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Inverter for 48v off grid - PIP xxxxx or something else?

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coulomb
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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by coulomb »

T1 Terry wrote:
Thu, 02 Apr 2020, 10:11
I think you are missing the fact an MPPT controller needs to see a battery voltage so it can determine the voltage/current that is the optimum output from the input it sees.
I assumed that it would just try larger and larger PWM duty cycle until the maximum current flowed. Maybe some would be surprised and refuse to start, but others would not care and it would just work.
This is where the Electrodacus MPPT unit comes in, it will change what ever is coming in to a useable output designed for heating via a resistive element.
Well, if there is a special product to drive a resistive element, that certainly seems as though I'm wrong.
Have you looked at the SBMS0, it is...
No, I have no interest in one at this stage. However, it may be suitable for @Honu.
Nissan Leaf 2012 with new battery May 2019.
5650 W solar, 2xPIP-4048MS inverters, 16 kWh battery.
1.4 kW solar with 1.2 kW Latronics inverter and FIT.
160 W solar, 2.5 kWh 24 V battery for lights.
Patching PIP-4048/5048 inverter-chargers.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

coulomb wrote:
Thu, 02 Apr 2020, 07:16
Ok, so that's rated for V = √(P·R) = √(2000 · 7) = 118 V. Perhaps it's designed for 120 VAC.
Hellooo,
For what i've read, a resistor got no working "voltage"... it got 2 parameters, internal resistance and Max power output it can accept, more then this power.. it blow out. (with perhaps a tolerance of .. let's say 10%)

My logic is that .. Ohm Law give me : V = R x I => R = V/I = 7 Ohm => V and I can only be on a straight line cause V/I = constant... dunno if my logic is clear ?
coulomb wrote:
Thu, 02 Apr 2020, 07:16
Your problem is that you're choosing both the voltage and the current; you can only choose one and the other will follow from Ohm's law.

If you apply 33 V to a 7 Ω element, the current will be I = V/R = 33/7 = 4.7 A. If you want to cause 2 A to flow, you'll need V = I·R = 2 · 7 = 14  V. For 4 A (double the current), you'll need double the voltage, or 28 V. You can figure out the power using any of the formulae. For example, applying 33 V: P = V²/R = 33²/7 = 156 W. Or P = V · I = 33 · 4.7 = 155 W (the 1 W difference is a rounding error). Finally, P = I² · R = (4.7)² · 7 = 155 W.
Ok now, i apply 33V to the 7 Ohm element, it flow 4.7 Amp => should i understand that IF i apply from the generator, a 33V and 10A, only 4.7A will flow through the resistor ?
If i want 2A to flow i need 14V in my 7 Ohm element => Does it mean a resistor element is driven by voltage not by intensity ? This would imply with the Ohm Law that V = R x I that all my couples (V,I) SHOULD be on a straight line, linearly depending on eah other to satisfy my resistance... if i go away from this line i lose power...

That's what is displayed here :
Image
coulomb wrote:
Thu, 02 Apr 2020, 07:16
Usually, the one that has no unknowns in it. Otherwise, you have to use another formula to find the unknown.
Allright, what you say seems to help a little, but .. i still got no response to

A : I'm applying 33V and 2A to my 7Ohm resistor, what is the power dissipated by the element ..?
A reply could be : A : The resistor is rated for 7 Ohm, under 33V it will draw at most I = V/R = 33/7 = 4.7 A, you throw 2A at him it will take them and generate : P = 7 x 2 x 2 = 28 Watts

B : I'm applying 33V and 100A to my 7Ohm resistor, what is the power dissipated by the element ..?
A reply could be : B : The resistor is rated for 7 Ohm, under 33V it will draw at most I = V/R = 33/7 = 4.7 A, you throw 100A at him, it will only take 4.7A and generate : P = 7 x 4.7 x 4.7 = 155 Watts
Meaning in this case some power is lost.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

T1 Terry wrote:
Thu, 02 Apr 2020, 10:11
I think you are missing the fact an MPPT controller needs to see a battery voltage so it can determine the voltage/current that is the optimum output from the input it sees.
This is where the Electrodacus MPPT unit comes in, it will change what ever is coming in to a useable output designed for heating via a resistive element. Have you looked at the SBMS0, it is basically a control device that monitors cell voltage and drives remote devices including an MPPT charge control as long as you design the on/off to work from a low current open/closed contact supplied by the SBM0.
If you are looking for a constant voltage then why not just add a simple square wave inverter powered by the battery bank, a resistive load is not fussed about wave shape, it is just a controlled short circuit designed to convert electrical energy into heat energy.
If you want efficiency then you need to look at a heat pump to heat the water rather than a resistive element, then you do need clean pure sine wave 240vac.

T1 Terry
The biggest problem with SBMS0 and DMPPT is that your panels got to be wired in a special way, you need a 1 panel string, a 2 panel strings, a 4 panels string ... a 8 panels string.. cause the DMPPT will "connect" those strings according to the resistive elements that are connected .. and that are THEMSELVES grouped with different resistive values.
It's kind having different size of potatoes and different sizes bags, the DMPPT match which potatoes goes into which bag, the idea is good .. BUT .. only 3kW of all you connect to this device can go to the SBMS0, the other device that charge the batteries and then be used by the inverter. That mean that if you got like me 7.3 Kw of panels, i can only have a 3kW inverter, all the remaining power will go to the "heating" DMPPT system... which is not acceptable cause in summer .. i near need NO hot water.
I've discussed this with Dacian, i asked him if i could use those resistive output to connect an inverter or a charger controller, he said those resistive output are not meant to be connected to this kind of device.
This special strings, with 1 panel, 2 ... 4.. 8 .. make it that i can't disconnect them from the DMPPT to connect them to let's say my inverter.. you see what i mean.. it's a mess.
Anyway, the device is meant for a precise purpose, i'm not in the scope of it... it's meant to heat your HOUSE with PVs... as i said to Dacian, near nobody will use it for this cause you need a hell lot of panels to heat a house in winter and that include a severe insulated house AND a very nice insulation in winter.. which .. most people do not have.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by coulomb »

Honu wrote:
Thu, 02 Apr 2020, 17:32
coulomb wrote:
Thu, 02 Apr 2020, 07:16
Ok, so that's rated for V = √(P·R) = √(2000 · 7) = 118 V. Perhaps it's designed for 120 VAC.
Hellooo,
For what i've read, a resistor got no working "voltage"... it got 2 parameters, internal resistance and Max power output it can accept, more then this power.. it blow out. (with perhaps a tolerance of .. let's say 10%)
Sure. But if you apply more than 120 V to this element, it will dissipate more than 2000 W. With a fixed resistance, a power limit also implies a voltage limit.
My logic is that .. Ohm Law give me : V = R x I => R = V/I = 7 Ohm => V and I can only be on a straight line cause V/I = constant...
Yes, fixed resistors come out as a straight line on a V versus I graph. But note that panels' graphs are decidedly not straight lines; they are not fixed resistors.
Ok now, i apply 33V to the 7 Ohm element, it flow 4.7 Amp => should i understand that IF i apply from the generator, a 33V and 10A, only 4.7A will flow through the resistor ?
A generator might have a 10 A limit, but if you plug in a small LED lamp, you won't get 10 A to flow. But yes, a 7 Ω resistor across a 33 V source will cause 4.7 A to flow, as long as the source can maintain that voltage at that current.
If i want 2A to flow i need 14V in my 7 Ohm element => Does it mean a resistor element is driven by voltage not by intensity ?
By "intensity" I'll assume you mean "current". Essentially, yes; the voltage establishes an electric field, which then induces the current. You don't usually think of "applying" a current to a resistor.
This would imply with the Ohm Law that V = R x I that all my couples (V,I) SHOULD be on a straight line, linearly depending on eah other to satisfy my resistance... if i go away from this line i lose power...
It's not that you'll lose power, it's that the current you thought would flow just doesn't happen, and the current demanded by Ohms law does.
Allright, what you say seems to help a little, but .. i still got no response to
A : I'm applying 33V and 2A to my 7Ohm resistor, what is the power dissipated by the element ..?
I thought I made it clear that you can't apply 33 V and 2 A to a 7 Ω resistor. Imagine a power supply with voltage and current limit controls. Initially, you set the voltage limit to 33 V, and the current limit to maximum. You connect the 7 Ω resistor; 4.7 A flows. Now you turn the current limit down to 2 A. You will find that the voltage drops to 14 V, to limit the current to 2 A.
A reply could be : A : The resistor is rated for 7 Ohm, under 33V it will draw at most I = V/R = 33/7 = 4.7 A, you throw 2A at him it will take them and generate : P = 7 x 2 x 2 = 28 Watts
Yes. But in the second case, the voltage won't be 33 V any more.
B : I'm applying 33V and 100A to my 7Ohm resistor, what is the power dissipated by the element ..?
Again, you're specifying voltage and current at the same time. It doesn't work that way.
A reply could be : B : The resistor is rated for 7 Ohm, under 33V it will draw at most I = V/R = 33/7 = 4.7 A,
It won't be "at most" 4.7 A, it will be 4.7 A, no more and no less. Otherwise, it won't be 33 V or it won't be 7 Ω.
you throw 100A at him, it will only take 4.7A and generate : P = 7 x 4.7 x 4.7 = 155 Watts
Meaning in this case some power is lost.
Current doesn't work like that. If you get 100 A into that element, it will blow up; the power would be P = I²R = (100)²·7 = 70 kW.

The only way to have 100 A coming in and only 4.7 A going into the resistor is if there is another path around the resistor, taking the other 95.3 A. Electric current (unlike water current) can't just leak out onto the ground without a conductive path.

The only way power can be "lost" in your example is if you had a source that was say 33 V and had the potential of providing 100 A. Then by loading that source with 7 Ω you are wasting 95.3% of that potential, so "losing" potential power in that sense.
Nissan Leaf 2012 with new battery May 2019.
5650 W solar, 2xPIP-4048MS inverters, 16 kWh battery.
1.4 kW solar with 1.2 kW Latronics inverter and FIT.
160 W solar, 2.5 kWh 24 V battery for lights.
Patching PIP-4048/5048 inverter-chargers.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

coulomb wrote:
Thu, 02 Apr 2020, 19:39
The only way power can be "lost" in your example is if you had a source that was say 33 V and had the potential of providing 100 A. Then by loading that source with 7 Ω you are wasting 95.3% of that potential, so "losing" potential power in that sense.
Yes that what i meant, cause each second MPPT got a power available on it's output side (depending on irradiation of the moment) and we got to make the best use of it, meaning not wasting any available Amp or volt.
So if i follow what you said, my graph make all sense .. we got to "move" our available power .. ON max dissipated power line which pass on the (0,0) point and on the max power point available on the panels, let's say 315W for one of our panel P = R x I x I => I = √(P/R) = √(315/7) = 6.7A and then V = R x I => V = 7x6.7 = 47V, second point is then (47,6.7)

It then mean that each couple (voltage, current) should be on this line passing by (0,0) and (47, 6.7) to make use of the full power. A device is needed that will first find the MPPT of the panel .. and convert that power on the "perfect" line.
coulomb wrote:
Thu, 02 Apr 2020, 19:39
I thought I made it clear that you can't apply 33 V and 2 A to a 7 Ω resistor. Imagine a power supply with voltage and current limit controls. Initially, you set the voltage limit to 33 V, and the current limit to maximum. You connect the 7 Ω resistor; 4.7 A flows. Now you turn the current limit down to 2 A. You will find that the voltage drops to 14 V, to limit the current to 2 A.
It was not clear, but now it seems to be, should i consider the resistor as a flow reducer that generate heat ? Like only 4 liters par min can go pass through, so it can be low pressure with high rate of flow or high pressure with low rate... but only 4l/min will pass through ? The 4 liters being the Resistor Ohm value ?
coulomb wrote:
Thu, 02 Apr 2020, 19:39
Honu wrote:
Thu, 02 Apr 2020, 19:39
you throw 100A at him, it will only take 4.7A and generate : P = 7 x 4.7 x 4.7 = 155 Watts
Meaning in this case some power is lost.
Current doesn't work like that. If you get 100 A into that element, it will blow up; the power would be P = I²R = (100)²·7 = 70 kW.
MMMm now i start to see .. we got to stay on the "line" but as the power rating of the resistor can't be exceeded, i knew it but i now start to see how it connect together. Thanks.

So in real world, i make available 100A to the resistor (7 Ohm, 2000W)... then i increase my voltage slowly ( I = V / R ), it will use to generate heat :
1 V => I = 1/7 A = 0.14A => 0.137W
2 V => I = 2/7 A = 0.28A
.
50 V => I = 50/7 A = 7.14A => 357W
.
100 V => I = 100/7 A = 14.3A => 1430W
.
.
118.32 V => I = 118.32/7 = 17.9A => 2000W (Max power : V = 7 x I and P = V x I => P = 7 x I x I => Imax = sqr(P/7) = sqr (2000/7) = 16.9A, since V = R x I => V = 7 x 16.9 = 118.32V)

=> each volt is a 0.14A increase (slope of the line)


So we can agree that to be efficient, we need a device that convert the panel power to couples of (Volt, Amp) that perfectly suit the resistor ?
OR can use capacitor to store MPPT output and then discharge by burst into the resistor by being always on the same point of the line... let's say the 50V/7.14A.
Does that make sense ?

Do those devices already exist ?
coulomb wrote:
Thu, 02 Apr 2020, 19:39
If i want 2A to flow i need 14V in my 7 Ohm element => Does it mean a resistor element is driven by voltage not by intensity ?
By "intensity" I'll assume you mean "current". Essentially, yes; the voltage establishes an electric field, which then induces the current. You don't usually think of "applying" a current to a resistor.
Ho sorry yes, in my mother tongue, we use :
U = R x I
I : Intensité du courant
R : Résistance
U : Voltage

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by coulomb »

Honu wrote:
Thu, 02 Apr 2020, 22:45
it seems to be, should i consider the resistor as a flow reducer that generate heat ? Like only 4 liters par min can go pass through, so it can be low pressure with high rate of flow or high pressure with low rate... but only 4l/min will pass through ? The 4 liters being the Resistor Ohm value ?
Um, that doesn't work for me.
So we can agree that to be efficient, we need a device that convert the panel power to couples of (Volt, Amp) that perfectly suit the resistor ?
OR can use capacitor to store MPPT output and then discharge by burst into the resistor by being always on the same point of the line... let's say the 50V/7.14A.
Does that make sense ?
Yes!
Do those devices already exist ?
Yes, they do. They are called buck converters. As found in most MPPT solar charge controllers.

Let's consider an example, where you have solar panels capable of say 150 V Vmp and 13.3 A Imp (2 kWmp). We'll use your 7 Ω element, rated at 2 kW. So we'd want the element current to rise to I = √(P/R) = √(2000/7) = 16.9 A. The average voltage would then be 118 V, as we've calculated before. Let's say the MPPT starts with a 40% duty cycle. When the buck switch is on, something like 120 V appears across the inductor, fluxing it up to some peak current, let's say it's 15 A. That's more than the panel can provide, but remember that there is a large capacitor at the input to the solar charge controller, providing the extra current. For the rest of the cycle (around 60% of the time), the catch diode keeps the current flowing into the load. Let's say the current ramps down to 10 A, and for simplicity let's say the average current is 12.5 A. That means an average power into the element, and out of the panel (neglecting losses for simplicity) of P = I²R = (12.5)²·7 = 1100 W. Only half what the panel can provide, so the MPPT algorithm increases the pulse width to say 50%. Now the inductor fluxes for longer, say it reaches 20 A at peak and 16 A at the end of the cycle. So the average current into the element is say 18 A, which is more than the 2 kW point for the element, in fact it's P = I²R = (18)²·7 = 2.27 kW. So this situation doesn't last long; the panel can't keep the capacitor at the MPPT input charged, and the panel voltage falls. The MPPT sees that it's gone too far, reduces the PWM ratio to say 45%, and gets closer to optimum power into the element.

Normally the MPPT is pushing its current into a battery, so the output voltage will stay fairly steady. With the resistive element, unless the MPPT has a ridiculously large capacitor at the output, is going to see a roughly sawtooth voltage. But that's fine (it may buzz a little, and the cables may generate some EMI, so you might have to encase the cable to the load in metal).

From the fact that the buck converter only decreases voltage (and increases current), it's clear that you need to choose the element such that at maximum power, it needs less voltage than the typical Vmp of the panels. Sometimes you want to be using PV power at only 20% of maximum sun power, but in those cases, you won't be able to extract anything like full element power, so the element voltage will be way less. In this example with a 118 V element at maximum power, you'd want your panels to be at least 3S, so that you'd have at least 120 V Vmp, possibly 4S or a more. For maximum efficiency, you don't want the typical panel voltage to be much more than about 150% of the element's maximum power voltage, because buck converters are more efficient and easier to control when the ratio of input divided by output voltage is lower (it can never be less than 1.0, obviously).

If you only want to be using "waste PV power" for the element, then there will be another MPPT competing for the PV power, supplying loads and possibly charging the battery. Without coordination, I suspect that the power will seesaw between the two MPPTs, not what you'd want. So you'd want to be able to control the element's MPPT PWM ratio with a small computer that was aware of the load power, and adjusted power to the element to not take too much power from the PV panels that the load (and possibly battery charger) actually needs.

Edit: at least, I realise the difference between a resistor load and a battery load. With a battery load, let's say it is a 100 V battery and 150 V PV input, you get 50 V across the inductor when fluxing it up. With a resistive load, and a modest capacitor at the output, you might have only 50 V at the output while the inductor is fluxing up. So then it's 100 V across the inductor, not 50 V (double the battery charging case). The output voltage, and hence this factor, will be highly dependent on element current. That could well cause the MPPT to perform rather differently, and the control constants might not be optimal. The MPPT might spend more time hunting to find the optimal power point. But I'm guessing that the effect would be minimal, and efficiency would still be reasonable. If as I suspect you'll have to end up controlling the PWM yourself, you can take this into account, optimising the control constants (proportional and integral) for the resistive load.
Nissan Leaf 2012 with new battery May 2019.
5650 W solar, 2xPIP-4048MS inverters, 16 kWh battery.
1.4 kW solar with 1.2 kW Latronics inverter and FIT.
160 W solar, 2.5 kWh 24 V battery for lights.
Patching PIP-4048/5048 inverter-chargers.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by T1 Terry »

Honu wrote:
Thu, 02 Apr 2020, 17:50

The biggest problem with SBMS0 and DMPPT is that your panels got to be wired in a special way, you need a 1 panel string, a 2 panel strings, a 4 panels string ... a 8 panels string.. cause the DMPPT will "connect" those strings according to the resistive elements that are connected .. and that are THEMSELVES grouped with different resistive values.
It's kind having different size of potatoes and different sizes bags, the DMPPT match which potatoes goes into which bag, the idea is good .. BUT .. only 3kW of all you connect to this device can go to the SBMS0, the other device that charge the batteries and then be used by the inverter. That mean that if you got like me 7.3 Kw of panels, i can only have a 3kW inverter, all the remaining power will go to the "heating" DMPPT system... which is not acceptable cause in summer .. i near need NO hot water.
I've discussed this with Dacian, i asked him if i could use those resistive output to connect an inverter or a charger controller, he said those resistive output are not meant to be connected to this kind of device.
This special strings, with 1 panel, 2 ... 4.. 8 .. make it that i can't disconnect them from the DMPPT to connect them to let's say my inverter.. you see what i mean.. it's a mess.
Anyway, the device is meant for a precise purpose, i'm not in the scope of it... it's meant to heat your HOUSE with PVs... as i said to Dacian, near nobody will use it for this cause you need a hell lot of panels to heat a house in winter and that include a severe insulated house AND a very nice insulation in winter.. which .. most people do not have.
I don't think that is how the SBMS0 works. It controls other peripheral devices and takes in information from shunts etc This gives a very rough view of the idea http://www.electrodacus.com/ but the solid state relays can be what ever you want to use, they don't have to be the DSSR that he makes. Say you used an SSR or contactor that could handle the 140vdc the PIP can handle, drive a DPDT relay with the SBMS0 so it opens the PIP supply and closes the DMPPT supply.
I think the DMPPT reads the incoming current and connects the required number of resistive elements to use that current effectively. There is no reason why those resistive elements can't be water heating elements rather than embedded in concrete, it's just a resistor after all.

T1 Terry
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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by T1 Terry »

coulomb wrote:
Thu, 02 Apr 2020, 11:57
T1 Terry wrote:
Thu, 02 Apr 2020, 10:11
I think you are missing the fact an MPPT controller needs to see a battery voltage so it can determine the voltage/current that is the optimum output from the input it sees.
I assumed that it would just try larger and larger PWM duty cycle until the maximum current flowed. Maybe some would be surprised and refuse to start, but others would not care and it would just work.
This is where the Electrodacus MPPT unit comes in, it will change what ever is coming in to a useable output designed for heating via a resistive element.
Well, if there is a special product to drive a resistive element, that certainly seems as though I'm wrong.
Have you looked at the SBMS0, it is...
No, I have no interest in one at this stage. However, it may be suitable for @Honu.
This is part of the nightmare we have with trying to control MPPT solar chargers that don't have separate battery voltage sensing connections. If we simply add an SSR or contactor after the MPPT controller the whole thing drops its bundle the moment the battery circuit is opened and won't restart unless the solar is disconnected, then the battery connected, then the solar reconnected. If we add a contactor before the MPPT controller, charging stops as soon as the solar input is cut, but it doesn't restart charging where it left off, it has to go through the whole ramp up output etc and if it senses a battery voltage close to the preset float voltage, it will restart in float mode if it restarts at all.

T1 Terry
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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

T1 Terry wrote:
Fri, 03 Apr 2020, 13:56
Honu wrote:
Thu, 02 Apr 2020, 17:50

The biggest problem with SBMS0 and DMPPT is that your panels got to be wired in a special way, you need a 1 panel string, a 2 panel strings, a 4 panels string ... a 8 panels string.. cause the DMPPT will "connect" those strings according to the resistive elements that are connected .. and that are THEMSELVES grouped with different resistive values.
It's kind having different size of potatoes and different sizes bags, the DMPPT match which potatoes goes into which bag, the idea is good .. BUT .. only 3kW of all you connect to this device can go to the SBMS0, the other device that charge the batteries and then be used by the inverter. That mean that if you got like me 7.3 Kw of panels, i can only have a 3kW inverter, all the remaining power will go to the "heating" DMPPT system... which is not acceptable cause in summer .. i near need NO hot water.
I've discussed this with Dacian, i asked him if i could use those resistive output to connect an inverter or a charger controller, he said those resistive output are not meant to be connected to this kind of device.
This special strings, with 1 panel, 2 ... 4.. 8 .. make it that i can't disconnect them from the DMPPT to connect them to let's say my inverter.. you see what i mean.. it's a mess.
Anyway, the device is meant for a precise purpose, i'm not in the scope of it... it's meant to heat your HOUSE with PVs... as i said to Dacian, near nobody will use it for this cause you need a hell lot of panels to heat a house in winter and that include a severe insulated house AND a very nice insulation in winter.. which .. most people do not have.
I don't think that is how the SBMS0 works. It controls other peripheral devices and takes in information from shunts etc This gives a very rough view of the idea http://www.electrodacus.com/ but the solid state relays can be what ever you want to use, they don't have to be the DSSR that he makes. Say you used an SSR or contactor that could handle the 140vdc the PIP can handle, drive a DPDT relay with the SBMS0 so it opens the PIP supply and closes the DMPPT supply.
I think the DMPPT reads the incoming current and connects the required number of resistive elements to use that current effectively. There is no reason why those resistive elements can't be water heating elements rather than embedded in concrete, it's just a resistor after all.

T1 Terry
The SBMS is just a charge controller, it got nothing special, yes the DSSR20 add some twist to it .. but still. The only "original" device is the DMPPT, nothing like it that i know exist on the market but still .. it do not fit my needs (or i missed something). Plus the SBMS0 can only handle 24V, only 8 cells .. and i got 16.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

coulomb wrote:
Fri, 03 Apr 2020, 08:32
Do those devices already exist ?
Yes, they do. They are called [url=https://en.wikipedia.org/wiki/Buck_c....the resistive load.
Hello,
The idea was to have the MPPT0 in the Pip and then an other MPPT1, both connected to the batteries. A SSR would disconnect MPPT1 output from the batteries to connect to the load when those are full. Your explanation is understandable to me (which is already .. a performance .. :lol: )

Now we got a potential ok solution, can we take the problem the other way, i keep the direct panels connection, no MPPT.. BUT i connect the perfect resistive element depending on the MPPT voltage/A given by the panels.

Image

Let's see here, let's say we got only ONE panel, to cover the 5 main irradiance, i need 5 resistive values. (not perfect, but will calculate the loss vs a perfect variable resistance)

So i got : 3.5, 4.4, 6, 9 and 18 Ohm, let's say i simplify to 3 values : 3, 6, 9 Ohm => 3, 6, 9, 12, 15, 18 are possible (i better have nice performance at low irradiance)

Question : If the panel at 1000W/m2 irradiance is connected to the perfect resistor that is at max power point (that is 3.5 Ohm and should provide 315W) .. will it provide the intended voltage and current (33V and 9.5 amp) ?

I could connect one resistor, starting with the 3 Ohm ... calculate power, go to 6 Ohm.. better power ? Yes, i go to the 9 .. better ? Yes, i go to 12 ..better power ? No, i go down to the 9. Then if irradiance changes drasticaly (could use a small panel dedicated to that), search again for the MHPT .. max heat power tracker.. :D

Third idea : Why not using a variable resistor ..?
Image

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by coulomb »

Honu wrote:
Fri, 03 Apr 2020, 19:37
Question : If the panel at 1000W/m2 irradiance is connected to the perfect resistor that is at max power point (that is 3.5 Ohm and should provide 315W) .. will it provide the intended voltage and current (33V and 9.5 amp) ?
Yes.
I could connect one resistor, starting with the 3 Ohm ... calculate power, go to 6 Ohm.. better power ? Yes, i go to the 9 .. better ? Yes, i go to 12 ..better power ? No, i go down to the 9. Then if irradiance changes drasticaly (could use a small panel dedicated to that), search again for the MHPT .. max heat power tracker.. :D
Cute, and workable, as long as you have enough DC rated SSRs or contactors to do all the switching, and a suitable small computer to control it all.
Third idea : Why not using a variable resistor ..?
A variable resistance element would be interesting. But an element has to thermally connect the resistance to the water, and that pretty much precludes a wiper.

Also, a mechanical control such as the shaft of a variable resistor is difficult to work with; you'd need a stepper motor or similar to drive the shaft.

I hope you're not thinking of a variable resistor in series with the element; that would get the current and total power correct, but in some cases most of the power would be wasted in the variable resistor, where it presumably doesn't heat the water. If you find a really high powered variable resistor and dunk it in water, you'd have all sorts of problems with electrolysis and electrochemical reactions, not to mention ordinary corrosion.
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1.4 kW solar with 1.2 kW Latronics inverter and FIT.
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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

coulomb wrote:
Fri, 03 Apr 2020, 21:10
Honu wrote:
Fri, 03 Apr 2020, 19:37
Question : If the panel at 1000W/m2 irradiance is connected to the perfect resistor that is at max power point (that is 3.5 Ohm and should provide 315W) .. will it provide the intended voltage and current (33V and 9.5 amp) ?
Yes.
Sweet .. ! Super sweet.
coulomb wrote:
Fri, 03 Apr 2020, 21:10
I could connect one resistor, starting with the 3 Ohm ... calculate power, go to 6 Ohm.. better power ? Yes, i go to the 9 .. better ? Yes, i go to 12 ..better power ? No, i go down to the 9. Then if irradiance changes drasticaly (could use a small panel dedicated to that), search again for the MHPT .. max heat power tracker.. :D
Cute, and workable, as long as you have enough DC rated SSRs or contactors to do all the switching, and a suitable small computer to control it all.
I'll try some diagrams to try optimising that. Arduino is the small computer, cost nothing, use near no power.. and very versatile.
coulomb wrote:
Fri, 03 Apr 2020, 21:10
Third idea : Why not using a variable resistor ..?
A variable resistance element would be interesting. But an element has to thermally connect the resistance to the water, and that pretty much precludes a wiper.

Also, a mechanical control such as the shaft of a variable resistor is difficult to work with; you'd need a stepper motor or similar to drive the shaft.

I hope you're not thinking of a variable resistor in series with the element; that would get the current and total power correct, but in some cases most of the power would be wasted in the variable resistor, where it presumably doesn't heat the water. If you find a really high powered variable resistor and dunk it in water, you'd have all sorts of problems with electrolysis and electrochemical reactions, not to mention ordinary corrosion.
[/quote]
Yea .. i thouthg about connecting it to the water, i though of a metal bar or somethinkg like that. No i'm not thinking of serialising it, would make no sense. And i got an other concern, those variable resistors are not intended to have their resistance changed all day long, the cable will wear at a certain point .. so .. not sure it's a smart idea, but still an other option.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

coulomb wrote:
Fri, 03 Apr 2020, 21:10
Cute, and workable, as long as you have enough DC rated SSRs or contactors to do all the switching, and a suitable small computer to control it all.
Image

Seems 3 SSR are enough, current will take the path of least resistance no ?

All closed => 0 Ohm
A Open, B, C Closed => 3 Ohm
A Closed, B Open, C Closed => 6 Ohm
.
.
All opened => 3+6+9 = 18 Ohm

Now to the efficiency, will try to calculate it.

NB : i tried to use them in parralel to give me even more options .. but .. it goes too low .. just the R = 1/(1/6+1/9) = 3.6 could be usefull .. and it make the schematics too complexe for certainly a too small gain => ruled out for now.

Ok, i've done something : https://docs.google.com/spreadsheets/d/ ... sp=sharing

Image

Irradiation : Fixed values
Power : I've seen that power efficiency is constant vs irradiation, power is linearly depending on irradiation : factor 3.3 (roughly)
Volt : Depending on number of Pv in string
Amp : Power / Volt
Perfect Ohm : Volt / Amp
Resistor Ohm : My discrete values : for a 3 panels string : 10, 15, 30 Ohm => 25, 40, 45, 55 are possible
Resistor Volt : Min of two possible values : Resistor Ohm x Pv Amps OR Pv Volt => i took a guess on this, if we put a too strong resistor, then the voltage will be driven by the amp available.

Ex : 99V, Amp 9.2, 10.8 Ohm => I use a 10 Ohm resistor => 92V or 99V => i use 92V.

Resistor Power disipated : Volt Resistor x Volt Resistor / Resistor Ohm
Efficiency : Resistor Power / PV Power

NB : My discrete values are perhaps not the best choices.. it certainly can be optimized, but any way it will depende on real world data, and specially when i need power the most...

Edit : If i add 2 more discrete values : 5 and 50 =>5, 10, 15, 30, 50 Ohm => 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 75...110 Ohm
Image
We are over 90% for most of it.

Comparing this to a fixed resistor of 20 Ohm : (15 Ogm give overall better perf).. we can see that with a fixed one we can only have a good performance on high irradiation or on low irradiation... or mid of course, while with the multiple discrete.. optimizing on many spot is possible.

Image

NB : Got to validate my wild guess on the Resistor Volt/Amp

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

I was watching the spreadsheet and, i can see that efficiency is linearly depending on the distance of the Ohm resistor vs perfect resistor. (Which is obvious by the formulae used..)
That would mean that to stay within 90% of efficiency we can't go further then 10% of the perfect resistance.

Ok i've builded a small optimiser (not perfect).

---- 3 resistors 88.9% efficiency ----
Image

---- 5 resistors 96% efficiency ----
Image
Last edited by Honu on Sun, 05 Apr 2020, 07:17, edited 1 time in total.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by coulomb »

Honu wrote:
Sat, 04 Apr 2020, 21:19
That would mean that to stay within 90% of efficiency we can't go further then 10% of the perfect resistance.
To help with that, instead of using resistances in a 1:2:3 ratio, perhaps use a 1:2:4 ratio instead. As it is now, you have two ways of choosing 9 ohms, 3+6+0 or 0+0+9. If you used 2.6, 5.2, and 10.4, or 2.5, 5, 10 for ease of maths, you have a binary number thing going on, with the choice of between 0 and 7 (000 and 111 binary) of 2.5 ohm steps. To result in 9 ohms for example, you'd calculate 9 / 2.5 = 3.6, closest integer is 4 or 100 binary, so you'd use just the 10 ohm resistance.
I note that if you use the 2.6 ohm set, 9/2.6 = 3.46, nearest integer is 3 or 011 binary, so you'd end up with 2.6 + 5.2 = 7.8 ohms, which is more than 10% away from the ideal 9 ohms.
Edit: I note that solar charge controllers are often around 98% efficient.
Edit 2: I haven't been following all the detail, but isn't about 112 ohms the minimum resistance you want, not 18 ohms the maximum? You can still use the same design, just with higher value resistors. But I think that the accuracy (which translates to efficiency for you) would be very poor at say 75% sun. Your two steps might be 112 or 224 ohms at the low end, when the ideal value might be say 168 ohms.
Edit 3: Though perhaps the parallel arrangement would reduce that problem. So something like 200, 400, and 800 ohm resistors, to be switched in parallel. Then you have even steps of current (for an approximately constant voltage), instead of even steps of resistance.
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5650 W solar, 2xPIP-4048MS inverters, 16 kWh battery.
1.4 kW solar with 1.2 kW Latronics inverter and FIT.
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Patching PIP-4048/5048 inverter-chargers.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

coulomb wrote:
Sun, 05 Apr 2020, 07:11
Honu wrote:
Sat, 04 Apr 2020, 21:19
That would mean that to stay within 90% of efficiency we can't go further then 10% of the perfect resistance.
To help with that, instead of using resistances in a 1:2:3 ratio, perhaps use a 1:2:4 ratio instead. As it is now, you have two ways of choosing 9 ohms, 3+6+0 or 0+0+9. If you used 2.6, 5.2, and 10.4, or 2.5, 5, 10 for ease of maths, you have a binary number thing going on, with the choice of between 0 and 7 (000 and 111 binary) of 2.5 ohm steps. To result in 9 ohms for example, you'd calculate 9 / 2.5 = 3.6, closest integer is 4 or 100 binary, so you'd use just the 10 ohm resistance.
I note that if you use the 2.6 ohm set, 9/2.6 = 3.46, nearest integer is 3 or 011 binary, so you'd end up with 2.6 + 5.2 = 7.8 ohms, which is more than 10% away from the ideal 9 ohms.
Edit: I note that solar charge controllers are often around 98% efficient.
Edit 2: I haven't been following all the detail, but isn't about 112 ohms the minimum resistance you want, not 18 ohms the maximum? You can still use the same design, just with higher value resistors. But I think that the accuracy (which translates to efficiency for you) would be very poor at say 75% sun. Your two steps might be 112 or 224 ohms at the low end, when the ideal value might be say 168 ohms.
Edit 3: Though perhaps the parallel arrangement would reduce that problem. So something like 200, 400, and 800 ohm resistors, to be switched in parallel. Then you have even steps of current (for an approximately constant voltage), instead of even steps of resistance.
I'll check tommorow with a binary approach but i think those result are pretty good since my system try many combination and give them a mark based on the resulting efficiency sum.
For this 3 resistor example i used : 12, 15 and 54 Ohm.
All those numbers are not real world cause i will certainly not use those exact panels and i still do not know how many strings will be connected, 1, 2 or 3... this is the next step, to get a good efficiency with 1, 2 or 3 strings (9 panels).

In this example, the lowest insulation need the higher resistance and it is 108Ohm. (for 100W/m2) The lowest being 10 Ohm for 1000 W/m2.

What i call efficiency is the what i got vs what would be a perfect variable resistor, can be called accuracy too.

MMmm interesting idea of the parralel resistors, i though of them but not with this approach .. will try tommorow.

Can someone validate this : My string generate 8.3 A at 99 V, a perfect resistor would be 12 Ohm.
But my resistor is not perfect, I put one of 10 Ohm => will the voltage be : V = 10 x 8.3 = 83 V ? I suppose no, both voltage and current will drop.. but where ?

I go to bed, i'll check that tommorow .. it's late, i'm on the other hemisphere and on the other side of the planet, time to sleep .. ;D

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by coulomb »

Honu wrote:
Sun, 05 Apr 2020, 10:02
Can someone validate this : My string generate 8.3 A at 99 V, a perfect resistor would be 12 Ohm.
But my resistor is not perfect, I put one of 10 Ohm => will the voltage be : V = 10 x 8.3 = 83 V ? I suppose no, both voltage and current will drop.. but where ?
That one is easiest to solve graphically (so: not so useful for an Arduino). It will be at the intersection of the graph for the 10 Ω resistor and the appropriate line (depending on sun strength) for the panel. It will indeed be less than Vmp and probably less than Imp.
Nissan Leaf 2012 with new battery May 2019.
5650 W solar, 2xPIP-4048MS inverters, 16 kWh battery.
1.4 kW solar with 1.2 kW Latronics inverter and FIT.
160 W solar, 2.5 kWh 24 V battery for lights.
Patching PIP-4048/5048 inverter-chargers.

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Re: Inverter for 48v off grid - PIP xxxxx or something else?

Post by Honu »

coulomb wrote:
Sun, 05 Apr 2020, 10:16
Honu wrote:
Sun, 05 Apr 2020, 10:02
Can someone validate this : My string generate 8.3 A at 99 V, a perfect resistor would be 12 Ohm.
But my resistor is not perfect, I put one of 10 Ohm => will the voltage be : V = 10 x 8.3 = 83 V ? I suppose no, both voltage and current will drop.. but where ?
That one is easiest to solve graphically (so: not so useful for an Arduino). It will be at the intersection of the graph for the 10 Ω resistor and the appropriate line (depending on sun strength) for the panel. It will indeed be less than Vmp and probably less than Imp.
Perfect, thanks, not it' not to solve it with the Arduino, but to calculate my efficiency wich is in fact false right now, cause i "solved" it the wrong way with V = 10 x 8.3 = 83 V
Ok i think i got it :

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