not have protection on the DC side..."

The above is probably a false assumption. Any laptop charger (perhaps excluding the worst of the knock-offs) would have current limiting on the output side.

 

[Paulusgnome]

It really does depend on the power supply in question.
For instance, I use plenty of 240VAC/24VDC switchmode supplies in the machines that we build, and these all have built-in short-circuit and overload protection which operates far more quickly than a circuit breaker. If I were to attempt to protect a 40A switchmode supply with a 40A CB, it would be clear from testing that the CB would never get the chance to trip, beaten every time by the PSU electronics.
However, if I were to attempt to use a cheap-and-nasty PSU without any short-circuit or overload protection, then it would be necessary to provide a CB of fuse to provide that protection, or risk frying both PSU and wiring loom under fault conditions.

 

[LiteYear]

Yep, in the absence of extra requirements, a switch/breaker on the 240V side and a fuse on the DC side would seem reasonable to me (in Aus land). The only reason to add a switch to the DC side is if the cord is long and integrated with the device it is powering. In that case it is convenient to have local on/off control.

 

[DRWeig]

Just echoing the others here -- your power supply probably already has protection on the low voltage side. Investigate it -- you should find a fuse, or, in the case of a smart switching supply, it's handled internally. Your power supply's data sheet should clue you in.

Best to you,

Goober Dave

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[faraway73]

Thanks all, the datasheet does state that the PSU has overload (constant current limiting) protection that limits the PSU to 150% of the rated output power. The units rated power is 75W and rated current at 24V is 3.2A. This means that in short circuit condition, the current is limited to 150% * 3.2A = 4.8A @24VDC. However, the breaker on the 240VAC side is 2A, should I size it to be smaller so that if there is a fault on the DC side, the breaker on the AC side should trip as well? 4.8A@24VDC = 115.2W => 0.48A @ 240VAC

The spec for the PSU:

http://www.meanwell.com/search/DR-75/DR-75-spec.pdf

 

[LiteYear]

However, the breaker on the 240VAC side is 2A, should I size it to be smaller so that if there is a fault on the DC side, the breaker on the AC side should trip as well?

Probably not - when the PSU is first energised there is probably an inrush current to charge the capacitors. The 2A fuse might be sized to accommodate this so anything less might lead to nuisance tripping.

 

[zappedagain]

Just remember to size your DC wiring large enough to handle the fault (short circuit) current. If you want to go to smaller wiring, you'll need a fuse on the DC side, and then you can use wire large enough to handle the fuse current.

It is a bit trickier on the AC side, at least here in the USA, as UL says you have to use 18 AWG or larger on the AC side, even after fuses. UL trumps logic around here.

Z