Hello Everybody

I am using 8x Power 95+ Gold LL1850FC power supplies, rated at 1850W but only Drawing 1400W max from each unit.
Im in the UK, so using AC 230v/240v, normal 13Amp rated plug sockets.

When I returned home 3 of the 8 had stopped working and the connectors were fried.

As I wasnt present I dont know specifically which part of the connector caught fire first.

What could have caused this?

Pics of my burnt stuff and the power supply attached.

 

138A output requires 25mm^2 cable (see 18th edition of the wiring regulations). None of those cables is anywhere near 25mm^2. If any cables are less than that size, they need individual fuses.
A proper supply with that capacity would have a M8 or larger stud for the output, not a selection of flimsy connectors.
The whole think looks like a fire hazard to me.

Meanwell knows how to do it properly:
https://www.farnell.com/datasheets/2724891.pdf
and that's why they can charge the price that they do.

 

It's virtually impossible to tell without a LOT more information, but the data plate says that the max output is 1650 W at 12 V, which is 138 A. I wouldn't trust that wire bundle to deliver 138 A, or anything close to it, continuously.

You say that you have eight of these. Are they all powering the same load? If so, how do you know that you are only drawing 1400 W from each unit? How are you load balancing them?

Have you checked reviews on those supplies and/or that manufacturer? Sometime the problem is the cheap crap that is becoming increasingly the norm -- which, given the wire sizes coming out of the unit, would seem more than a bit likely.

 

Re 25mm^2
Would the PSU not be distributing the power over the smaller sections of wire.

re load, The 8 PSU were powering 4 crypto rigs, 2 per rig.
4 double sockets, every 2 sockets had 4mm^2 twin earth

 

Regarding the 25mm^2
Would the PSU not be distributing the power over the smaller sections of wire.

The 8 PSU were powering 4 crypto rigs, 2 per rig.
4 double sockets, every 2 sockets had 4mm^2 twin earth

No. Why would it do that? The power is going to head towards the lowest load resistance.
Those cables are 16AWG or 1.5mm^2, rated about 16A.
Not sure about the sockets - they look like Molex microfit 3.0 and they are rated less than that.
From the photo of the pcb there are no fuses at all on the outputs.
if I were your house insurer, I would be invalidating your policy if i knew you had something so dangerous.

 

No. Why would it do that?

Per PSU there are 9 separate Molex connectors, each having 6 wires per, IE 54 total individual wires coming out of the power unit.(evidenced in first picture)

Hence...

Would the PSU not be distributing the power over the smaller sections of wire.

 

8 x 1400W = 11,200W (power needed)

13A x 240V = 3120W (power available)

How is that supposed to work?

 

Per PSU there are 9 separate Molex connectors(evidenced in first picture), each having 6 wires per, IE 54 total individual wires coming out of the power unit.
Hence...

No. The power is going to head towards the lowest load resistance.

 

No. The power is going to head towards the lowest load resistance.

so then why have they bothered putting 9 molex connectors on the PSU?

 

8 x 1400W = 11,200W (power needed)

13A x 240V = 3120W (power available)

How is that supposed to work?

I answered that already above your post.

4 double sockets, every 2 sockets had 4mm^2 twin earth

ie

4mm^2 -------Twin 13A socket----------Twin 13A socket
4mm^2 -------Twin 13A socket----------Twin 13A socket

 

so then why have they bothered putting 9 molex connectors on the PSU?

Because they have no concept of safety and compliance with regulations.

 

No. The power is going to head towards the lowest load resistance.

So how are the other four PSU's still working to power my other two loads if all power is only heading towards one molex connector?

Because they have no concept of safety and compliance with regulations.

Have you studied the schematics of the PSU to rule out components inside are not limiting power output per molex?

 

So how are the other four PSU's still working to power my other two loads if all power is only heading towards one molex connector?

No one is saying that all of the current goes to one connector, only that unless steps are taken in the design, they will not share the current equally and there isn't a lot of margin available.

Have you studied the schematics of the PSU to rule out components inside are not limiting power output per molex?

You asked strangers on the Internet to try to tell you what might have happened. We have done that the best we can given the very limited information. The wires and connectors that were damaged carried far more current than they are rated for. The supplies powering them are capable of delivering far more current than they are rated for. Whether the PSU design made any attempt to balance the load and/or limit the output per wire is largely beside the point, as it is pretty evident that it didn't actually do so.

Have you checked reviews/complaints about those supplies or that manufacturer?

Have you contacted the manufacturer and asked them what might have happened and how to prevent it from happening again?

 

No one is saying that all of the current goes to one connector, only that unless steps are taken in the design, they will not share the current equally and there isn't a lot of margin available.

This is where you're completely wrong, if you read my convo above with Ian0 thats exactly what he is saying, and in multiple posts.
Which Ive been disputing all along because it sounds incorrect.

You asked strangers on the Internet to try to tell you what might have happened. We have done that the best we can given the very limited information. The wires and connectors that were damaged carried far more current than they are rated for. The supplies powering them are capable of delivering far more current than they are rated for. Whether the PSU design made any attempt to balance the load and/or limit the output per wire is largely beside the point, as it is pretty evident that it didn't actually do so.

Have you checked reviews/complaints about those supplies or that manufacturer?

Have you contacted the manufacturer and asked them what might have happened and how to prevent it from happening again?

And I appreciate the help, however when something doesn't sound right it makes sense to challenge it.

Going back to the point in question, Your post made sense to me as you described the wires in a "bundle", an accurate representation, the power is obviously intended, and does, flow through the bundled wires out through the numerous molex connectors (despite Ian0s opposing view)

Also, There are numerous things that can go wrong in such PSU to cause a fire, the most obvious is the wire ratings and ampage, but I posted here to see other suggestions.(maybe faulty ac/dc conversion, poor soldering, etc)

I will reach out to the manufacturer now, good idea.
And also will open one up and get some inside photos to see if load balancing exists.

 

I think it's pretty obvious from this image that there are no protection circuits for each output. If there were, they would be visible on this picture. All I can see is a very large track commoning all the outputs.
There may be fuses on the other side of the board, but if there were, why didn't they work?

This is where you're completely wrong, if you read my convo above with Ian0 thats exactly what he is saying, and in multiple posts.
Which Ive been disputing all along because it sounds incorrect.

I said it takes the path of lowest resistance. If that low resistance is a short circuit, you have all the current down one cable, and 135A down 2.5mm^2 cable equals a fire, as you found out.
Think of your house consumer unit, with a 100A mains supply on 16mm^2 or 25mm^2 cable. Would you wire the sockets on 2.5mm^2 cable without a separate fuse for each circuit?

 

This is where you're completely wrong, if you read my convo above with Ian0 thats exactly what he is saying, and in multiple posts.
Which Ive been disputing all along because it sounds incorrect.

That's not what he is saying, it's what you are hearing.

This is actually a common misunderstanding/miscommunication.

There is the common phrase that "current follows the path of least resistance."

This does NOT mean that if you have one thousand paths and one of them has slightly less resistance than all of the others, that all of the current is going to go down that one single path that has just slightly less resistance. In that regard, it is not a correct statement. But it's a lot easier to say than, "given multiple paths, a larger fraction of the total current will flow through a path having less resistance than another path."

Ian0 is merely saying that if you put all of those connectors and wires in parallel, that it is not automatic that they will share the current equally and that more current will go down those paths that have lower resistance.

In a later post, you indicated that there are two power supplies for each crypto rig. This virtually guarantees that one of the power supplies will supply the bulk of the total current.

Let's use a simplified example to illustrate the point.

Imagine you have a load that is 50 mΩ and it supposed to be powered by 12 V. That's a total current of 240 A.

Now let's say that you don't have a single supply that can deliver that much current, but you have two supplies that can each deliver 150 A. So you put them in parallel via cables that have 1 mΩ of resistance.

What happens if the output voltage of the two supplies are not exactly the same? Let's say that one of them is 12.00 V and the other is 12.05 V (which is actually a very tight tolerance, just 1% difference).

The voltage on the load will end up being 11.905 V (with about 0.1 V dropped across the cables).

But how much will each supply be delivering?

The 12.00 V supply will be putting out still end up being 12 V, but the 12.1 V supply will be delivering 95.25 A while the 12.05 V supply will be pumping out 145.25 A.

The situation is made worse either by having even lower resistance cables or by having more difference between the supplies. In fact, you can quickly get to a situation in which the higher output supply is putting out not only all of the current taken by the load, but also driving substantial current into the other supply as it tries to hold its output voltage down while the other supply is trying to drag it up.

My guess is that the underlying problem isn't load balancing coming out of the multiple wires of a single supply, it's contention between two supplies powering the same load resulting in one supply being overloaded.

 

138A output requires 25mm^2 cable (see 18th edition of the wiring regulations). None of those cables is anywhere near 25mm^2. If any cables are less than that size, they need individual fuses.
A proper supply with that capacity would have a M8 or larger stud for the output, not a selection of flimsy connectors.
The whole think looks like a fire hazard to me.

Meanwell knows how to do it properly:
https://www.farnell.com/datasheets/2724891.pdf
and that's why they can charge the price that they do.

It may seem a bit dodgy but PC supplies have been doing this almost from the start.
A later response says there are 54 wires and another post says they are 1.5sqmm each. So assuming 27 positives and 27 negatives: that is 40sqmm for each and honour is satisfied.
To the OP: it is telling that the burns are all at the connector ends. I see the following possibilities, 1. the connectors are forming 'hot loose joints' which can be caused by 2. excessive number of connector mating cycles (check the connector data sheet) or 3. they are just underrated or 4. the heat is coming up from the pcb (this would be obvious because the pcb would be burned). Or it could be a combination of these factors. If the connectors do get hot then connectors generally begin to fail in as much as their contact resistance increases, they get hotter, the R goes up further, etc etc thermal runaway and excessive carbon formation results.
138A is a lot to carry in pcb traces or planes. If your pcb is not super heavy Cu weight (and I mean >10oz) I think there would need to be some Cu bar to carry the current. Using the pcb to route the individual connector points to the CU bar should be ok but the combined current should get some extra sq mm to flow in.
Fuses on these wires is unlikely to help. Try to see if you can find a fuse that does not drop excessive volts, does not melt under normal load and yet will still stop the wires from melting. There is a basis for Murphy's Law (#?) that says, an expensive appliance will always protect a 30c fuse by burning out first. In truth, fuses are for hard faults, not overloads. And even then, they only work well if the fault current is huge.

 

View attachment 318519I think it's pretty obvious from this image that there are no protection circuits for each output. If there were, they would be visible on this picture. All I can see is a very large track commoning all the outputs.
There may be fuses on the other side of the board, but if there were, why didn't they work?

I said it takes the path of lowest resistance. If that low resistance is a short circuit, you have all the current down one cable, and 135A down 2.5mm^2 cable equals a fire, as you found out.
Think of your house consumer unit, with a 100A mains supply on 16mm^2 or 25mm^2 cable. Would you wire the sockets on 2.5mm^2 cable without a separate fuse for each circuit?

Actually, the current is shared according to the individual wire (etc) resitances. A lower resistance does not immediately turn all the current off the other wires.
And a small series resistance in the output of parallel power supplies is commonly used to have the power supplies share the load. That trick has been around for a very very long time.

 

And a small series resistance in the output of parallel power supplies is commonly used to have the power supplies share the load. That trick has been around for a very very long time.

Yes, but I don't see any indication that these supplies have ballast resistors. If the crypto boxes the TS is using were intended to be powered by multiple supplies (which I rather doubt), then MAYBE they incorporate ballast resistors. But at these currents, they will have to be pretty small and if they aren't well matched, that will cause its own problems.

 

That's not what he is saying, it's what you are hearing.

This is actually a common misunderstanding/miscommunication.

There is the common phrase that "current follows the path of least resistance."

This does NOT mean that if you have one thousand paths and one of them has slightly less resistance than all of the others, that all of the current is going to go down that one single path that has just slightly less resistance. In that regard, it is not a correct statement. But it's a lot easier to say than, "given multiple paths, a larger fraction of the total current will flow through a path having less resistance than another path."

Ian0 is merely saying that if you put all of those connectors and wires in parallel, that it is not automatic that they will share the current equally and that more current will go down those paths that have lower resistance.

In a later post, you indicated that there are two power supplies for each crypto rig. This virtually guarantees that one of the power supplies will supply the bulk of the total current.

Let's use a simplified example to illustrate the point.

Imagine you have a load that is 50 mΩ and it supposed to be powered by 12 V. That's a total current of 240 A.

Now let's say that you don't have a single supply that can deliver that much current, but you have two supplies that can each deliver 150 A. So you put them in parallel via cables that have 1 mΩ of resistance.

What happens if the output voltage of the two supplies are not exactly the same? Let's say that one of them is 12.00 V and the other is 12.05 V (which is actually a very tight tolerance, just 1% difference).

The voltage on the load will end up being 11.905 V (with about 0.1 V dropped across the cables).

But how much will each supply be delivering?

The 12.00 V supply will be putting out still end up being 12 V, but the 12.1 V supply will be delivering 95.25 A while the 12.05 V supply will be pumping out 145.25 A.

The situation is made worse either by having even lower resistance cables or by having more difference between the supplies. In fact, you can quickly get to a situation in which the higher output supply is putting out not only all of the current taken by the load, but also driving substantial current into the other supply as it tries to hold its output voltage down while the other supply is trying to drag it up.

My guess is that the underlying problem isn't load balancing coming out of the multiple wires of a single supply, it's contention between two supplies powering the same load resulting in one supply being overloaded.

WBahn is correct about the current flow and load sharing. Differences in output resistance does not act like a magic current switch, Ohms law applies under those conditions. I don't agree about the supplies being 2 or more quadrant devices though. These power supplies will not be anything more than single quadrant outputs, ie they supply a positive voltage and a positive current, ie they will not sink current (unless they have a cro bar circuit and the voltage is high enough to trip that).
And output resistance in power supplies is routinely used to have parallel supplies share the load. Sometimes the output resistance is a real resistance, like in this case where it is the resistance of the wires that cause the wires to share the load current, or, a common and much smarter method is to tweak the control loop of the power supply to mimic an output resistance (wouldn't help in this case though, where the wires, where we want the load current shared amongst them, are all connected to the same power supply).
If there is a small difference in the output voltage settings of two power supplies with outputs connected together, and the power supplies are not designed to be paralleled, then there could be the answer to the problem taken in combination with other possibilities, such as the connectors are already at their limit or the pcb is already under Cu weght. Remembering that resistive power loss is I squared, so it don't take much when you are already close to the edge.