PIP-4048MS inverter

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PIP-4048MS inverter

Post by paulvk » Sat, 14 Jan 2017, 14:22

As I have a PIP apart I can see that the inductor is potted at the bottom so not going to be easy to take apart.

Looking at the SCC board it has the two connections to the inverter marked as "To Axpert" so we know who came first.

On the subject of mosfets it would be nice to know what ones and how many in the 80amp versions.
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Post by coulomb » Sat, 14 Jan 2017, 14:35

offgridQLD wrote: Weber, I have a spare pip that hasn't been installed yet. I am happy to try it on. Thanks for doing all the ground work. I will order the litz wire and update when it arrives.

Yes, thanks, Kurt! If it can be done with low enough pain, it could become another automatic modification, like a lot of us invert the fan direction.

I have an older model main board here for parts, so in the unlikely event that your efforts completely fail, the inductor from this unit should work. This main board has no processor daughter board or case, so it's difficult to test with.
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Post by coulomb » Sat, 14 Jan 2017, 14:59

They have made some attempt at reducing the skin effect; the inductor in question is wound "three in hand" (tri-filar, three wires wound together and in parallel):

Image     Image

It's a little difficult to see from the first photo whether it's two or three in hand, but the second photo from underneath the board makes it very clear.

Edit: you can also see some of the black potting material Paulvk mentioned at the bottom of the first photo. Unfortunately, it's the rock hard type, not the stiff rubbery kind. But Kurt is a resourceful fellow, so I'm sure he'll figure out a way to remove the potting material without mangling the wire too much. A heat gun might be the first line of attack.
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PIP-4048MS inverter

Post by offgridQLD » Sat, 14 Jan 2017, 17:07

Thanks for the pics coulomb.

I assume the potting material was used just to mechanically fix it to the board and limit the inductor from resonating to much.

I guess I will need to obtain some kind of potting material to achieve the same once it's rewound.

I notice the original solid wire has a resin coating. Is the litz wire in the ebay listing raw. Edit: I see now it's listed as enamel coated.

Kurt.
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Post by weber » Sat, 14 Jan 2017, 18:42

3-in-hand makes a lot more sense. Thanks Coulomb. I was wondering how they could get away with putting 20 amps through 1.3 mm² in such a packed winding. But in fact it's 5 mm². They may not have done it with skin effect in mind. They would need to do that just so it's flexible enough to physically do the winding.

So Kurt, that means you should use 100 strand litz wire, like this:
http://www.ebay.com/itm/Litz-Wire-100-3 ... zOqCTn1BhA

That's only 4 mm² in total copper cross section [Edit: actually it's only 3.1 mm², as I point out in a later post], but if you used 120 strands (to get nearer 5 mm²) I don't think you would fit the same number of turns as the original, due to every strand being individually enameled. As it is, you will need to wind it firmly and evenly. But if we succeed in significantly reducing the heating due to PWM then we'll have more heat dissipation capacity remaining to cope with the maximum 50 Hz load.
offgridQLD wrote:I assume the potting material was used just to mechanically fix it to the board and limit the inductor from resonating to much.

I guess I will need to obtain some kind of potting material to achieve the same once it's rewound.
Yes. It's purely mechanical fixing. If it was needed to stop the inductor from emitting sound, it would fill the whole winding. Yes, you can worry about that if and when it works. But any two-part epoxy such as Araldite will do.
I notice the original solid wire has a resin coating. Is the litz wire in the ebay listing raw. Edit: I see now it's listed as enamel coated.

Yes. It wouldn't be litz wire if every strand wasn't individually insulated from the others. It must also be arranged so that no strand spends more time near the centre of the bundle. So they can't just twist the lot. They have to twist them in groups of 5 or less first, then twist 5 or less of those bundles together and so on.
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PIP-4048MS inverter

Post by offgridQLD » Sat, 14 Jan 2017, 19:33

Dam it I ordered the original ebay listing.

Edit: ok I see it's from the same seller see if they can emend that.

Kurt
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Post by coulomb » Sat, 14 Jan 2017, 22:01

I thought of a way of estimating the length of wire needed for the PIP's large inductor, before anyone pulls one apart. Presuming Weber's claim of 1.3 mm² wire (about 1.3 mm diameter as it happens), and using Wikipedia's tables for 16 AWG wire (13.17 mΩ/m each wire, or 13.17/3 = 4.39 mΩ/m for three in hand), so each metre of trifilar wound wire would have 4.39 mΩ of resistance. I guessed the toroid's cross section at 30 x 15 mm, for a perimeter of 90 mm. Call it 100 mm with thick wire bending radii. So ten turns to the metre, or 0.439 mΩ per turn.

I put three amps through the coil with my bench power supply, and measured just over 58 mV voltage drop. So that's a resistance of 58/3 = 19.3 mΩ. So that's 19.3 / 0.439 = 44 turns. That seems reasonable. So those 44 turns are going to need some 4.4 metres of wire. There are many approximations in the above, so the result would be +- 25%.

Now to convert to caveman units; I believe that's some 14.4 feet. So Kurt, I hope you ended up with 20' of the 100 strand litz wire, or 50' of the 35 strand wire (which you could wind three in hand).

I didn't see any mention of the actual length of wire you are getting for the stated price in the second Ebay listing.

For the record, my old Digitech multimeter measures the inductance at 0.39 mH (390 μH).

[ Edit: added some context at the start. ]
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PIP-4048MS inverter

Post by paulvk » Sat, 14 Jan 2017, 22:26

Before taking the inductor apart I would measure the inductance.
It may also be simpler to just get a core and make a new inductor with so much room even an ferrite "E" core could be used

Also after the inductor there is an AC current measuring sensor then what appears to be a 20uF 550V capacitor after this a common mode choke.
As my 2 year warranty's are now up I will be changing the capacitors I found some 85deg 3300uF 80V, 100V surge, use switching power supplies only 2mm larger in diameter on Element14 when I get back to my large system I will post the part number
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Post by weber » Sun, 15 Jan 2017, 01:36

Oops! I messed up. 100 strands of 0.2 dia isn't 4 mm² it's 3.1 mm².

[Edit: I earlier wrote here that 120 strands would be required, but Coulomb convinced me that 3.1 mm² is sufficient and 120 strands won't fit.]

On the eBay listing, there's a "length:" menu just above the price and quantity.

Well done calculating the length Coulomb! So a 30 ft roll will probably do two inductors. [Edit: turned out that's wrong. Keep reading]
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PIP-4048MS inverter

Post by offgridQLD » Sun, 15 Jan 2017, 03:57

At $50 for a rewire and teir 1 PV now avalable at 75 cents a watt it's starting to look less atractive.

Though if your running a small battery or multiple pips then it might be worth a shot.
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Post by offgridQLD » Sun, 15 Jan 2017, 19:36

I pulled the front cover off my two pip4048 inverters today to familiarize myself with the inductor in question.

Way larger than I had imagined. I see perhaps one potentual issue when rewiring it with litz wire that will take up more space.

On my 2013 model (with the solar charge controller up top large black heat sink) there is plenty of free space surrounding the inductor.

On on the late 2016 model (solar charge controller down the middle of the unit) the inductor sits very close to the inverters heat sink.

Could this be a issue?

you can see in this pic taken sideways that the silver heat sink is just a few mm away from the inductor wire in its stock form.
Image

Image

Image

Perhaps best to try it on a early model (black heat sink up top) PIP4048 first. Im sure if your keen after you could trim some off the heatsink to create room. Or perhaps lean over the inductor a little? Will all depend on how much it grows when rewound.

Kurt

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Post by weber » Sun, 15 Jan 2017, 20:36

offgridQLD wrote:I see perhaps one potentual issue when rewiring it with litz wire that will take up more space.
...
On on the late 2016 model ... the inductor sits very close to the inverters heat sink.

Could this be a issue?

Yes it could. Well spotted.

Fortunately, Coulomb convinced me last night by email, that my original suggestion of 100 strands of 0.2 mm will be sufficient to carry the maximum continuous current of the inverter without overheating (dissipating 10 W at 21.7 A), and that any greater number of strands is likely to completely fill the hole in the middle of the toroid before achieving the required number of turns.

But even with the 100 strand wire, the inductor will probably end up a little larger than it was.

I just measured the clearance that you mention, as 4 mm. There is a clearance of 10 mm above the toroid, so I think we should offset it upward to maintain that 4 mm clearance if possible, before we epoxy it in place.
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Post by Northland » Sun, 15 Jan 2017, 23:24

Could one of you explain slowly the theory here? I understand the inductor filtering out the frequency part. But why is the factory way wrong / inefficient? Is it cost issue? Space?

I have a spare inductor so I could replicate this
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Post by andys » Mon, 16 Jan 2017, 00:48

I'm curious what you will discover!

Every inverter I've seen (that can do >4kW) wastes alot of power when idle, except the fancy models which seem to be able to do dynamic power down of a high power circuit until it is needed.

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Post by mattwrightzen » Mon, 16 Jan 2017, 01:01

Hi, a friend has a 5kW and 8.2kW fronius primo inverter with 17kW of panels. When he loses the grid due to blackout can he tie those inverters to the mpp pip4048

The fronius supports a ramp down of its output as grid frequency rises from 51-52.7hz

http://www.fronius.com/cps/rde/xchg/SID ... Hs3_GlQaEf

Does anyone know how the mpp 4048 will behave if there is grid feed in on the load side

Ie
1 voltage rises
2 batteries keep charging and then overcharge
3 frequency rises
4 batteries get charged until they reach float then voltage rises
5 batteries get charged until they reach float then frequency rises
6 something else

Appreciate any knowledge on this that can be shared
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Post by weber » Mon, 16 Jan 2017, 03:39

Northland wrote: Could one of you explain slowly the theory here? I understand the inductor filtering out the frequency part. But why is the factory way wrong / inefficient? Is it cost issue? Space?

We might well be missing something, and the factory method might turn out to be almost as good or better. But the theory we're basing this on can be read here:
http://en.wikipedia.org/wiki/Skin_effect
http://en.wikipedia.org/wiki/Proximity_ ... magnetism)
http://en.wikipedia.org/wiki/Litz_wire
Litz wire does cost more and take up more space.
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Post by offgridQLD » Mon, 16 Jan 2017, 05:20

After looking at my two pip4048 inverters side by side today. I am sure cost is playing big part in the design and material choice of the inverter.

This new design with the little heat sink for the SCC jammed between the two inverter heatsinks has cost cutting written all over it. I see no other reason for the redesign.


Allthough the pip works well as a standalone (offgrid) inverter perhaps when it was first designed they were thinking the pip would be used more as a grid conected battery backup power. If you have the grid feeding the pip overnight that 50w vs say 25w idle consumption isn't so critical.

The results will be interesting.



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Post by coulomb » Mon, 16 Jan 2017, 16:03

mattwrightzen wrote: Hi, a friend has a 5kW and 8.2kW fronius primo inverter with 17kW of panels.
Impressive.
When he loses the grid due to blackout can he tie those inverters to the mpp pip4048
So presumably he also has a separate PIP system with batteries that he can switch the loads to.
The fronius supports a ramp down of its output as grid frequency rises from 51-52.7hz
As I indicate below, I don't believe that this will come into play with the PIP.
Does anyone know how the mpp 4048 will behave if there is grid feed in on the load side
I assume you mean what happens if he connects his grid tie inverter(s) to the PIP outputs, which have loads connected, during a blackout only. I don't think that there is a practical and safe way to connect the PIP output to the mains/utility.

My wild guess, expanded below, is
6 something else
Firstly, although I've worked on a research microgrid project, I was not there for the design, and took virtually no part in the design. I did have to think about power flows and AV vector diagrams a bit, so I have a little knowledge in this area.

I'm confident enough that I think we can take as facts:
* The PIP inverter makes no attempt to synchronise with the AC output terminals. In other words, it's not a grid interactive inverter.
* The PIP inverter always seems to be aiming for 230 VAC at the output. I'm assuming "battery mode", where the inverter hardware is configured as an actual inverter (as opposed to being configured as an AC charger, when in "line mode").
* When there is AC present at the AC input terminal, and if and when it passes some quality checks, the inverter synchronises to the AC input terminals. My guess is that this is to make easier the switching of the loads to the AC input (either by going to line mode, using its internal relay, or using an external relay or contactor).

I'm fairly confident of this point:
* When there is no AC present at the AC input terminals, the PIP inverter operates at 50.0 Hz (or 60 Hz if so configured), and makes no attempt to synchronise with anything else.

So now to my guess as to what will happen if a grid interactive inverter is connected to the PIP's output, when powering a load. For the sake of example, let's say the load is 2.0 kW, all supplied by the PIP from a battery. Let's say the grid interactive inverter has 3.0 kW of power to provide, but we'll imagine that it ramps up the power slowly (say 200 W per second), starting at zero power.

While there is less than 2.0 kW from the grid interactive inverter, it seems to me that the phase of the power to the loads will shift slightly (and no-one will notice that unless making very careful measurements), the load AC voltage will remain very close to 230 V, and the PIP will gradually supply less and less power to the loads. It won't find anything unexpected (ignoring the very real possibility of higher frequency spikes mucking up its load voltage and current measurements), it will just look like the load is gradually reducing.

At the point where the grid interactive inverter is providing 2.0 kW, all the power to the load will come from the grid interactive inverter, and the PIP will feel as though it has no load.

Of course, the interesting part is what happens next, as the grid interactive inverter attempts to provide more power to its output terminals.

The PIP is targetting 230 VAC at its output. So it looks like a low impedance 230 VAC source with an inductor (very likely the one we're talking about in other posts recently) between that AC source and its output terminals. With say 2.1 kW coming from the grid interactive inverter, and only 2.0 kW absorbed by the loads, the extra 0.1 kW will have to go somewhere. It seems to me that the load voltage will increase only slightly (due to voltage drop across the inductor), but it will only be a volt or two at most, much less while there is only 0.1 kW of excess power. But the PIP in targeting 230 VAC will experience a rise in the voltage across its output full bridge. In other words, the excess power will tend to charge the capacitor feeding the output full bridge. In other words, the power supply for the output MOSFETs or IGBTs that make up the full bridge making the 50 Hz 230 VAC. I'll refer to this capacitor (there may be more than one, but they would be equivalent to one) as the "output capacitor" for brevity. This is not a capacitor doing output filtering; its filtering the power supply for the full bridge.

The PIP has an unusual architecture, in that there is a battery side full bridge (50 V to 400 V) feeding a buck converter feeding the output full bridge. There is a "bus voltage" as reported by the QPIGS (general status query) command of the PIP likely refers to the voltage at the input to the buck converter. It always seems to be quite close to 8 times the battery voltage, so that means the battery side converter has a transformer with a 1:8 turns ratio.

Here is the important thing: the buck converter is unidirectional. The output full bridge, and the battery side full bridge are both bidirectional; they can push or pull power in either direction (to/from the battery, and in theory, to/from the load terminals). So I think what will happen is that the 100 W of excess power from the grid interactive inverter will very quickly charge up the capacitors at the output of the buck converter, and there will be nowhere else for that energy (over a few cycles) to go. The PIP monitors "bus voltage", but it seems to me that this could be on the input side of the buck converter. So maybe it won't notice, the output capacitor will go to 600 V, destroying the output bridge and the output capacitor. Or maybe it will notice, and will shut down the inverter. If it just stops switching the output full bridge, this won't help, as the free-wheeling diodes of those devices will still pump up the output capacitor. [ Edit: actually, but not very much. ] If it disconnects the output relay, the grid interactive inverter will cause the load AC voltage to rise. Probably the load voltage will quickly exceed a limit and they will trip off. Perhaps the PIP will switch to line mode, which will at least protect the inverter hardware, but in a blackout situation, there will be nothing to power the loads other than the grid interactive inverter. In all these scenarios, the loads get too much voltage or no power at all.

So unfortunately, I can't see a way that this will work, unless there is a "zero export" system in place, so the grid interactive inverter never provides too much power for the load. Those things are always a bit tricky, and tend to under or overshoot with sudden load changes, so the zero export system would have to be very good not to cause overvoltage of the PIP's output capacitor.

Finally, the anti-islanding provisions of the grid interactive inverters would have to regard the PIP's output as being the presence of a grid. I hear rules of thumb like "the grid interactive inverter has to be no higher in rated power than that of the stand alone inverter", but I'm dubious about this.

Edit: When trying to find a diagram to illustrate the topology, I realise I've misremembered how the buck converter works. It seems that it's only active in battery charge mode (line mode). In battery (inverter) mode, it's effectively just an inductor. So that affects my answer; see my next post.

[ Edit: minor rewording, energy -> power, battery side inverter -> battery side converter, etc. ]
[ Edit: integral diode -> free-wheeling diode ]
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Post by coulomb » Mon, 16 Jan 2017, 17:21

[ My apologies if this posts twice. Server crash. ]

Here is the power flow topology for the PIP-4048, from "Axpert MKS-4000/KS-5000 Service manual", which I found on the web:

Image

As actually implmented:

Image

The battery is at the left, at the terminals marked P1. The AC loads appear at the right, at the terminals marked L and N (Line and Neutral). Note that one of the "switches" is actualy a diode, and the buck switch is actually in the bus negative rail, rather than the positive rail.

The two full bridge converters at the left convert the ~50 V of the battery to ~400 V. This flows through inductor 15 and S12 (S10 is off when inverting) to Vbus. The "output capacitor" I mentioned in my previous post is marked Vbus. The full bridge on the right converts this ~400 VDC to 230 VAC. The full bridge appears to be a generic diagram, not intended to represent specifics of the PIP inverter (which uses IGBTs not MOSFETs, and one not two large inductors L1/L2).

So the question remains: what happens if a grid interactive inverter is connected across the PIP's loads, and generates more power than the loads absorb. Again, Vbus will rise in voltage, but now there is a power flow path from Vbus to the battery. I don't know what will happen. My guess is that the PIP will prevent power flow, even unintended power flow, from Vbus to the battery (in battery mode). If it did allow this power flow, then all of the excess power from the grid interactive inverter (and there could be many kilowatts of this at times) will flow to the battery, with no way of regulating the power flow or the battery voltage.

So my answer remains much the same: I can't see a way of making this work unless there is a good system for ensuring zero power is "exported" to the PIP inverter. Even then, I'd be dubious that it could be made to work, but there may be a way I haven't thought of.

[ Edit: Inductors 13 and 15 seem to be combined into a single inductor L1 in place of 15. Switch S10 does not exist, and S12 is a paralleled pair of either MOSFETs or IGBTs (Q31/Q32) in the negative side of the power supply, not the positive side as shown. Inductors L1 and L2 of the diagram are combined into a single inductor L4 on the actual hardware. ]
[ Edit: The large (470 μF 500 V) electrolytic capacitors (C40, C41) are on the high voltage side of the DC/DC converter, in parallel with two 0.47 μF film capacitors (C32, C34) near the associated IGBTs. The 230 V inverter IGBTs have two film capacitors, 0.68 μF and 0.47 μF, across their DC bus, which is separated from the actual DC bus by the buck transistor(s). The large film capacitor C33 (20 uF) is part of the 230 VAC line LC filter, along with L4. ]
[ Edit: Added second diagram for comparison. Removed the "suspect buck converter is back to front"comment. ]
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Post by mattwrightzen » Mon, 16 Jan 2017, 18:13

I wonder if the hybrid series would be any different as they seem to support AC charging.

http://www.mppsolar.com/v3/mpi-hybrid-series-2/

[ Edited Coulomb: made link clickable ]

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Post by paulvk » Mon, 16 Jan 2017, 19:02

More on the SCC:
I like semiconductors in my own designs to not get hot and definitely not so hot that I can not touch them, the heat sink on the SCC gets very hot with no forced air cooling so I sat fans on top of two of my units even at low speeds the fans keep the heatsink to warm.
So not wanting to run them 24/7 as they do at present and only when the SCC is active I have been looking at the SCC board to see where I can get a signal or power.
I have found 2 ways:
1. connect to relay coils and run 12v fan from there but involves removing from unit!
2. there is a 6 pin header on the board for CPU debug with + & - volts on it so use this to run an opto coupler (for isolation) then switch a transistor on to run the fan when SCC has power.

The connection from the inverter to the SCC is isolated with 3 opto couplers and is serial rs232 should not be to hard to record what is going on!
Note there is an unused serial & panel socket on the board.

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Post by solamahn » Mon, 16 Jan 2017, 19:48

There is a socket marked CN12 (FAN3) at the top of the main PCB in the middle. Having the 2 main fans mounted upside down keeps the scc heat sink a bit cooler. Having the solar panel array working voltage closer to 60 keeps the scc heat sink temperature lower but solar charging using a higher string voltage seems to work better.
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Post by weber » Mon, 16 Jan 2017, 19:59

Solamahn, Which direction of air flow are you calling "upside down"? I assume you are talking about the newer models that have the SCC between the two main-board heatsinks. Is that correct?

Paulvk, I assume you are talking about the older models that have the black SCC heatsink on the outside top. Is that correct?
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PIP-4048MS inverter

Post by coulomb » Mon, 16 Jan 2017, 20:00

When looking at the power flow topology diagram I posted (five posts ago now), it struck me that it looked rather like a diagram you'd find in a patent application. I did a quick search, and it looks like the diagrams come from this patent, though I'm not sure that it applies to the PIP-4048:

https://www.google.com/patents/US20140268892

Image

I find it confusing. They seem to be claiming a buck/boost capability with S12 and S10, but I only see boost (taking power flow from left to right). They mention a resonant circuit, but I only see Ls and Cs that are not meant to resonate. Still they might have something novel there; I'm no expert on these converters.
Last edited by coulomb on Mon, 16 Jan 2017, 09:04, edited 1 time in total.
Learning how to patch and repair PIP-4048 inverter-chargers and Elcon chargers.

paulvk
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PIP-4048MS inverter

Post by paulvk » Mon, 16 Jan 2017, 21:02

Yes I have the older PIPs with nice big black heat sink!
I am also going to look for some 18mm x 10mm copper bar to replace the aluminum under the FETs I have found it greatly reduces the package temperature.

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